Présentée Par Mehrdad

Nasernejad Intitulé

M Baptiste

Mme Viviana

M Jürgen

Herzog Professeur

Professeur M Ahmed Laghribi

M André

Professeur M Adam Van Tuyl

Mehrdad Nasernejad

MÉMOIRE présenté pour obtenir L'HABILITATION À DIRIGER DES RECHERCHES de

Keywords: Monomial ideals, Associated primes, Persistence property, Normally torsion-freeness, Normality, Graphs, Hypergraphs. will always denote the set of positive integers (respectively, nonnegative integers, rational numbers, nonnegative rational numbers)

monomiaux en vue de la combinatoire

Introduction

Let R be a commutative Noetherian ring and I be an ideal of R. A well-known result of Brodmann [21] showed that the sequence {Ass R (R/I k )} k≥1 of associated prime ideals is stationary for large k. That is, there exists a positive integer k 0 such that Ass R (R/I k ) = Ass R (R/I k 0 ) for all k ≥ k 0 . The minimal such k 0 is called the index of stability of I and Ass R (R/I k 0 ) is called the stable set of associated prime ideals of I, which is denoted by Ass ∞ (I). A natural question arises in the context of Brodmann's Theorem is as follows:

( * ) Is it true that Ass R (R/I)

⊆ Ass R (R/I 2 ) ⊆ • • • ⊆ Ass R (R/I k ) ⊆ • • • ?
McAdam [START_REF] Mcadam | Asymptotic Prime Divisors[END_REF] presented an example which says, in general, the above question has negative answer. We say that an ideal I of R satisfies the persistence property if it holds true in ( * ), i.e., Ass R (R/I k ) ⊆ Ass R (R/I k+1 ) for all k ≥ 1. Along this argument, an ideal I of R has the strong persistence property if (I k+1 : R I) = I k for all k ≥ 1. Ratliff showed in [START_REF] Ratliff | On prime divisors of I n , n large[END_REF] that (I k+1 : R I) = I k for all large k. Note that the strong persistence property implies the persistence property, however, the converse is not true.

Suppose now that I is a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K and x 1 , . . . , x n are indeterminates. Question ( * ) does not have an affirmative answer for all monomial ideals, see [START_REF] Herzog | The depth of powers of an ideal[END_REF] and [START_REF] Morey | Edge ideals: algebraic and combinatorial properties[END_REF] for counterexamples. Moreover, Kaiser, Stehl ík, and Škrekovski [START_REF] Kaiser | Replication in critical graphs and the persistence of monomial ideals[END_REF] showed that not all square-free monomial ideals have the persistence property. However, recently, by applying combinatorial methods, several papers have been published for finding the classes of monomial ideals which satisfy the persistence property, refer to [START_REF] Hà | Embedded associated primes of powers of square-free monomial ideals[END_REF], [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF], and [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF] for more information. So, we have the following subquestion:

( †) Which classes of monomial ideals have the persistence property? Question ( †) has affirmative answers in the following cases:

(1) I is the edge ideal of a simple graph (see [START_REF] Martinez-Bernal | Associated primes of powers of edge ideals[END_REF]Theorem 2.15]).

(2) I is the cover ideal of a perfect graph (see [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Corollary 5.11]).

(3) I is a polymatroidal ideal (see [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF]Proposition 3.3]). ix (4) I is the edge ideal of a finite graph with loops (see [START_REF] Reyes | On the strong persistence property for monomial ideals[END_REF]Corollary 2]). Furthermore, it is known by [START_REF] Martinez-Bernal | Associated primes of powers of edge ideals[END_REF] that all edge ideals of finite simple graphs satisfy the strong persistence property. It has already been shown in [START_REF] Herzog | Persistence and stability properties of powers of ideals[END_REF]Proposition 2.4] that every polymatroidal ideal has the strong persistence property. In addition, according to [4], [3], and [START_REF] Nasernejad | Associated primes of powers of cover ideals under graph operations[END_REF], the cover ideals of the following imperfect graphs are normal and satisfy the strong persistence property:

(1) Cycle graphs of odd orders.

(2) Wheel graphs of even orders.

(3) Helm graphs of odd orders with greater than or equal to 5.

Along this argument, the concept of symbolic strong persistence property was presented in [START_REF] Reyes | On the strong persistence property for monomial ideals[END_REF]. An ideal I in a commutative Noetherian ring R has the symbolic strong persistence property if (I (k+1) : R I (1) ) = I (k) for all k ≥ 1, where I (k) = p∈Min(I) (I k R p ∩ R) denotes the k-th symbolic power of I. It has been shown in [START_REF] Reyes | On the strong persistence property for monomial ideals[END_REF] that the strong persistence property implies the symbolic strong persistence property, however, little is known for the other classes of monomial ideals which satisfy the symbolic strong persistence property.

The edge ideal corresponding to a graph G has been introduced by Villarreal [START_REF] Villarreal | Cohen-Macaulay graphs[END_REF] and is generated by the monomials x i x j , where {i, j} is an edge of G, and the Alexander dual of it, denoted by I(G) ∨ , is called the cover ideal corresponding to G. Next, path ideals of graphs were first introduced by Conca and De Negri [30] in the context of monomial ideals of linear type. An ideal I in a commutative Noetherian ring R is called normally torsion-free if Ass(R/I k ) ⊆ Ass(R/I) for all k ≥ 1. A few examples of normally torsion-free monomial ideals appear from graph theory. Already, in [START_REF] Simis | On the ideal theory of graphs[END_REF], it has been proved that a finite simple graph G is bipartite if and only if its edge ideal is normally torsion-free. Moreover, by [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF], it is well-known that the cover ideals of bipartite graphs are normally torsion-free. In addition, in [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF], it has been verified that every transversal polymatroidal ideal is normally torsionfree. However, normally torsion-free square-free monomial ideals have been studied in [START_REF] Hà | Embedded associated primes of powers of square-free monomial ideals[END_REF][START_REF] Simis | On the ideal theory of graphs[END_REF][START_REF] Sullivant | Combinatorial symbolic powers[END_REF], but little is known for the normally torsion-free monomial ideals which are not square-free, for example see [START_REF] Olteanu | Normally torsion-free lexsegment ideals[END_REF].

This manuscript has 11 chapters. Chapter 1 is devoted to introducing the required definitions and notations of commutative algebra and (hyper)graph theory used throughout the text to keep this work as self-contained as possible.

In Chapter 2, we argue on the persistence property of monomial ideals. In fact, several results related to stable sets and index of stability of monomial and squarefree monomial ideals are given. In particular, we investigate the relation between persistence property and the Alexander dual of square-free monomial ideals. We x also discuss the properties of two classes of monomial ideals, that is, unisplit and separable monomial ideals, such that we show that every monomial ideal in these two classes (i) has the (strong) persistence property, (ii) has a strongly superficial element, and (iii) is Ratliff-Rush closed. We close this chapter with a discussion about persistence property for associated primes of a family of ideals.

In Chapter 3, we are interested in studying the strong persistence property of monomial ideals. We start with looking at the associated primes of powers of cover ideals of three imperfect graphs, i.e., odd cycle graphs, wheel graphs of even orders, and helm graphs of odd orders with greater than or equal to 5, such that we show their cover ideals have the strong persistence property. Next, we introduce the notion of superficial ideals. We continue with investigating the relation between superficiality and normality. We terminate this chapter with giving a class of primary monomial ideals do not satisfy the strong persistence property.

The purpose of Chapter 4 is to probe the concept of symbolic strong persistence property. We first state several results on the relation between symbolic strong persistence property and monomial operations such as weighting, contraction, and monomial localization. After that, we investigate some consequences related to the relation between the strong persistence property and symbolic strong persistence property. In particular, we prove that every square-free monomial ideal satisfies the symbolic strong persistence property.

The focus of Chapter 5 is to identify some classes of normally torsion-free monomial ideals. Indeed, we first study the normally torsion-freeness of the path ideals associated to directed trees. After that, we consider the relation between the normally torsion-freeness and the Alexander dual of square-free monomial ideals. Next, we examine the normally torsion-freeness under some monomial operations such as deletion, contraction, expansion, monomial localization, and weighting. We finally finish this chapter with a discussion related to normally torsion-freeness of ideals arising from neighborhood hypergraphs and their transversal hypergraphs. In particular, we inspect the normally torsion-freeness of dominating ideals of cycles.

The aim of Chapter 6 is to settle the other classes of nearly normally torsionfree monomial ideals. To do this, we first explore the graphs whose cover ideals are nearly normally torsion-free monomial ideals. We next turn our attention to studying the t-spread principal Borel ideals, and present several classes of t-spread principal Borel ideals which are nearly normally torsion-free. We close this chapter with an argument on the nearly normally torsion-freeness of closed neighborhood ideals and dominating ideals of some classes of trees and cycles.

In Chapter 7, we concentrate on the normality of monomial ideals. In the first step, we show that the cover ideals of three classes of imperfect graphs are normal, that is, odd cycle graphs, wheel graphs of even orders, and helm graphs of odd orders xi with greater than or equal to 5. We next consider the normality of monomial ideals under some monomial operations like deletion, contraction, monomial localization, expansion, and weighting. We also present an argument on the normality of linear combinations of normal monomial ideals. After that, we verify the cover ideals of Jahangir's graphs are normal. We next show that the cover ideals of theta graphs are normal. We enrich this chapter with giving some results regarding to the normality of closed neighborhood ideals and dominating ideals of complete bipartite graphs. This chapter will be terminated by a discussion about the normality of dominating ideals of h-wheel graphs.

Chapter 8 deals with the combinatorial aspects. In fact, we are going to answer two questions. Suppose that V = {1, . . . , n} is a non-empty set of n elements, A = {A 1 , . . . , A m } a non-empty family of m non-empty subsets of V , and 2 ≤ d ≤ n a positive integer. In the first question, we investigate the question arises whether there exists a d-uniform hypergraph H with V (H) = V where A is the set whose elements are the minimal vertex cover sets of H. Now, assume that V = {1, . . . , n} is a non-empty set of n elements, S = {S 1 , . . . , S m } a non-empty set of m non-empty subsets of V. In the second question, we inspect the question arises whether there exists an undirected finite simple graph G with V (G) = V where S is the set whose elements are the minimal dominating sets of G.

Chapter 9 is concerned with max-path ideals and matroidal path ideals. In the first part, we start introducing the definitions of maximal paths and path cover sets. Also, we state several properties of these notions. Next, we present the definition of max-path ideals, and argue on the graphs which their max-path ideals have certain heights. In the second part, we consider a class of polymatroidal ideals, that is, matroidal path ideals of graphs. In particular, we explore the Cohen-Macaulay property and limit depth of matroidal path ideals.

In Chapter 10, we first provide a necessary and sufficient condition whether the graded maximal ideal appears in the set of associated prime ideals of a monomial ideal such that we only employ the unique minimal set of monomial generators. Next, we turn our attention to studying the nature of corner-elements of monomial ideals and discussing the supports of corner-elements related to monomial ideals. As application, we verify that under certain conditions, every unmixed König ideal is normally torsion-free. Furthermore, we show that every square-free transversal polymatroidal ideal is normally torsion-free.

Chapter 11 offers a lot of open questions in this field. In fact, our aim is to provide a perspective for this scope for future research.

Throughout this text, K is a field and R = K[x 1 , . . . , x n ] is a polynomial ring in the variables x 1 , . . . , x n . For a monomial ideal I in R, we denote the unique minimal set of monomial generators of I by G(I). The symbol N (respectively, N 0 , Q, Q ≥0 )

Introduction ix

Chapter 1

Preliminaries

In this chapter, we collect all basic and crucial materials for this manuscript coming from commutative algebra and combinatorics. We summarize them in the following sections.

Commutative Algebra

In this section, we recall some definitions and notions from commutative algebra, which can be found in [START_REF] Herzog | Monomial Ideals[END_REF], [START_REF] Villarreal | Monomial algebras[END_REF], and [START_REF] Sharp | Steps in commutative algebra[END_REF].

Let R be a commutative Noetherian ring and I be an ideal of R. A prime ideal p ⊂ R is an associated prime of I if there exists an element v in R such that p = (I : R v), where (I : R v) = {r ∈ R | rv ∈ I}. The set of associated primes of I, denoted by Ass R (R/I), is the set of all prime ideals associated to I. When there is no confusion about the underlying ring, we will denote the set of associated primes of I simply by Ass(R/I) or Ass(I), see [START_REF] Sharp | Steps in commutative algebra[END_REF]Definition 9.32 and Remarks 9.33].

A presentation of an ideal I as intersection I = k i=1 q i with each q i is a primary ideal is called a primary decomposition of I. Let √ q i = p i for each i = 1, . . . , k.

The primary decomposition is called a minimal primary decomposition if none of the q i can be omitted in this intersection, and also p i = p j for all 1 ≤ i = j ≤ k. If I = k i=1 q i is a minimal primary decomposition of I, then we have Ass R (R/I) = {p 1 , . . . , p k }. The minimal members of Ass R (R/I) are called the minimal primes of I, and Min(I) denotes the set of minimal prime ideals of I. Moreover, the associated primes of I which are not minimal are called the embedded primes of I, see [START_REF] Sharp | Steps in commutative algebra[END_REF]Definition 4.15] for more information.

In particular, if I is a square-free monomial ideal, then Ass R (R/I) = Min(I), see [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.3.6].

Let I be an ideal of R and Min(I) = {p 1 , . . . , p r }. Given an integer n ≥ 1, the n-th symbolic power of I is defined to be the ideal I (n) = q 1 ∩ • • • ∩ q r , where q i is the primary component of I n corresponding to p i , see [START_REF] Villarreal | Monomial algebras[END_REF]Definition 4.3.22].

A well-known result of Brodmann [21] showed that the sequence {Ass R (R/I k )} k≥1 of associated prime ideals is stationary for large k. That is, there exists a positive integer k 0 such that Ass R (R/I k ) = Ass R (R/I k 0 ) for all k ≥ k 0 . The minimal such k 0 is called the index of stability of I, is denoted by astab(I), and Ass R (R/I k 0 ) is called the stable set of associated prime ideals of I, which is denoted by Ass ∞ (I), refer to [START_REF] Hoa | Stability of associated primes of monomial ideals[END_REF] in order to see some examples.

An ideal I ⊂ R satisfies the persistence property if Ass R (R/I k ) ⊆ Ass R (R/I k+1 ) for all positive integers k.

An ideal I ⊂ R has the strong persistence property if (I k+1 : R I) = I k for all positive integers k.

If I and J are two ideals in R, then we say that J is a superficial ideal for I if the following conditions are satisfied:

(i) G(J) ⊆ G(I), where G(L) denotes the unique minimal set of monomial generators of a monomial ideal L;

(ii) (I k+1 : R J) = I k for all positive integers k.

It should be noted that an ideal has a superficial ideal if and only if it has the strong persistence property, refer to [START_REF] Rajaee | Superficial ideals for monomial ideals[END_REF] for more information.

An ideal I ⊂ R satisfies the symbolic strong persistence property if (I (k+1) : R I (1) ) = I (k) for all positive integers k, where I (k) denotes the k-th symbolic power of I, see [START_REF] Reyes | On the strong persistence property for monomial ideals[END_REF].

An ideal I ⊂ R is called normally torsion-free if Ass R (R/I k ) ⊆ Ass R (R/I) for all k ≥ 1, see [START_REF] Herzog | Monomial Ideals[END_REF]Definition 1.4.5]. In particular, if I is a square-free monomial ideal, then I is normally torsion-free if and only if I k = I (k) for all k ≥ 1 (cf. [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.6]).

A monomial ideal I in a polynomial ring R = K[x 1 , . . . , x n ] over a field K is called nearly normally torsion-free if there exist a positive integer k and a monomial prime ideal p such that Ass R (R/I m ) = Min(I) for all 1 ≤ m ≤ k, and Ass R (R/I m ) ⊆ Min(I) ∪ {p} for all m ≥ k + 1, see [9,Definition 2.1].

Let R be a unitary commutative ring and I an ideal in R. An element f ∈ R is integral over I, if there exists an equation

f k + c 1 f k-1 + • • • + c k-1 f + c k = 0 with c i ∈ I i .
The set of elements I in R which are integral over I is the integral closure of I. The ideal I is integrally closed, if I = I, and I is normal if all powers of I are integrally closed, refer to [START_REF] Herzog | Monomial Ideals[END_REF]Definition 1.4.1].

The notion of integrality for a monomial ideal I can be described in a simpler way as following: a monomial u ∈ R = K[x 1 , . . . , x n ] is integral over I ⊂ R if and only if there exists an integer k such that u k ∈ I k , see [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.2].

Let R = K[x 1 , . . . , x n ] be a polynomial ring over a field K. An ideal I ⊂ R is called a monomial ideal if it admits a generating set consisting of monomials. Furthermore, I is said to be a square-free monomial ideal if it is generated by squarefree monomials.

Let G(I) be the unique minimal generating set of a monomial ideal I of a polynomial ring R. The support of a monomial u ∈ R, denoted by supp(u), is the set of variables that divide u (cf. [START_REF] Villarreal | Monomial algebras[END_REF]Definition 6.1.5]). Moreover, for a monomial ideal I, we set supp(I) = u∈G(I) supp(u). Given a square-free monomial ideal I ⊂ R, the Alexander dual of I, denoted by I ∨ , is given by

I ∨ = u∈G(I) (x i : x i ∈ supp(u)),
refer to [START_REF] Villarreal | Monomial algebras[END_REF]Definition 6.3.38 and Theorem 6.3.39].

Graph theory

In this section, we provide some necessary definitions related to graph theory, which can be found in [18] and [START_REF] West | Introduction to graph theory[END_REF].

Let G = (V (G), E(G)) be a finite simple undirected graph with the vertex set V (G) and the edge set E(G). The degree of a vertex v ∈ V (G) is denoted by deg(v) and it represents the number of vertices adjacent to v. A subgraph of G is a graph

H with V (H) ⊆ V (G)

and E(H) ⊆ E(G).

A subgraph H of G is called an induced subgraph if it contains all the edges of G that have both vertices in H. A set T ⊆ V (G) is called a vertex cover of G if it intersects every edge of G non-trivially. A vertex cover is called minimal if it does not properly contain any other vertex cover of G. For each vertex v ∈ V (G), the closed neighborhood of v in G is defined as follows:

N G [v] = {u ∈ V (G) : {u, v} ∈ E(G)} ∪ {v}.
When there is no confusion about the underlying graph, we will denote

N G [v] simply by N [v]. A subset S ⊆ V (G) is called a dominating set of G if S ∩ N [v] = ∅ for all v ∈ V (G).
A dominating set is called minimal if it does not properly contain any other dominating set of G. A minimum dominating set of G is a minimal dominating set with the smallest size. The dominating number of G, denoted by γ(G), is the size of its minimum dominating set, that is, γ(G) = min{|S| : S is a minimal dominating set of G}.

The dominating sets and domination numbers of graphs are well-studied topics in graph theory. Further information on dominations in graphs can be found in [START_REF] Haynes | Fundamentals of Domination in Graphs[END_REF].

In general, square-free monomial ideals can be associated with graphs in many different ways. We recall some commonly known definitions in this context. Let V (G) = {x 1 , x 2 , . . . , x n } and R = K[x 1 , . . . , x n ] be a polynomial ring in n variables over a field K. The edge ideal of G, denoted by I(G), is

I(G) = (x i x j : {x i , x j } ∈ E(G)).
The cover ideal of G, denoted by J(G), is

J(G) = ( x i ∈T
x i : T is a minimal vertex cover of G).

In [START_REF] Sharifan | Closed neighborhood ideal of a graph[END_REF], the closed neighborhood ideal of G has been introduced as

N I(G) = ( x j ∈N [x i ] x j : x i ∈ V (G)).
Moreover, in [START_REF] Sharifan | Closed neighborhood ideal of a graph[END_REF], the dominating ideal of G is defined as

DI(G) = ( x i ∈S x i : S is a minimal dominating set of G).
It is a well-recognized fact that J(G) is the Alexander dual of I(G) (cf. [START_REF] Herzog | Monomial Ideals[END_REF]Lemma 9.1.4]). It is shown in [START_REF] Sharifan | Closed neighborhood ideal of a graph[END_REF]Lemma 2.2

] that DI(G) is the Alexander dual of N I(G).

A path in G = (V (G), E(G)) is a sequence of distinct vertices x i 1 , . . . , x it such that {x i j , x i j+1 } ∈ E(G) for j = 1, . . . , t -1. The length of a path is the number of edges in it. If a graph G itself is a path on n vertices, then it is denoted by P n . A cycle, denoted by C n , is a graph with V (C n ) = {x 1 , . . . , x n } and

E(C n ) = {{x i , x i+1 } : i = 1, . . . , n -1} ∪ {{x 1 , x n }}.
A graph G is said to be chordal if it has no induced cycles of length greater than three. Equivalently, a graph G is chordal if and only if it admits a perfect elimination ordering of its vertices. A perfect elimination ordering of G is an ordering v 1 , . . . , v n of V (G) with the property that for each i, j, and l,

if i < j, i < l with v l , v j ∈ N G [v i ], then v l ∈ N G [v j ].
Note that chordal graphs are sometimes referred to as triangulated graphs.

Let G = (V (G), E(G)) be a finite simple graph. The graph G is called perfect if for any induced subgraph G S , with S ⊆ V (G), we have χ(G S ) = ω(G S ), where ω(H) denotes the size of the largest clique of a graph H, see [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Definition 5.1]. An imperfect graph is a graph that is not perfect.

In fact, according to [18,Theorem 14.18], a graph is perfect if and only if it contains no odd cycle of length at least five, or its complement, as an induced subgraph. Families of imperfect graphs include:

(1) Cycle graphs C 2n+1 ;

(2) Fullerenes (which by definition contain an odd 5-cycle);

(3) King graphs K m,n with min{m, n} ≥ 4;

(4) Helm graphs H n for odd n ≥ 5;

(5) Wheel graphs W 2n .

Hypergraph theory

In this section, we recall some notions and concepts from hypergraph theory, which can be found in [15] and [START_REF] Villarreal | Monomial algebras[END_REF]. A finite hypergraph H = (V (H), E(H)) on a vertex set V (H) = {x 1 , x 2 , . . . , x n } is a collection of edges E(H) = {E 1 , . . . , E m } with E i ⊆ V (H) for all i = 1, . . . , m. The incidence matrix of H is a n × m matrix A = (a ij ) such that a ij = 1 if x i ∈ E j , and a ij = 0 otherwise. A hypergraph H is called simple, if E i ⊆ E j implies i = j. Simple hypergraphs are also known as clutters. Moreover, if |E i | = d for all i = 1, . . . , m, then H is called a d-uniform hypergraph. A 2-uniform hypergraph H is just a finite simple graph.

Let V (H) = {x 1 , x 2 , . . . , x n } and R = K[x 1 , . . . , x n ] be a polynomial ring in n variables over a field K. The edge ideal of H is given by

I(H) = ( x j ∈E i x j : E i ∈ E(H)).
A subset W ⊆ V (H) is a vertex cover or transversal of H if W ∩ E i = ∅ for all i = 1, . . . , m. A transversal W is minimal if no proper subset of W is a transversal of H. The family of minimal transversals of H constitutes a simple hypergraph on V (H) called the transversal hypergraph of H, and denoted by T r (H). The cover ideal of the hypergraph H, denoted by J(H), is given by J(H) = (

x i ∈W x i : W is a minimal transversal of H).
In combinatorial commutative algebra, the transversals of a hypergraph are referred to as covers, and that is why the ideal J(H) is usually called the cover ideal of H. Moreover, similar to the case of edge ideal of graphs, the cover ideal J(H) is the Alexander dual of I(H), that is, J(H) = I(H) ∨ , consult [START_REF] Villarreal | Monomial algebras[END_REF]Theorem 6.3.39].

Let G be a graph with V (G) = {x 1 , . . . , x n }. The neighborhood hypergraph of G, denoted by N (G), is a hypergraph on the vertex set V (G) with edges N [x i ] for all i = 1, . . . , n. Note that the edge ideal of N (G) is the closed neighborhood ideal

N I(G).

A reader may find a confusing that we use the word "closed" with ideal but not with hypergraphs. It is because we want to keep the standard notations given in [15] and the naming of ideal given in [START_REF] Sharifan | Closed neighborhood ideal of a graph[END_REF]. The t-path hypergraph of G, denoted by H t (G), is a hypergraph on the vertex set V (G) whose edges are all distinct t-paths in G. The edge ideal of H t (G) is called the t-path ideal of G and is denoted by I t (G). When t = 2, then I 2 (G) is simply the edge ideal I(G) of G. If there is no t-path in G, then we set I t (G) = 0.

A simplicial complex ∆ on a set of vertices V (∆) = {x 1 , . . . , x n } is a collection of subsets of V (∆), with the property that if F ∈ ∆, then all subsets of F are also in ∆, including the empty set. An element of ∆ is called a face of ∆, and the maximal faces of ∆ with respect to inclusion are called facets of ∆, see [60, Page 15]. The facet ideal of ∆ is given by

I(∆) = ( x i ∈F x i : F is a facet of ∆).
The set of all facets of ∆ can be viewed as a simple hypergraph. In this case, the definition of the edge ideal of a hypergraph consisting of all the facets of ∆ is the same as the facet ideal of ∆.

Chapter 2

On the persistence property of monomial ideals

Stable set and index of stability of a monomial ideal

Let R be a commutative Noetherian ring and I be an ideal of R. A well-known result of Brodmann [21] showed that the sequence {Ass R (R/I k )} k≥1 of associated prime ideals is stationary for large k. That is, there exists a positive integer k 0 such that Ass R (R/I k ) = Ass R (R/I k 0 ) for all k ≥ k 0 . The minimal such k 0 is called the index of stability of I and Ass R (R/I k 0 ) is called the stable set of associated prime ideals of I, which is denoted by Ass ∞ (I). In recent researches, by using combinatorial tools, many papers have been published in order to describe the stable set of associated prime ideals of a monomial ideal and a square-free monomial ideal, see [14], [27], [START_REF] Hà | Embedded associated primes of powers of square-free monomial ideals[END_REF], and [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF] for more information. In this section, we study these notions. It should be noted that the results of this section can be found in [START_REF] Khashyarmanesh | On the stable set of associated prime ideals of monomial ideals and square-free monomial ideals[END_REF] and [START_REF] Khashyarmanesh | Some results on the associated primes of monomial ideals, Southeast Asian Bull[END_REF].

On the stable set of monomial ideals

The following result is a key lemma for us to show our main results. Notice that throughout this section, R = K[x 1 , . . . , x n ] stands for a polynomial ring over a field K. . We will prove that Ass R (R/I) = {p 1 , . . . , p m }. In order to achieve this claim, firstly, we show that m j=1,j̸ =i p 2 j-1 d j ⊈ p 2 i-1 d i for all i = 1, . . . , m. Suppose to the contrary that there exists x kt with x kt ∈ G(p i ) for all t = 1, . . . , 2 i-1 . Since x ℓ j / ∈ G(p i ) for all j = i + 1, . . . , m and u divides v, we have

i ∈ N with 1 ≤ i ≤ m such that
2 i-1 d t=1 x kt divides x d ℓ 1 x 2d ℓ 2 • • • x 2 i-2 d ℓ i-1 .
Then deg(

2 i-1 d t=1 x kt ) ≤ deg(x d ℓ 1 x 2d ℓ 2 • • • x 2 i-2 d ℓ i-1 )
, and so

2 i-1 d ≤ d + 2d + • • • + 2 i-2 d.
Hence,

2 i-1 ≤ 1 + 2 + • • • + 2 i-2
, which is a contradiction. for all i = 1, . . . , m, and hence

Case 3. Let

p d 1 ∩ p 2d 2 ∩ p 4d 3 ∩ • • • ∩ p 2 m-1 d m ,
is an irredundant presentation of I. Also, in view of [START_REF] Villarreal | Monomial algebras[END_REF]Corollary 6.1.8], every p id i is a p i -primary ideal for i = 1, . . . , m. This implies that

p d 1 ∩ p 2d 2 ∩ p 4d 3 ∩ • • • ∩ p 2 m-1 d
m is a minimal primary decomposition of I, and hence

Ass R (R/I) = Ass R (R/p d 1 ∩ p 2d 2 ∩ p 4d 3 ∩ • • • ∩ p 2 m-1 d m ) =
Ass R (R/p d 1 ∩ p 2d 2 ∩ p 4d 3 ∩ • • • ∩ p 2 m-1 d m ) = {p 1 , . . . , p m }.
This means there exist infinite monomial ideals with associated prime {p 1 , . . . , p m }.

Proposition 2.1.3. Let A = {p 1 , . . . , p m } and B = {p ′ 1 , . . . , p ′ t } be two sets of monomial prime ideals of R such that B ⊆ A. Then there exist monomial ideals I and J of R with the following properties:

(i) Ass R (R/I) = A, Ass R (R/J) = B; and, (ii) I ⊆ J, Ass R (J/I) = A\B.

Proof. Set

I := p 1 ∩ p 2 2 ∩ p 4 3 ∩ • • • ∩ p 2 m-1 m
. By virtue of Remark 2.1.2, we obtain

Ass R (R/I) = {p 1 , . . . , p m } = A. Now, let be the set of all R-submodules R/I of the form K/I, where Ass R (K/I) ⊆ A\B and K ⊇ I is a monomial ideal of R. Clearly, = ∅, and hence has a maximal element. Suppose that J is a monomial ideal of R such that J/I is a maximal element in .

We first show that Ass R (R/J) ⊆ B. To do this, suppose that p is an arbitrary element of Ass R (R/J). Then there exists a monomial v of R such that p = (J : R v). Thus, R/p ∼ = (Rv + J)/J. Set F := Rv + J. It is clear that F is a monomial ideal and that Ass R (F/J) = {p}. On the other hand, the following exact sequence 0 → J/I → F/I → F/J → 0, induces the following inclusions Ass R (F/I) ⊆ Ass R (J/I) ∪ Ass R (F/J) ⊆ (A\B) ∪ {p}.

Suppose, on the contrary, that p ∈ A\B. Then Ass R (F/I) ⊆ A\B. Hence, F/I ∈ . By maximality of J/I, we have F/I = J/I. Thus, v ∈ J, which is a contradiction. Therefore, p / ∈ A\B, and so p ∈ Ass R (F/I). Thus, p ∈ A. Since p / ∈ A\B, we get p ∈ B. Therefore, Ass R (R/J) ⊆ B. Now, we show that A\B ⊆ Ass R (J/I). Assume that q is an arbitrary element in A\B. Then q / ∈ B, and hence q / ∈ Ass R (R/J). On the other hand, the exact sequence 0 → J/I → R/I → R/J → 0, implies the inclusion Ass R (R/I) ⊆ Ass R (J/I) ∪ Ass R (R/J). Thus, q ∈ Ass R (J/I). Therefore, A\B ⊆ Ass R (J/I).

Finally, we show that B ⊆ Ass R (R/J). Suppose that b is an arbitrary element in B. Since B ⊆ A, we deduce that b / ∈ A\B, and thus b / ∈ Ass R (J/I). On the other hand, we have Ass R (R/I) ⊆ Ass R (J/I) ∪ Ass R (R/J). Thus, b ∈ Ass R (R/J). We therefore obtain B ⊆ Ass R (R/J), and the proof is complete. The following theorem has been proved also in [14,Theorem 1.2].

Theorem 2.1.5. Let p 1 , . . . , p m be nonzero monomial prime ideals of R. Then there exists a monomial ideal I of R such that Ass ∞ (I) = {p 1 , . . . , p m }.

Proof. By virtue of [START_REF] Herzog | Symbolic powers of monomial ideals and vertex cover algebras[END_REF]Corollary 2.2], there exists some d ∈ N such that for I :=

p d 1 ∩ p 2d 2 ∩ p 4d 3 ∩ • • • ∩ p 2 m-1 d m
, we have

I k = p kd 1 ∩ p 2kd 2 ∩ p 4kd 3 ∩ • • • ∩ p 2 m-1 kd m ,
for all k ∈ N. Now, by Remark 2.1.2, we obtain

Ass R (R/I k ) = Ass R (R/p kd 1 ∩ p 2kd 2 ∩ p 4kd 3 ∩ • • • ∩ p 2 m-1 kd m ) = {p 1 , . . . , p m },
for all k ∈ N. In fact, the index of stability of I is equal to 1. Therefore, Ass ∞ (I) = {p 1 , . . . , p m }, as desired.

Proposition 2.1.6. Let A = {p 1 , . . . , p m } and B = {p ′ 1 , . . . , p ′ t } be two sets of monomial prime ideals of R such that B ⊆ A. Then there exist monomial ideals I and J of R with the following properties:

(i) Ass ∞ (I) = A, Ass R (R/J) = B; and, (ii) I ⊆ J, Ass R (J/I) = A\B.

Proof. By Theorem 2.1.5, there exists an ideal I such that Ass ∞ (I) = Ass R (R/I) = A. Now, by using a method similar to the one has been used in the proof of Proposition 2.1.3, one can complete the proof. Corollary 2.1.7. Let A = {p 1 , . . . , p m } and B = {p ′ 1 , . . . , p ′ t } be two arbitrary sets of monomial prime ideals of R. Then there exist monomial ideals I and J of R such that (i) Ass ∞ (I) = A ∪ B, Ass R (R/J) = B; and, (ii) I ⊆ J, Ass R (J/I) = A\B.

Proof. Set D := A ∪ B. By Theorem 2.1.5, there exists an ideal I such that Ass ∞ (I) = Ass R (R/I) = D. Now, by using a method similar to the one has been used in the proof of Theorem 2.1.4, one can complete the proof.

On the stable set of square-free monomial ideals

In this subsection, we investigate the stable set of a square-free monomial ideal. Proposition 2.1.8. Let p 1 , . . . , p m be nonzero monomial prime ideals of R such that they are generated by disjoint non-empty subsets of {x 1 , . . . , x n }. Then there exists a square-free monomial ideal I such that Ass R (R/I k ) = {p 1 , . . . , p m } for all k ∈ N. Moreover, Ass ∞ (I) = {p 1 , . . . , p m }.

Proof. Clearly, in the polynomial ring R = K[x 1 , . . . , x n ], the sequence x 1 , . . . , x n is a permutable regular sequence. Hence, by [START_REF] Ratliff | On prime divisors of I n , n large[END_REF]Remark 2.8], for all positive integers k 1 , . . . , k m , we have the following equalities Ass R (R/p k 1 1 . . .

p km m ) = m i=1
Ass R (R/p i ) = {p 1 , . . . , p m }.

Also, by virtue of [START_REF] Ratliff | On prime divisors of I n , n large[END_REF]Theorem 2.7], we have

p k 1 1 • • • p km m = p k 1 1 ∩ • • • ∩ p km m for all positive integers k 1 , . . . , k m . Set I := p 1 • • • p m = p 1 ∩ • • • ∩ p m .
Then I is a squarefree monomial ideal and Ass R (R/I k ) = {p 1 , . . . , p m } for all k ∈ N. Moreover, we have Ass ∞ (I) = {p 1 , . . . , p m }. Remark 2.1.9. Let p 1 , . . . , p m be prime ideals as in Proposition 2.1.8. Then I := p 1 • • • p m is a square-free monomial ideal such that Ass R (R/I k ) = {p 1 , . . . , p m } for all k ∈ N and Ass ∞ (I) = {p 1 , . . . , p m }.

Example 2.1.10. Assume that p 1 = (x 1 , x 3 ), p 2 = (x 2 , x 4 , x 5 ), and p 3 = (x 6 ) are monomial prime ideals of R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ]. Put I := p 1 p 2 p 3 = (x 1 x 2 x 6 , x 1 x 4 x 6 , x 1 x 5 x 6 , x 2 x 3 x 6 , x 3 x 4 x 6 , x 3 x 5 x 6 ).

Then, by Proposition 2.1.8, we deduce that Ass R (R/I k ) = {p 1 , p 2 , p 3 } for all k ∈ N. Also, one has Ass ∞ (I) = {p 1 , p 2 , p 3 }. Theorem 2.1.11. Let A = {p 1 , . . . , p m } be a set of nonzero monomial prime ideals of R such that they are generated by disjoint non-empty subsets of {x 1 , . . . , x n }. Also, suppose that {A 1 , . . . , A r } is a partition of A. Then there exist square-free monomial ideals I 1 , . . . , I r such that, for all positive integers k 1 , . . . , k r , d,

(i) Ass R (R/I k i i ) = A i ; (ii) Ass R (R/I k 1 d 1 • • • I krd r ) = {p 1 , . . . , p m }; and, (iii) Ass ∞ (I k 1 1 • • • I kr r ) = Ass ∞ (I k 1 1 ) ∪ • • • ∪ Ass ∞ (I kr r ).
Proof. For all i = 1, . . . , r, suppose that A i = {p i,1 , . . . , p i,t i }. Hence, we get r i=1

A i = {p 1 , . . . , p m }. Set I i := p i,1 • • • p i,t i for all i = 1, . . . , r. By Proposition 2.1.8, we have Ass R (R/I k i i ) = {p i,1 , . . . , p i,t i } for all k i ∈ N and i = 1, . . . , r. Also, we derive that

I k 1 1 • • • I kr r = r i=1 t i j=1
p k i i,j . Thus, by [START_REF] Ratliff | On prime divisors of I n , n large[END_REF]Remark 2.8], we obtain

Ass R (R/(I k 1 1 • • • I kr r ) d ) = Ass R (R/ r i=1 t i j=1 p k i d i,j ) = r i=1 t i j=1 Ass R (R/p k i d i,j )
= {p 1 , . . . , p m }, for all positive integers k 1 , . . . , k r , d. It follows now from Proposition 2.1.8 that Ass ∞ (I k i i ) = {p i,1 , . . . , p i,t i } for all i = 1, . . . , r. So, the other assertion is obvious and the proof is complete.

Associated primes of square-free monomial ideals

We begin with the following lemma, which will be used in the proof of Theorem 2.1.15 Lemma 2.1.12. Let S be a commutative ring and let a 1 , . . . , a m be elements constituting a permutable S-sequence. Let J be an ideal generated by monomials in a t+1 , . . . , a m for some t ∈ N with 1 ≤ t ≤ m -1. Then (J : S a n 1 1 a n 2 2 • • • a nt t ) = J for all n 1 , n 2 , . . . , n t ∈ N.

Proof. We proceed by induction on t. In the case in which t = 1 follows from [77, Lemma 3]. Now, suppose, inductively, that t > 1 and that the result has been proved for t. Then, by using Lemma 3 in [START_REF] Kaplansky | R-sequences and homological dimension[END_REF] in conjunction with the inductive hypothesis, we obtain the following equalities

(J : S a n 1 1 a n 2 2 • • • a n t+1 t+1 ) = ((J : S a n 1 1 a n 2 2 • • • a nt t ) : S a n t+1 t+1 ) = (J : S a n t+1 t+1 ) = J.
This completes the inductive step and the proof.

Corollary 2.1.13. Let I be a monomial ideal of R with G(I) = {u 1 , . . . , u m } and J be an arbitrary ideal of R. Also, suppose that x ℓ j / ∈ m i=1 supp(u i ) for all j = 1, . . . , t with 1 ≤ t ≤ n -1. If v := x n 1 ℓ 1 • • • x nt ℓt with n 1 , n 2 , . . . , n t ∈ N, then (vJ + I : R v) = J + I.

Proof. By virtue of Lemma 2.1.12, one has (I : R v) = I. Hence, we deduce that (vJ + I : R v) = (vJ : R v) + (I : R v) = J + I, and the proof is over. Proposition 2.1.14. Let I and J be (not necessary monomial) ideals of a commutative Noetherian ring S. Also, suppose that I is a radical ideal of S. Then (I : S J) is a radical ideal and (I : S J) = (I : S √ J).

Proof. It is clear (I : S J) ⊆ (I : S J). For the reverse inclusion, suppose that h ∈ (I : S J). Then h ℓ J ⊆ I for some ℓ ∈ N. Assume that J = (y 1 , . . . , y k ). Then h ℓ y i ∈ I for all i = 1, . . . , k. Hence, (hy i ) ℓ ∈ I for all i = 1, . . . , k. Since I is a radical ideal, it follows that hJ ⊆ I, and so h ∈ (I : S J). Therefore, (I : S J) = (I : S J). Now, we prove that (I : S J) = (I : S √ J). Clearly, (I : S √ J) ⊆ (I : S J). Suppose that h is an arbitrary element in (I : S J). Then hJ ⊆ I. Assume that √ J = (z 1 , . . . , z r ). Then there exist β 1 , . . . , β r ∈ N such that z β i i ∈ J for all i = 1, . . . , r. Hence, hz β i i ∈ I for all i = 1, . . . , r. Therefore, (hz i ) β i ∈ I for all i = 1, . . . , r. Since I is a radical ideal, we have √ I = I. Hence, h √ J ⊆ I, and so h ∈ (I : S √ J). Consequently, (I : S J) = (I : S √ J).

In the following theorem we study the associated prime ideals of a square-free monomial ideal of R = K[x 1 , . . . , x n ]. Theorem 2.1.15. Let I be a square-free monomial ideal of R, G(I) = {u 1 , . . . , u m }, and p ∈ Ass R (R/I). Then there exists a square-free monomial v of R with supp(v) ⊆ m i=1 supp(u i ) such that p = (I : R v).

Proof. On account of [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.3.10], there exists a monomial w of R such that p = (I : R w). In view of Lemma 2.1.12, we may assume that w = x α 1 ℓ 1 • • • x αt ℓt , where the x ℓ i 's are distinct, α 1 , . . . , α t are positive integers, and supp(w) ⊆ m i=1 supp(u i ). By setting v := x ℓ 1 • • • x ℓt , we show that p = (I : R v). Clearly, supp(w) = supp(v). Since I is a square-free monomial ideal, by Proposition 2.1.14, we have the following equalities p = (I : R w) = (I : R √ w) = (I : R v), as claimed.

The next example shows that Theorem 2.1.15 can fail for monomial ideals in general.

Proposition 2.1.17. Let I and J be two square-free monomial ideals of R such that Ass R (R/I) = Ass R (R/J). Then I = J.

Proof. Since I and J are square-free monomial ideals, by [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.3.6], we have Min(I) = Ass R (R/I) = {p 1 , . . . , p t } = Ass R (R/J) = Min(J). We therefore obtain I = t i=1 p i = J.

Note that Proposition 2.1.17 can fail for monomial ideals in general.

Example 2.1.18. Consider the ideals I = (x 2 1 , x 3 2 , x 4 3 ) and J = (x 4 1 , x 3 2 , x 2 3 ) in the polynomial ring R = K[x 1 , x 2 , x 3 ], where K is a field and x 1 , x 2 , x 3 are indeterminates. It follows from [START_REF] Herzog | Monomial Ideals[END_REF]Proposition 1.3.7] that Ass R (R/I) = Ass R (R/J) = {(x 1 , x 2 , x 3 )}, while I = J.

Suppose that A = {p 1 , . . . , p m } and B = {p ′ 1 , . . . , p ′ t } are two sets of monomial prime ideals of R such that the minimal elements of these sets with respect to inclusion are the same. Now, one can ask the question whether there exist squarefree monomial ideals I and J of R such that Ass R (R/I) = A and Ass R (R/J) = B?

In the following, we provide a partial answer to the above question.

Proposition 2.1.19. Let A = {p 1 , . . . , p m } and B = {p ′ 1 , . . . , p ′ t } be two sets of monomial prime ideals of R such that the elements in each of these sets are minimal with respect to inclusion and that B ⊆ A. Then there exist unique square-free monomial ideals I and J of R with the following properties: Ass R (K/I) ⊆ A\B and K ⊇ I is a square-free monomial ideal of R. Clearly, = ∅, and hence has a maximal element. Suppose that J is a square-free monomial ideal of R such that J/I is a maximal element in . We first show that Ass R (R/J) ⊆ B. To do this, suppose that p is an arbitrary element of Ass R (R/J). Then, by Theorem 2.1.15, there exists a square-free monomial v of R such that p = (J : R v). Thus, R/p ∼ = (Rv + J)/J. Set F := Rv + J. It is clear that F is a square-free monomial ideal and that Ass R (F/J) = {p}. Now, in view of Proposition 2.1.3, we have Ass R (R/J) = B and Ass R (J/I) = A\B.

Set

J 1 := p ′ 1 ∩ • • • ∩ p ′ t . Since J 1 = √ J 1
, again by [60, Corollary 1.2.5], we obtain J 1 is a square-free monomial ideal such that Min(J 1 ) = Ass R (R/J 1 ) = {p ′ 1 , . . . , p ′ t }. On account of Ass R (R/J) = Ass R (R/J 1 ) = B, Proposition 2.1.17 implies that

J = p ′ 1 ∩ • • • ∩ p ′ t
, and the proof is done. Example 2.1.20. Let B = {(x 2 ), (x 4 )} and A = {(x 2 ), (x 4 ), (x 1 , x 3 )} be two sets of monomial prime ideals of the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 ], where K is a field and x 1 , x 2 , x 3 , x 4 are indeterminates. Now, by setting I := (x 2 ) ∩ (x 4 ) ∩ (x 1 , x 3 ) = (x 1 x 2 x 4 , x 2 x 3 x 4 ), and J := (x 2 ) ∩ (x 4 ) = (x 2 x 4 ), we obtain Ass R (R/I) = {(x 2 ), (x 4 ), (x 1 , x 3 )}, Ass R (R/J) = {(x 2 ), (x 4 )}, and Ass R (J/I) = {(x 1 , x 3 )}. Now, we state the main result of this subsection in the next theorem.

Theorem 2.1.21. Let A = {p 1 , . . . , p m } and B = {p ′ 1 , . . . , p ′ t } be two arbitrary sets of monomial prime ideals of R such that the elements of A ∪ B are minimal with respect to inclusion. Then there exist unique square-free monomial ideals I and J of R with the following properties:

(i) Ass R (R/I) = A ∪ B, Ass R (R/J) = B; and, (ii) I ⊆ J, Ass R (J/I) = A\B.

Proof. Set D := A ∪ B. Then, by Proposition 2.1.19, there exist unique square-free monomial ideals

I := p 1 ∩ • • • ∩ p m ∩ p ′ 1 ∩ • • • ∩ p ′ t and J := p ′ 1 ∩ • • • ∩ p ′ t of R such that I ⊆ J, Ass R (R/I) = D = A ∪ B, Ass R (R/J) = B,
and Ass R (J/I) = D\B. Since D\B = A\B, we deduce that Ass R (J/I) = A\B. This finishes the proof.

Example 2.1.22. Let B = {(x 4 ), (x 1 , x 3 ), (x 2 , x 3 )} and A = {(x 4 ), (x 5 ), (x 1 , x 3 )} be two sets of monomial prime ideals in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 ], where K is a field and x 1 , x 2 , x 3 , x 4 , x 5 are indeterminates. Now, set J := (x 4 ) ∩ (x 1 , x 3 ) ∩ (x 2 , x 3 ) = (x 3 x 4 , x 1 x 2 x 4 ), and I := (x 4 ) ∩ (x 5 ) ∩ (x 1 , x 3 ) ∩ (x 2 , x 3 ) = (x 1 x 2 x 4 x 5 , x 3 x 4 x 5 ).

As the elements of A ∪ B = {(x 4 ), (x 5 ), (x 1 , x 3 ), (x 2 , x 3 )} are minimal with respect to inclusion, by Theorem 2.1.21, we get Ass R (R/J) = {(x 4 ), (x 1 , x 3 ), (x 2 , x 3 )}, Ass R (R/I) = {(x 4 ), (x 5 ), (x 1 , x 3 ), (x 2 , x 3 )}, and Ass R (J/I) = {(x 5 )}.

Associated prime ideals under squeezing

In this subsection, we investigate the relation among associated prime ideals of three nested ideals. Suppose that J 1 ⊆ I ⊆ J 2 are three ideals of R such that Ass R (R/J 1 ) = Ass R (R/J 2 ). Then can we conclude that Ass R (R/J 1 ) = Ass R (R/I) = Ass R (R/J 2 )? As we shall show in Example 2.1.25, there is no general result in this context even for monomial ideals. Although, there exists a similar result for minimal prime ideals which we state in Theorem 2.1.23. We also show that if J 1 and J 2 are two square-free monomial ideals of R and J 1 ⊆ I ⊆ J 2 , then the answer of the above question is affirmative in this situation which we give in Proposition 2.1.24.

Theorem 2.1.23. (Squeeze theorem for minimal prime ideals) Let I, J 1 , and J 2 be ideals of a commutative Noetherian ring S such that J 1 ⊆ I ⊆ J 2 and that Min(J 1 ) = Min(J 2 ) = {p 1 , . . . , p t }. Then Min(I) = {p 1 , . . . , p t }.

Proof. Let I = Q 1 ∩ • • • ∩ Q m be
a minimal primary decomposition of I such that √ Q i = q i for all i = 1, . . . , m, and also let Min(I) = {q i 1 , . . . , q is }. Suppose to the contrary that s = t. Then t < s or s < t. First assume that t < s. Since √ I = s j=1 q i j and √ J 1 = √ J 2 = t i=1 p i , we have s j=1 q i j = t i=1 p i . Also, since s j=1 q i j ⊆ p 1 , there exists j ∈ N with 1 ≤ j ≤ s such that q i j ⊆ p 1 . It follows from the minimality of p 1 that p 1 = q i j . Without loss of generality, we may assume that p 1 = q i 1 . Also, s j=1 q i j ⊆ p 2 . Hence, there exists 1 ≤ j ≤ s such that q i j ⊆ p 2 . Suppose to the contrary that q i j = q i 1 . Then p 1 ⊆ p 2 , and, by the minimality of p 2 , we obtain p 1 = p 2 , which is a contradiction. Hence, q i j = q i 1 . Without loss of generality, we may assume that q i j = q i 2 . Then p 2 = q i 2 . Continuing this procedure yields p j = q i j , where j = 1, . . . , t. Because s j=1 q i j = t i=1 p i , this implies that q i 1 ∩ • • • ∩ q it ⊆ q i t+1 ∩ • • • ∩ q is . Hence, there exists 1 ≤ k ≤ t such that q i k ⊆ q is , and, by the minimality of q is , we deduce that q i k = q is , which is a contradiction. In a similar way, we conclude that the hypothesis s < t leads to a contradiction. We therefore get s = t, and so Min(I) = {q i 1 , . . . , q is } = {p 1 , . . . , p t }. Proposition 2.1.24. Let J 1 and J 2 be square-free monomial ideals of R and I be an ideal of R such that J 1 ⊆ I ⊆ J 2 and that Ass R (R/J 1 ) = Ass R (R/J 2 ) = {p 1 , . . . , p t }.

Then

Ass R (R/I) = Min(I) = {p 1 , . . . , p t }.

Proof. According to Proposition 2.1.17, it follows that J 1 = J 2 . This yields that I = J 1 = J 2 , and so Ass R (R/I) = Min(I) = {p 1 , . . . , p t }.

Note that Proposition 2.1.24 can fail for monomial ideals in general.

Example 2.1.25. Consider the ideals J 1 = (x 1 , x 2 , x 4 ) 3 , J 2 = (x 1 , x 2 , x 4 ), and I = (x 1 , x 2 , x 4 ) 3 + (x 1 , x 2 , x 3 , x 4 )(x 1 , x 2 , x 4 ) 2 , in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 ], where K is a field and x 1 , x 2 , x 3 , x 4 are indeterminates. Then J 1 ⊆ I ⊆ J 2 and, by computation with Macaulay2, we have Ass R (R/J 1 ) = Ass R (R/J 2 ) = {(x 1 , x 2 , x 4 )}, while Ass R (R/I) = {(x 1 , x 2 , x 4 ), (x 1 , x 2 , x 3 , x 4 )}.

Some recursive formulas for associated prime ideals

We begin with the following lemma, which will be used in the proof of Theorem 2.1.27. Lemma 2.1.26. Let R be a polynomial ring with coefficients in a field K and indeterminates x 1 , . . . , x n . Also, let Q = (x a 1 i 1 , . . . , x ar ir ) be a monomial ideal of R and h = x b 1 j 1 • • • x bs js be a monomial of R such that x i 1 , . . . , x ir , x j 1 , . . . , x js ∈ {x 1 , . . . , x n } and a 1 , . . . , a r , b 1 , . . . , b s ∈ N. Then Ass R (R/hQ) = {(x i 1 , . . . , x ir ), (x j 1 ), . . . , (x js )}.

Proof. Since Ass R (R/Q) = {(x i 1 , . . . , x ir )}, we have (x i 1 , . . . , x ir ) = (Q : R v) for some monomial v in R. Since (hQ : R hv) = (Q : R v) and (hQ : R hv) ∈ Ass R (R/hQ), we deduce that (x i 1 , . . . , x ir ) ∈ Ass R (R/hQ). In addition, since (x j k ) ∈ Min(hQ) for all k = 1, . . . , s, we obtain (x j 1 ), . . . , (x js ) ∈ Ass R (R/hQ). Therefore, one has {(x i 1 , . . . , x ir ), (x j 1 ), . . . , (x js )} ⊆ Ass R (R/hQ).

For completing the proof, we show the reverse inclusion. To do this, set L := supp(h) ∩ {x i 1 , . . . , x ir }. Suppose that c k := a k + b k when x i k ∈ L, and also c k := a k when x i k / ∈ L, for all k = 1, . . . , r. Now, put Q ′ := (x c 1 i 1 , x c 2 i 2 , . . . , x cr ir ). We verify that hQ = hR ∩ Q ′ . It is easy to see that the unique minimal set of monomial generators of hQ is G(hQ) = {hx a 1 i 1 , . . . , hx ar ir }. Also, the unique minimal set of monomial generators of hR

∩ Q ′ is G(hR ∩ Q ′ ) = {lcm(h, x c 1 i 1 )
, lcm(h, x c 2 i 2 ), . . . , lcm(h, x cr ir )}.

If x i k / ∈ L, for some 1 ≤ k ≤ r, then c k = a k , and so lcm(h,

x c k i k ) = hx a k i k . Also, if x i k ∈ L, for some 1 ≤ k ≤ r, then c k = a k + b k , and so lcm(h, x c k i k ) = hx a k i k . Thus, G(hR ∩ Q ′ ) = G(hQ), and therefore hQ = hR ∩ Q ′ . Since the following R-homomorphism θ : R/hR ∩ Q ′ → R/hR ⊕ R/Q ′ ,
given by θ(r

+ hR ∩ Q ′ ) = (r + hR, r + Q ′ ) for r ∈ R, is monomorphism, we deduce that Ass R (R/hR ∩ Q ′ ) ⊆ Ass R (R/hR) ∪ Ass R (R/Q ′ ),
and also

Ass R (R/hQ) ⊆ {(x i 1 , . . . , x ir ), (x j 1 ), . . . , (x js )}.

This finishes the proof.

Theorem 2.1.27. Let I be a monomial ideal of R with G(I) = {u 1 , . . . , u m }. Also, assume that there exists a monomial h = x b 1 j 1 • • • x bs js such that h|u i for all i = 1, . . . , m. By setting J := (u 1 /h, . . . , u m /h), we have Ass R (R/I) = Ass R (R/J) ∪ {(x j 1 ), . . . , (x js )}.

Proof. Since h|u i for all i = 1, . . . , m, we have I = hJ. Assume that J = Q 1 ∩• • •∩Q k is a minimal primary decomposition of J such that √ Q i = p i for all i = 1, . . . , k. Then we get I = h( k i=1 Q i ) = k i=1 (hQ i ), and so the following R-homomorphism

ϕ : R/I → R/hQ 1 ⊕ • • • ⊕ R/hQ k ,
given by ϕ(r + I) = (r + hQ 1 , . . . , r + hQ k ) for r ∈ R, is monomorphism. Therefore,

Ass R (R/I) ⊆ Ass R (R/hQ 1 ) ∪ • • • ∪ Ass R (R/hQ k ).
Now, Lemma 2.1.26 implies that Ass R (R/hQ i ) = {p i , (x j 1 ), . . . , (x js )} for all i = 1, . . . , k. Thus, we obtain Ass R (R/I) ⊆ {p 1 , . . . , p k , (x j 1 ), . . . , (x js )}.

For completing the proof, we need to show that {p 1 , . . . , p k , (x j 1 ), . . . , (x js )} ⊆ Ass R (R/I).

To do this, choose p ∈ Ass R (R/J). Then p = (J : R v) for some monomial v in R.

Since (hJ : R hv) = (J : R v) and also (hJ : R hv) ∈ Ass R (R/I), it is easy to see that {p 1 , . . . , p k } ⊆ Ass R (R/I).

On the other hand, since for all ℓ ∈ {1, . . . , s}, (x j ℓ ) ∈ Min(I), one can conclude that {(x j 1 ), . . . , (x js )} ⊆ Ass R (R/I), and the proof is done.

Corollary 2.1.28. Let I be a monomial ideal of R with G(I) = {u 1 , . . . , u m }. Also, assume that there exists a monomial h = x α 1 j 1 • • • x αt jt in R such that h|u i for all i = 1, . . . , m. By setting J := (u 1 /h, . . . , u m /h), we have Min(I) = {(x j 1 ), (x j 2 ), . . . , (x jt )} ∪ {p ∈ Min(J) | supp(p) ∩ {x j 1 , . . . , x jt } = ∅}. Remark 2.1.29. ( [START_REF] Villarreal | Monomial algebras[END_REF]Exercise 6.1.26]) Let R ′ = K[x 2 , . . . , x n ] and R = R ′ [x 1 ] be polynomial rings over a field K and x 1 , . . . , x n are indeterminates. If I ′ is an ideal of R ′ and p ∈ Ass R (R/(I ′ , x 1 )), then (a) p = x 1 R + p ′ R, where p ′ is a prime ideal of R ′ ; and, (b) p ′ is an associated prime of R ′ /I ′ .

In the following theorem we provide a generalization of Remark 2.1.29. Theorem 2.1.30. Suppose that I with G(I) = {u 1 , . . . , u m-1 , x α 1 j 1 • • • x αr jr } is a monomial ideal of R such that x jt / ∈ supp(u i ) for all i = 1, . . . , m -1 and t = 1, . . . , r. By setting J := (u 1 , . . . , u m-1 ), we have Ass R (R/I) = {p | p = q + x jt R for all t = 1, . . . , r and q ∈ Ass R (R/J)}.

Proof. Let J = Q 1 ∩ • • • ∩ Q s be
a minimal primary decomposition of J such that √ Q i = p i for all i = 1, . . . , s. For simplicity, we put h := x α 1 j 1 • • • x αr jr . Then we have

I = J + hR = hR + s i=1 Q i = s i=1 (hR + Q i ).
Also, there exists an R-monomorphism θ : R/I →

s i=1 R/hR + Q i ,
given by θ(r + I) = (r + hR + Q 1 , . . . , r + hR + Q s ) for r ∈ R. Hence,

Ass R (R/I) ⊆ s i=1
Ass R (R/hR + Q i ).

Next, fix 1 ≤ i ≤ s, and let Q i = (x c 1 i 1 , . . . , x c k i k ) with c 1 , . . . , c k ∈ N. Since

Q i + hR = Q i + r t=1 (x αt jt ) = r t=1 (Q i + (x αt jt )),
there exists an R-monomorphism

ϕ : R/Q i + hR → r t=1 R/Q i + (x αt jt ),
given by ϕ(r + Q i + hR) = (r + Q i + (x α 1 j 1 ), . . . , r + Q i + (x αr jr )) for r ∈ R, and so

Ass R (R/Q i + hR) ⊆ r t=1
Ass R (R/Q i + (x αt jt )).

Since Q i + (x αt jt ) = (x c 1 i 1 , . . . , x c k i k , x αt jt ) is a (x i 1 , . . . , x i k , x jt )-primary ideal, for all t = 1, . . . , r, it is easy to see that

Ass R (R/Q i + (x αt jt )) = (x i 1 , . . . , x i k , x jt ) = p i + (x jt ).

Therefore, we get

Ass R (R/I) ⊆ {p | p = q + x jt R for all t = 1, . . . , r and q ∈ Ass R (R/J)}.

For the reverse inclusion, suppose that q ∈ Ass R (R/J). Then there exists monomial v in R such that q = (J : R v). On account of Lemma 2.1.12, we may assume that supp(v) ⊆ m-1 i=1 supp(u i ). Fix 1 ≤ t ≤ r and put h

′ := x α 1 j 1 • • • x α t-1 j t-1 x αt-1 jt x α t+1 j t+1 • • • x αr jr . Since x js / ∈ m-1 i=1 supp(u i )
, where 1 ≤ s ≤ r, by Lemma 2.1.12, it is easy to see that (J : R v) = (J : R vh ′ ) and (x jt ) = (x jt : R v). We therefore get the following equalities q + (x jt ) = (J : R v) + (x jt : R v) = (J : R v) + (x jt h ′ : R vh ′ ) = (J : R v) + (h : R vh ′ ) = (J : R vh ′ ) + (h : R vh ′ ) = (J + hR : R vh ′ ) = (I : R vh ′ ). Also, since (I : R vh ′ ) ∈ Ass R (R/I), we derive that q + (x jt ) ∈ Ass R (R/I). Thus, {p | p = q + x jt R for all t = 1, . . . , r and q ∈ Ass R (R/J)} ⊆ Ass R (R/I).

Corollary 2.1.31. Suppose that I with G(I) = {u 1 , . . . , u m-1 , w 1 , . . . , w r } is a monomial ideal of R such that supp(w i ) ∩ supp(w j ) = ∅ for all 1 ≤ i = j ≤ r and supp(w i ) ∩ supp(u j ) = ∅ for all 1 ≤ i ≤ r and 1 ≤ j ≤ m -1. Then the set of associated prime ideals of I is equal to the following set {q + r i=1 z i R | for all q ∈ Ass R (R/J) and z i ∈ supp(w i ) with i = 1, . . . , r}, where J := (u 1 , . . . , u m-1 ).

Proof. We proceed by induction on r. In the case in which r = 1, the claim can be deduced from Theorem 2.1.30. Now, suppose, inductively, that r > 1 and that result has been proved for r. Set J 1 := (u 1 , . . . , u m-1 , w 1 , . . . , w r-1 ). Theorem 2.1.30 implies that Ass R (R/I) = {q ′ + z r R | for all q ′ ∈ Ass R (R/J 1 ) and z r ∈ supp(w r )}.

Moreover, our inductive hypothesis yields that the set Ass R (R/J 1 ) is equal to {q + r-1 i=1 z i R | for all q ∈ Ass R (R/J) and z i ∈ supp(w i ) with i = 1, . . . , r -1}.

This completes the inductive step and the proof.

Example 2.1.32. Consider the monomial ideal I = (x 3 1 x 2 2 x 3 , x 2 1 x 2 2 x 3 x 4 , x 2 1 x 4 2 x 2 3 x 5 , x 1 x 2 2 x 3 6 x 7 , x 1 x 2 2 x 8 x 2 9 ), in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 , x 8 , x 9 ], where K is a field and x 1 , . . . , x 9 are indeterminates. Now, put h := x 1 x 2 2 and J := (x 2 1 x 3 , x 1 x 3 x 4 , x 1 x 2 2 x 2 3 x 5 , x 3 6 x 7 , x 8 x 2 9 ).

By Theorem 2.1.27, we can conclude that

Ass R (R/I) = Ass R (R/J) ∪ {(x 1 ), (x 2 )}.

Next, set J 1 := (x 2 1 x 3 , x 1 x 3 x 4 , x 1 x 2 2 x 2 3 x 5 ), w 1 := x 3 6 x 7 , and w 2 := x 8 x 2 9 . Corollary 2. 1.31 shows that Ass R (R/J) is given by {(q, x 6 , x 8 ), (q, x 6 , x 9 ), (q, x 7 , x 8 ), (q, x 7 , x 9 ) | for all q ∈ Ass R (R/J 1 )}.

Also, put h := x 1 x 3 and J 2 := (x 1 , x 4 , x 2 2 x 3 x 5 ). By Theorem 2.1. 27

, one derives that

Ass R (R/J 1 ) = Ass R (R/J 2 ) ∪ {(x 1 ), (x 3 )}.

Again, set J 3 := (x 1 , x 4 ). By Corollary 2.1.31, we obtain Ass R (R/J 2 ) = {(q ′ , x 2 ), (q ′ , x 3 ), (q ′ , x 5 ) | for all q ′ ∈ Ass R (R/J 3 )}.

Finally, Ass R (R/J 3 ) = {(x 1 , x 4 )}, and so the procedure has been terminated. Then we have

Ass R (R/J 2 ) = {(x 1 , x 4 , x 2 ), (x 1 , x 4 , x 3 ), (x 1 , x 4 , x 5 )}, and Ass R (R/J 1 ) = {(x 1 , x 4 , x 2 ), (x 1 , x 4 , x 3 ), (x 1 , x 4 , x 5 ), (x 1 ), (x 3 )}.

Therefore, Ass R (R/J) is given by Ass R (R/J) = {(x 1 , x 4 , x 2 , x 6 , x 8 ), (x 1 , x 4 , x 3 , x 6 , x 8 ), (x 1 , x 4 , x 5 , x 6 , x 8 ), (x 1 , x 6 , x 8 ), (x 3 , x 7 , x 8 ), (x 3 , x 6 , x 8 ), (x 1 , x 4 , x 2 , x 6 , x 9 ), (x 1 , x 4 , x 3 , x 6 , x 9 ), (x 1 , x 4 , x 5 , x 6 , x 9 ), (x 1 , x 6 , x 9 ), (x 3 , x 6 , x 9 ), (x 1 , x 4 , x 2 , x 7 , x 8 ), (x 1 , x 4 , x 3 , x 7 , x 8 ), (x 1 , x 4 , x 5 , x 7 , x 8 ), (x 1 , x 7 , x 8 ), (x 1 , x 4 , x 5 , x 7 , x 9 ), (x 1 , x 7 , x 9 ), (x 3 , x 7 , x 9 ), (x 1 , x 4 , x 2 , x 7 , x 9 ), (x 1 , x 4 , x 3 , x 7 , x 9 )}.

Ultimately, we conclude that Ass R (R/I) is given by Ass R (R/I) = {(x 1 , x 4 , x 2 , x 6 , x 8 ), (x 1 , x 4 , x 3 , x 6 , x 8 ), (x 1 , x 4 , x 5 , x 6 , x 8 ), (x 1 ), (x 2 ), (x 1 , x 6 , x 8 ), (x 3 , x 6 , x 8 ), (x 1 , x 4 , x 2 , x 6 , x 9 ), (x 1 , x 4 , x 3 , x 6 , x 9 ), (x 1 , x 4 , x 5 , x 6 , x 9 ), (x 1 , x 6 , x 9 ), (x 3 , x 6 , x 9 ), (x 1 , x 4 , x 2 , x 7 , x 8 ), (x 1 , x 4 , x 3 , x 7 , x 8 ), (x 1 , x 4 , x 5 , x 7 , x 8 ), (x 1 , x 7 , x 8 ), (x 1 , x 4 , x 5 , x 7 , x 9 ), (x 1 , x 7 , x 9 ), (x 3 , x 7 , x 9 ), (x 1 , x 4 , x 2 , x 7 , x 9 ), (x 1 , x 4 , x 3 , x 7 , x 9 ) (x 3 , x 7 , x 8 )}.

Moreover, direct computation with Macaulay2 shows that our result is true.

Persistence property and the Alexander dual

We first recall some definitions from [START_REF] He | Algebraic properties of the path ideal of a tree[END_REF] and [START_REF] West | Introduction to graph theory[END_REF] which will be needed in the following. It should be noted that the results of this section can be found in [START_REF] Nasernejad | On the Alexander dual of the path ideals of rooted and unrooted trees[END_REF].

Definition 2.2.1.

A directed edge of a graph is an assignment of a direction to an edge of a graph. If {w, u} is an edge, we write (w, u) to denote the directed edge, where the direction is from w to u. A graph is a directed graph if each edge has been assigned a direction. A path of length t in a directed graph is a sequence of vertices i 1 , . . . , i t+1 such that e j = (i j , i j+1 ) is a directed edge for j = 1, . . . , t. Fix a positive integer t and a directed graph G. The path ideal of G of length t is the following monomial ideal

I t (G) := (x i 1 • • • x i t+1 : i 1 , . . . , i t+1 is a path of G of length t),
and also the Alexander dual of I t (G) is defined as follows:

I t (G) ∨ := i 1 ,...,i t+1 is a path of G of length t (x i 1 , . .

. , x i t+1 ).

A tree T can be considered as a directed graph by choosing any vertex of T to be the root of the tree, and assigning to each edge the direction "away" from the root. Since T is a tree, the assignment of a direction will always be possible. A rooted tree T is a tree with one vertex chosen as root. If tree T has no such root, then we say that T is unrooted.

As an example, consider T 1 = (V (T 1 ), E(T 1 )) and T 2 = (V (T 2 ), E(T 2 )), where

V (T 1 ) = V (T 2 ) = {v 1 , v 2 , v 3 , v 4 , v 5 }, E(T 1 ) = {{v 1 , v 2 }, {v 1 , v 3 }, {v 2 , v 4 }, {v 2 , v 5 }}, and E(T 2 ) = {(v 1 , v 2 ), (v 1 , v 3 ), (v 2 , v 4 ), (v 2 , v 5
)} in the following graphs. Then the tree T 1 , the left graph in Figure 1, is an example of a tree which is not rooted, while the tree T 2 , the right graph in Figure 1, is rooted at the vertex v 1 .

v 1 v 3 v 2 v 5 v 4 v 1 v 3 v 2 v 5 v 4 Figure 1
It has already been proved in [16,Theorem 5.8] and [17, Theorem 3.1] (when t = 2) the following theorem. Theorem 2.2.2. Let K 1,n be a star graph with the vertex set {z, x 1 , . . . , x n } and centre z. Let J 2 be the corresponding Alexander dual of

I 2 (K 1,n ). Then (1) (z, x 1 , . . . , x n ) ∈ Ass(R/J s 2 ) for s ≥ n -1; and, (2) (z, x 1 , . . . , x n ) / ∈ Ass(R/J s 2 ) for s < n -1, where (z, x 1 , . . . , x n ) ⊂ K[z, x 1 , . . . , x n ] is the maximal ideal.
Next, in [16,Conjecture 6.5], made up the following conjecture.

Conjecture 2.2.3. Let G be an unrooted tree on the vertex set

V G = {z, x 1 , . . . , x n } with the edge set E G which is {zx 1 , zx 2 , . . . , zx k , x 1 x k+1 , x 2 x k+2 , . . . , x k x 2k , x k+1 x 2k+1 , x k+2 x 2k+2 , . . . , x 2k x 3k , . . .}. Then (1) (z, x 1 , . . . , x n ) ∈ Ass(R/J s 2ℓ ) for s ≥ k -1; and, (2) (z, x 1 , . . . , x n ) / ∈ Ass(R/J s 2ℓ ) for s < k -1,
where 2ℓ is the length of the longest even path of G.

In this section, we provide an affirmative answer to this conjecture. To do this, we begin with the following definition.

Definition 2.2.4.

A tree is said to be starlike if exactly one of its vertices has degree greater than two. This vertex is called the center of the starlike.

We also say that T is a symmetric starlike tree if T is a starlike tree and

d(u, v) = d(u ′ , v ′ ) for all u, u ′ , v, v ′ ∈ L T with u = v and u ′ = v ′ .
For instance, consider the following trees. Then the left graph in Figure 2 is an example of a starlike tree with center z, while the right graph in Figure 2 is an example of a symmetric starlike tree with center z. z In the following, we probe the persistence property of the Alexander dual of path ideals. For this purpose, we need the following lemma. Lemma 2.2.5. Suppose that I is a square-free monomial ideal in a polynomial ring

R = K[x 1 , . . . , x n ], G(I) = {u 1 , . . . , u m }, and Ass R (R/I) = {p 1 , . . . , p s }. Suppose also that there exists a variable x t ∈ {x 1 , . . . , x n } such that x t ∈ p i for all i = 1, . . . , s. Then there exists i ∈ N with 1 ≤ i ≤ m such that u i = x t .
Proof. Since I is a square-free monomial ideal, by [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.3.6], it follows that I = p 1 ∩ • • • ∩ p s , and hence x t ∈ I. On account of [START_REF] Herzog | Monomial Ideals[END_REF]Proposition 1.1.5], there exists i ∈ N with 1 ≤ i ≤ m such that u i ∈ G(I) and u i divides x t . Because u i is a monomial in R, this implies that u i = x t , as claimed. Theorem 2.2.6. Let T be an unrooted starlike tree on the vertex set {z, 1, . . . , n} with center z. Also, let I be the monomial ideal corresponding to T which is generated by the paths of maximal lengths, and corresponding Alexander dual J. Then the ideal J has the persistence property.

Proof. Suppose that |L T | = k. It is routine to check that the path i 1 , . . . , i t has maximal length in T if and only if i 1 , i t ∈ L T . This implies that

I = (x i 1 • • • x i j x z x i j+1 • • • x it : i 1 , i t ∈ L T ).
We note that I has exactly k 2 minimal generators. Thus, one can easily show that

J = i 1 ,it∈L T (x i 1 , . . . , x i j , x z , x i j+1 , . . . , x it ),
which is a minimal primary decomposition of J in R ′ = K[x z , x 1 , . . . , x n ] over a field K. In view of Lemma 2.2.5, we deduce that x z ∈ G(J). Assume that

G(J) = {x z , u 1 , . . . , u s }.
Since G(J) is a minimal generating set for J, we obtain gcd(x z , u j ) = 1 for all j = 1, . . . , s. Now, let t ∈ N be an arbitrary element and pick p ∈ Ass R ′ (R ′ /J t ). Hence, there exists a monomial v in R ′ such that p = (J t : R ′ v).

Our aim is now to establish p = (J t+1 : R ′ x z v). In the light of

J = x z R ′ + s j=1 u j R ′ , it follows that J t+1 = J t x z R ′ + J t s j=1 u j R ′ ,
and so

(J t+1 : R ′ x z v) = (J t x z : R ′ x z v) + ( s j=1 J t u j : R ′ x z v).
Since (J t x z : R ′ x z v) = (J t : R ′ v) and ( s j=1 J t u j : R ′ x z v) = s j=1 (J t u j : R ′ x z v), we can derive the following equality

(J t+1 : R ′ x z v) = (J t : R ′ v) + s j=1 (J t u j : R ′ x z v).
For completing the proof, it is enough to prove that (J t u j : R ′ x z v) ⊆ p for any j = 1, . . . , s. To do this, fix j with 1 ≤ j ≤ s and consider the following two cases:

Case 1. u α 1 1 • • • u αs s ∈ G(J t ) with α 1 + • • • + α s = t and α 1 , . . . , α s ∈ N 0 . Due to (u j , x z ) = 1 for all j = 1, . . . , s, one has (u α 1 1 • • • u αs s u j : R ′ x z v) = (u α 1 1 • • • u αs s u j : R ′ v). Thus, (u α 1 1 • • • u αs s u j : R ′ x z v) ⊆ p. Case 2. u α 1 1 • • • u αs s x β z ∈ G(J t ) with β ∈ N, α 1 +• • •+α s +β = t, and α 1 , . . . , α s ∈ N 0 . It follows that (u α 1 1 • • • u αs s x β z u j : R ′ x z v) = (u α 1 1 • • • u αs s x β-1 z u j : R ′ v). Hence, (u α 1 1 • • • u αs s x β z u j : R ′ x z v) ⊆ p. Thus, p ∈ Ass R ′ (R ′ /J t+1
), and so the ideal J has the persistence property, as desired.

In the next example we describe the major goal of Theorem 2.2.6.

Example 2.2.7. Suppose that T is the unrooted starlike tree with center z, as the left graph in Figure 2. By using the notations similar that we used in the proof of Theorem 2.2.6, we obtain

I = (x 4 x 3 x 2 x 1 x z x 8 x 9 x 10 , x 4 x 3 x 2 x 1 x z x 7 , x 5 x 6 x z x 7 , x 4 x 3 x 2 x 1 x z x 5 x 6 ,
x 5 x 6 x z x 8 x 9 x 10 , x 7 x z x 8 x 9 x 10 ), and so

J = (x 4 , x 3 , x 2 , x 1 , x z , x 8 , x 9 , x 10 ) ∩ (x 4 , x 3 , x 2 , x 1 , x z , x 7 ) ∩ (x 5 , x 6 , x z , x 7 ) ∩ (x 4 , x 3 , x 2 , x 1 , x z , x 5 , x 6 ) ∩ (x 5 , x 6 , x z , x 8 , x 9 , x 10 ) ∩ (x 7 , x z , x 8 , x 9 , x 10 ).
According to Theorem 2.2.6, the ideal J has the persistence property. Corollary 2.2.8. Suppose that T is an unrooted symmetric starlike tree on the vertex set V (T ) = {z, 1, . . . , n} with center z and the following edge set

E(T ) = {{z, i}, {kj + i, kj + k + i} | i = 1, . . . , k, j = 0, . . . , m -1}, such that n = k(m + 1) for some k ∈ N and m ∈ N 0 . Suppose also that I 2m+2 (T )
is the path ideal corresponding to T of length 2m + 2 and corresponding Alexander dual J 2m+2 . Then the ideal J 2m+2 has the persistence property.

Let us illustrate Corollary 2.2.8 with an example. Suppose that T is the unrooted symmetric starlike tree with center z, as shown in Figure 2, on the vertex set V (T ) = {z, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Then we have

I 6 (T ) = (x 7 x 4 x 1 x z x 2 x 5 x 8 , x 7 x 4 x 1 x z x 3 x 6 x 9 , x 8 x 5 x 2 x z x 3 x 6 x 9 ).
It follows that the Alexander dual of I 6 (T ), denoted by J 6 , is given by

J 6 = (x 7 , x 4 , x 1 , x z , x 2 , x 5 , x 8 ) ∩ (x 7 , x 4 , x 1 , x z , x 3 , x 6 , x 9 ) ∩ (x 8 , x 5 , x 2 , x z , x 3 , x 6 , x 9 ) = (x z , x 8 x 9 , x 7 x 9 , x 5 x 9 , x 4 x 9 , x 2 x 9 , x 1 x 9 , x 7 x 8 , x 6 x 8 , x 4 x 8 , x 3 x 8 , x 1 x 8 , x 6 x 7 , x 5 x 7 , x 3 x 7 , x 2 x 7 , x 5 x 6 , x 4 x 6 , x 2 x 6 , x 1 x 6 , x 4 x 5 , x 3 x 5 , x 1 x 5 , x 3 x 4 , x 2 x 4 , x 2 x 3 , x 1 x 3 , x 1 x 2 ).
Now, Corollary 2.2.8 implies that the ideal J 6 has the persistence property. On the other hand, one can see this result in another way. In fact, the ideal L = (x 8 x 9 , x 7 x 9 , x 5 x 9 , x 4 x 9 , x 2 x 9 , x 1 x 9 , x 7 x 8 , x 6 x 8 , x 4 x 8 , x 3 x 8 , x 1 x 8 , x 6 x 7 , x 5 x 7 , x 3 x 7 , x 2 x 7 , x 5 x 6 , x 4 x 6 , x 2 x 6 , x 1 x 6 , x 4 x 5 , x 3 x 5 , x 1 x 5 ,

x 3 x 4 , x 2 x 4 , x 2 x 3 , x 1 x 3 , x 1 x 2 ),
is generated by quadratic square-free monomials, and so it is the edge ideal of another graph. By [START_REF] Martinez-Bernal | Associated primes of powers of edge ideals[END_REF]Theorem 2.15], one can conclude that L has the persistence property, and [START_REF] Hà | Embedded associated primes of powers of square-free monomial ideals[END_REF]Lemma 3.4] yields that J 6 has the persistence property.

We now explore when the unique homogenous maximal ideal appears. To achieve this, we first recall the definition of the expansion operator on monomial ideals which has been stated in [13], and then apply it as a criterion for the persistence property of monomial ideals. Definition 2.2.9. Let K be a field and R be the polynomial ring over a field K in the variables x 1 , . . . , x n . Fix an ordered n-tuple (i 1 , . . . , i n ) of positive integers, and consider the polynomial ring R (i 1 ,...,in) over K in the variables x 11 , . . . , x 1i 1 , x 21 , . . . , x 2i 2 , . . . , x n1 , . . . , x nin . Let p j be the monomial prime ideal (x j1 , x j2 , . . . , x ji j ) ⊆ R (i 1 ,...,in) for all j = 1, . . . , n. Attached to each monomial ideal I ⊂ R with a set of monomial generators {x a 1 , . . . , x am }, where n) and a i (j) denotes the j-th component of the vector a i = (a i (1), . . . , a i (n)) for all i = 1, . . . , m. We define the expansion of I with respect to the n-tuple (i 1 , . . . , i n ), denoted by I (i 1 ,...,in) , to be the monomial ideal

x a i = x 1 a i (1) • • • x n a i (
I (i 1 ,...,in) = m i=1 p a i (1) 1 • • • p a i (n) n ⊆ R (i 1 ,...,in) .
We simply write R * and I * , respectively, rather than R (i 1 ,...,in) and I (i 1 ,...,in) .

For example, consider R = K[x 1 , x 2 , x 3 ] and the ordered 3-tuple (1, 3, 2). Then we have p 1 = (x 11 ), p 2 = (x 21 , x 22 , x 23 ), and p 3 = (x 31 , x 32 ). So, for the monomial ideal I = (x 1 x 2 , x 2 3 ), the ideal ).

I * ⊆ K[x
To prove Lemma 2.2.12, one needs frequently use the following proposition and lemma. We state them here for ease of reference. (vii) If the monomial ideal Q is p-primary, then Q * is p * -primary.

Proposition 2.2.10. ([13, Proposition 1.2]) Let I be a monomial ideal, and consider an (irredundant) primary decomposition

I = Q 1 ∩ • • • ∩ Q m of I. Then I * = Q 1 * ∩ • • • ∩ Q m * is an (irredundant)
To prove Theorem 2.2.13 we require the following auxiliary lemma. ). This implies that I has the persistence property. Sufficiency follows from in a similar way, and the proof is complete.

We are now ready to establish Theorem 2.2.13.

Theorem 2.2.13. Let T be an unrooted starlike tree on the vertex set {z, 1, . . . , n} with center z. Also, let I be the monomial ideal corresponding to T which is generated by the paths of maximal lengths, and corresponding Alexander dual J.

If deg T z = k, then (1) (x z , x 1 , . . . , x n ) ∈ Ass R ′ (R ′ /J s ) for s ≥ k -1; and, (2) (x z , x 1 , . . . , x n ) / ∈ Ass R ′ (R ′ /J s ) for s < k -1,
where

(x z , x 1 , . . . , x n ) is the unique homogeneous maximal ideal in the polynomial ring R ′ = K[x z , x 1 , . . . , x n ].
Proof. Since deg T z = k, the graph T \ {z} has exactly k connected components, say L 1 , . . . , L k , where each component is a path graph with |V (L i )| = h i for each i = 1, . . . , k. Put h 0 := 0 and for all i = 1, . . . , k, let

V (L i ) := {h 1 + • • • + h i-1 + j : j = 1, . . . , h i }.
This implies that E(T ) is given by

{{z, i-1 t=1 h t + 1}, { i-1 t=1 h t + j, i-1 t=1 h t + j + 1} : i = 1, . . . , k, j = 1, . . . , h i -1}. Now, set p i := (x h 1 +•••+h i-1 +j , x h 1 +•••+h i-1 +j+1 : j = 1, . . . , h i -1) for all i = 1, . . . , k.
Then one can conclude that

J = i,j∈{1,...,k},i̸ =j (p i + x z R ′ + p j ),
and so J = x z R ′ + i,j∈{1,...,k},i̸ =j (p i + p j ). It is routine to check that i,j∈{1,...,k},i̸ =j

(p i + p j ) = k j=1 p 1 ∩ • • • ∩ p j-1 ∩ p j ∩ p j+1 ∩ • • • ∩ p k ,
where p j means that this term is omitted. As p i and p j , for every i, j ∈ {1, . . . , k} with i = j, are generated by disjoint sets, it follows that p i ∩ p j = p i p j , and so

J = x z R ′ + k j=1 p 1 • • • p j-1 p j p j+1 • • • p k .
Now, consider the following monomial ideal

a := x z R ′ + k j=1 (x 1 • • • x j-1 x j x j+1 • • • x k )R ′ ,
where x j means that this term is omitted. In order to use Lemma 2.2.12, set p k+1 := x z R ′ . This implies that J is the expansion of a with respect to the (k + 1)tuple (h Corollary 2.2.14. Suppose that T is an unrooted symmetric starlike tree on the vertex set V (T ) = {z, 1, . . . , n} with center z and the following edge set

E(T ) = {{z, i}, {kj + i, kj + k + i} | i = 1, . . . , k and j = 0, . . . , m -1}, such that n = k(m + 1) for some k ∈ N and m ∈ N 0 . Suppose also that I 2m+2 (T )
is the path ideal corresponding to T of length 2m + 2 and corresponding Alexander dual J 2m+2 . Then (1)

(x z , x 1 , . . . , x n ) ∈ Ass R ′ (R ′ /J s 2m+2 ) for s ≥ k -1; and, (2) (x z , x 1 , . . . , x n ) / ∈ Ass R ′ (R ′ /J s 2m+2 ) for s < k -1,
where (x z , x 1 , . . . , x n ) is the unique homogeneous maximal ideal in the polynomial ring Hence, we obtain the path ideal I 6 (T ) is given by

R ′ = K[x z , x 1 , . . . , x n ].
(x 11 x 6 x 1 x z x 2 x 7 x 12 , x 11 x 6 x 1 x z x 3 x 8 x 13 , x 11 x 6 x 1 x z x 4 x 9 x 14 , x 11 x 6 x 1 x z x 5 x 10 x 15 ,
x 12 x 7 x 2 x z x 3 x 8 x 13 , x 12 x 7 x 2 x z x 4 x 9 x 14 , x 12 x 7 x 2 x z x 5 x 10 x 15 , x 13 x 8 x 3 x z x 4 x 9 x 14 , x 13 x 8 x 3 x z x 5 x 10 x 15 , x 14 x 9 x 4 x z x 5 x 10 x 15 ).

It follows that the Alexander dual of I 6 (T ) is given by

J 6 = (x 11 , x 6 , x 1 , x z , x 2 , x 7 , x 12 ) ∩ (x 11 , x 6 , x 1 , x z , x 3 , x 8 , x 13 ) ∩ (x 11 , x 6 , x 1 , x z , x 4 , x 9 , x 14 ) ∩ (x 11 , x 6 , x 1 , x z , x 5 , x 10 , x 15 ) ∩ (x 12 , x 7 , x 2 , x z , x 3 , x 8 , x 13 ) ∩ (x 12 , x 7 , x 2 , x z , x 4 , x 9 , x 14 ) ∩ (x 12 , x 7 , x 2 , x z , x 5 , x 10 , x 15 ) ∩ (x 13 , x 8 , x 3 , x z , x 4 , x 9 , x 14 )
∩ (x 13 , x 8 , x 3 , x z , x 5 , x 10 , x 15 ) ∩ (x 14 , x 9 , x 4 , x z , x 5 , x 10 , x 15 ). Due to Theorem 2.2.13, one can conlcude that (1)

(x z , x 1 , . . . , x 15 ) ∈ Ass R ′ (R ′ /J s 6 ) for s ≥ 4; and, (2) (x z , x 1 , . . . , x 15 ) / ∈ Ass R ′ (R ′ /J s 6 ) for s < 4,
where (x z , x 1 , . . . , x 15 ) is the unique homogeneous maximal ideal in the polynomial ring

R ′ = K[x z , x 1 , . . . , x 15 ].
We conclude this argument with giving another class of graphs which the path ideals generated by path of length two have the persistence property. To do this, we first recall the definition of strongly superficial elements and then introduce a class of graphs which are called centipede graphs. Moreover, it should be noted that the following results can be found in [START_REF] Nasernejad | Asymptotic behaviour of associated primes of monomial ideals with combinatorial applications[END_REF]. Definition 2.2.16. Let I be an ideal in a commutative ring S, and let k ∈ N. An element x in S is called a superficial element of degree k for I if x ∈ I k and there exists c ∈ N such that (I n+k : S x) ∩ I c = I n for all n ≥ c. If (I n+k : S x) = I n for all n ∈ N, we say that x is a strongly superficial element of degree k for I (see [90, 4.1.2]).

The following proposition is essential for us in order to complete the proof of Theorem 2.2.19. Proposition 2.2.17. Suppose that I is an arbitrary ideal in a commutative Noetherian ring S such that has a strongly superficial element of degree one. Then I has the persistence property.

Proof. Let u be a strongly superficial element of degree one for I. Then (I n+1 : S u) = I n for all n ∈ N. Since (I n+1 : S I) ⊆ (I n+1 : S u), it follows that (I n+1 : S I) = I n for all n ∈ N. For completing the proof, assume that I = (v 1 , . . . , v t ). Choose an arbitrary m ∈ N and consider p ∈ Ass S (S/I m ). Then there exists an element c in S such that p = (I m : S c). This implies that p = ((I m+1 : S I) : S c)

= (I m+1 : S Ic) = t i=1 (I m+1 : S v i c).
One can conclude that there exists a positive integer 1 ≤ j ≤ t such that p = (I m+1 : S v j c), and so p ∈ Ass S (S/I m+1 ). Therefore, I has the persistence property, as claimed.

In the next definition, we present a class of graphs which we will use in Theorem

E(W n ) = {{a i , b i } : 1 ≤ i ≤ n} ∪ {{b j , b j+1 } : 1 ≤ j ≤ n -1}. a 1 a 2 a 3 a n-1 a n b 1 b 2 b 3 b n-1 b n
We now state another main result of this section in the following theorem. Theorem 2.2.19. Let W n , for some n ∈ N with n ≥ 2, be a centipede graph with corresponding path ideal I 2 (W n ). Then I 2 (W n ) has the persistence property. Proof. To simplify the notation, we consider the following centipede graph on the vertex set V (W n ) = {1, 2, 3, . . . , 2n -2, 2n -1, 2n} and also set J := I 2 (W n ). It is routine to verify that

G(J) ={x 2k-1 x 2k x 2k+1 , x 2k-1 x 2k+1 x 2k+2 : k = 1, . . . , n -1} ∪{x 2r-1 x 2r+1 x 2r+3 : r = 1, . . . , n -2}.
Clearly, the ideal J has exactly 3n -4 distinct minimal generators. Assume that

G(J) = {u 1 , . . . , u k } with k := 3n -4, such that u 1 := x 1 x 2 x 3 , u 2 := x 1 x 3 x 4 , u 3 := x 1 x 3 x 5 , u 4 := x 3 x 5 x 7 , u 5 := x 3 x 5 x 6
, and u 6 := x 3 x 4 x 5 . Here and in the following, our aim is to show that (J m+1 : R u 1 ) = J m for all m ∈ N. To do this, choose an arbitrary m ∈ N. Then we have the following equalities

(J m+1 : R u 1 ) =(J m J : R u 1 ) = k i=1 (J m u i : R u 1 ) =J m + k i=2
(J m u i : R u 1 ).

For completing the proof, we show that (J m u i : R u 1 ) ⊆ J m for all i = 2, . . . , k. To achieve this, choose an arbitrary element v ∈ (J m u i : R u 1 ) for some 2 ≤ i ≤ k. Hence, vu 1 ∈ J m u i , and so there exists a generator of J m such as

u a 1 1 • • • u a k k with a 1 + • • • + a k = m, such that u i u a 1 1 • • • u a k k divides vu 1 . So, there is a monomial w in R such that vu 1 = wu i u a 1 1 • • • u a k k . If a 1 ≥ 1, then v = wu i u a 1 -1 1 • • • u a k k . By u i u a 1 -1 1 • • • u a k k ∈ J m , one can conclude that v ∈ J m
. Thus, we assume that a 1 = 0, and so

a 2 + • • • + a k = m. Since x 2 / ∈ u j for all j = 2, . . . , k, it follows that vx 1 x 3 = w ′ u i u a 2 2 • • • u a k k such that w = x 2 w ′ for some monomial w ′ in R.
Hence, one may consider the following cases:

Case 1.

u i = u 2 or u i = u 3 . Accordingly, v = w ′ x 4 u a 2 2 • • • u a k k or v = w ′ x 5 u a 2 2 • • • u a k k , and so v ∈ J m . Case 2. u i = u 4 . Then vx 1 = w ′ x 5 x 7 u a 2 2 • • • u a k k .
Hence, we have to consider the following subcases: Subcase 2.1.

a 2 ≥ 1 or a 3 ≥ 1. Therefore, v = w ′ x 4 u 4 u a 2 -1 2 u a 3 3 • • • u a k k or v = w ′ x 5 u 4 u a 2 2 u a 3 -1 3 • • • u a k k . This implies that v ∈ J m . Subcase 2.2. a 2 = 0 and a 3 = 0. Then vx 1 = w ′ x 5 x 7 u a 4 4 • • • u a k k . Due to x 1 / ∈ u j for all j = 4, . . . , k, it follows that v = w ′′ x 5 x 7 u a 4 4 • • • u a k k such that w ′ = x 1 w ′′ for some monomial w ′′ in R. One can conclude that v ∈ J m . Case 3. u i = u r with r ≥ 5. Then vx 1 x 3 = w ′ u r u a 2 2 • • • u a k k .
We now consider the following subcases: Subcase 3.1. a 2 ≥ 1 or a 3 ≥ 1. The proof is similar to Subcase 2.1. Subcase 3.2. a 2 = 0 and a 3 = 0. This implies that vx

1 x 3 = w ′ u r u a 4 4 • • • u a k k . On account of x 1 / ∈ u j for all j = 4, . . . , k, it follows that vx 3 = w ′′ u r u a 4 4 • • • u a k k such that w ′ = x 1 w ′′ for some monomial w ′′ in R. If a 4 ≥ 1 or a 5 ≥ 1 or a 6 ≥ 1, then v = w ′′ x 5 x 7 u r u a 4 -1 4 • • • u a k k or v = w ′′ x 5 x 6 u r u a 4 4 u a 5 -1 5 • • • u a k k or v = w ′′ x 4 x 5 u r u a 4 4 u a 5 5 u a 6 -1 6 • • • u a k k , and so v ∈ J m . If a i = 0 for all i = 4, 5, 6, according to x 3 / ∈ u j for all j = 7, . . . , k, one can conclude that v = w 1 u a 7 7 • • • u a k k such that w ′′ u r = x 3 w 1 for some monomial w 1 in R. Thus, v ∈ J m .
Consequently, we obtain (J m+1 : R u 1 ) = J m for all m ∈ N, and so u 1 is a strongly superficial element of degree one for the ideal J. Due to Proposition 2.2.17, one can derive that J has the persistence property, as desired.

Condition (♯), and unisplit and separable monomial ideals

Our main aim in this section is to introduce the other classes of monomial ideals with the persistence property which call them unisplit and separable monomial ideals. In fact, we show that every ideal I in this class has the following properties:

(i) I has the (strong) persistence property.

(ii) I has a strongly superficial element.

(iii) I is Ratliff-Rush closed.

To do this, we state condition (♯) on monomial ideals and show that a monomial ideal I is unisplit (or separable) if and only if it satisfies condition (♯). Moreover, we prove that monomial ideals with condition (♯) have the properties (i)-(iii). In addition, note that the results of this section can be found in [START_REF] Nasernejad | Persistence property for some classes of monomial ideals of a polynomial ring[END_REF].

Throughout this section, R = K[x 1 , . . . , x n ] stands for a polynomial ring over a field K. We first begin with the following notions. Let G = (V (G), E(G)) be a finite simple graph on the vertex set [n] := {1, . . . , n}. Recall that the edge ideal associated to G is the monomial ideal

I(G) = (x i x j : {i, j} ∈ E(G)) ⊂ R = K[x 1 , . . . , x n ],
and the cover ideal associated to G is the monomial ideal

J(G) = {i,j}∈E(G) (x i , x j ) ⊂ R = K[x 1 , . . . , x n ].
Let I be a monomial ideal of R generated in one degree. We say that I is polymatroidal if the following "exchange condition" is satisfied: For monomials u =

x a 1 1 • • • x an n and v = x b 1 1 • • • x bn n belonging to G(I)
and, for each i with a i > b i , one has j with a j < b j such that x j u/x i ∈ G(I) (see [START_REF] Herzog | Monomial Ideals[END_REF]Definition 12.6.1]).

In what follows, we present two classes of monomial ideals such that are none of them, but we will show that satisfying the persistence property.

We define unisplit monomial ideals in the following definition.

Definition 2.3.1. Let I be a monomial ideal of R with G(I) = {u 1 , . . . , u m }. Then we say that I is a unisplit monomial ideal if there exists i ∈ N with 1 ≤ i ≤ m such that each monomial u j has no common factor with u i for all j ∈ N with 1 ≤ j ≤ m and j = i. We call u i as split generator.

Example 2.3.2. Consider the ideal

I = (x 2 3 x 5 x 3 6 , x 3 1 x 2 2 x 4 4 , x 6 1 x 3 2 x 4 7 , x 2 2 x 4 7 x 5 4 ) in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 ]. It is easy to see that I is a unisplit monomial ideal of R with x 2 3 x 5 x 3 6 is the split generator.
We next give the definition of separable monomial ideals.

Definition 2.3.3. Let I be a monomial ideal of R with G(I) = {u 1 , . . . , u m }. Then we say that I is a separable monomial ideal if there exist i ∈ N with 1 ≤ i ≤ m and monomials g and w in R such that w = 1, u i = wg, gcd(w, g) = 1, and for all

j ∈ N with 1 ≤ j = i ≤ m, gcd(u j , u i ) = w.
Example 2.3.4. Consider the following monomial ideal

I = (x 1 x 2 x 3 3 x 5 4 , x 2 1 x 3 2 x 4 3 x 3 5 x 5 6 , x 2 1 x 2 x 5 3 x 2 5 x 6 , x 3 1 x 2 2 x 6 3 x 4 6 ), in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ]
. Then, by setting

u 1 := x 1 x 2 x 3 3 x 5 4 , u 2 := x 2 1 x 3 2 x 4 3 x 3 5 x 5 6 , u 3 := x 2 1 x 2 x 5 3 x 2 5 x 6 , u 4 := x 3 1 x 2 2 x 6 3 x 4 6 , i := 1 and w := x 1 x 2 x 3 3 , one can easily check that I is a separable monomial ideal of R.
It is necessary to note that the set of unisplit monomial ideals and the set of separable monomial ideals are disjoint.

Here, we study some properties of monomial ideals which we need in the following. We begin with the remark below.

Remark 2.3.5. Let I 1 , . . . , I n be monomial ideals of R. Then it is easy to see that

(I 1 + • • • + I n : R u) = (I 1 : R u) + • • • + (I n : R u) for all monomials u in R.
Proposition 2.3.6. Let I and J be two ideals of a commutative Noetherian ring S and u be an arbitrary element in S. Then (uJ + I : S u) = J + (I : S u). Moreover, if u is a non-zerodivisor in S, then (uJ : S u) = J.

Proof. Suppose that h is an arbitrary element in (uJ + I : S u). Then hu ∈ uJ + I. Hence, there exist α ∈ J and β ∈ I such that hu = αu + β. Thus, h -α ∈ (I : S u), and so h ∈ J + (I : S u). Therefore, (uJ + I : S u) ⊆ J + (I : S u). For the reverse inclusion, let h be an arbitrary element in J + (I : S u). Then h = α + β for some α ∈ J and β ∈ (I : S u). Since hu = αu + βu, it follows that hu ∈ uJ + I. Hence, h ∈ (uJ + I : S u). Therefore, J + (I : S u) ⊆ (uJ + I : S u), and so the proof is complete. Now, suppose that I = 0 and u is a non-zerodivisor in S. Then (I : S u) = 0, and so (uJ : S u) = J, as required.

In the next definition, we introduce condition (♯) on monomial ideals and we show that any monomial ideal that satisfying condition (♯) has the persistence property and is Ratliff-Rush closed. Also, we prove that it has a strongly superficial element.

Definition 2.3.7. Suppose that I is a monomial ideal of R with G(I) = {u 1 , . . . , u m }.
We say that I satisfies condition (♯) if there exists a positive integer i with

1 ≤ i ≤ m such that (u α 1 1 • • • u α i-1 i-1 u α i i u α i+1 i+1 • • • u αm m u j : R u i ) = u α 1 1 • • • u α i-1 i-1 u α i i u α i+1 i+1 • • • u αm m (u j : R u i ),
for all j = 1, . . . , m with j = i and α 1 , . . . , α m ∈ N 0 , where u α i i means that this term is omitted and N 0 is the set of nonnegative integers. Now, we state the first main result of this section. To accomplish this, one has to frequently apply the following proposition.

(I : J) = v∈G(J) (I : v). Moreover, {u/gcd(u, v) : u ∈ G(I)} is a set of generators of (I : v).
Theorem 2.3.9. Let I be a monomial ideal satisfies condition (♯). Then I is either a unisplit monomial ideal or a separable monomial ideal.

Proof. Without loss of generality, we may assume that G(I) = {u 1 , . . . , u m } is the unique minimal set of monomial generators of I such that

(u α 2 2 • • • u αm m u j : R u 1 ) = u α 2 2 • • • u αm m (u j : R u 1 ),
for all j = 2, . . . , m and α 2 , . . . , α m ∈ N 0 . Consider k, j ∈ {2, . . . , m}. By [60, Proposition 1.2.2], the generator of (u k u j : R u 1 ) has the form

u k u j gcd(u k u j , u 1 )
. On the other hand, we have (u k u j : R u 1 ) = u k (u j : R u 1 ) = u j (u k : R u 1 ). Due to [60, Proposition 1.2.2], the generator of (u j : R u 1 ) (respectively, (u k : R u 1 )) has the form u j gcd(u j , u 1 )

(respectively,

u k gcd(u k , u 1 )
). This implies that

u k u j gcd(u k u j , u 1 ) = u k u j gcd(u j , u 1 ) = u j u k gcd(u k , u 1 )
.

Accordingly, for each k, j ∈ {2, . . . , m}, it follows that gcd(u k u j , u 1 ) = gcd(u j , u 1 ) = gcd(u k , u 1 ).

Thus, there exists a fixed monomial such as w in R such that gcd(u j , u 1 ) = w for all j = 2, . . . , m. So, we have the following two cases:

Case 1. w = 1.
Then based on Definition 2.3.1, one can deduce that I is a unisplit monomial ideal.

Case 2. w > 1. Assume that u 1 = wg for some monomial g in R. By Definition 2.3.3, in order to show that I is a separable monomial ideal, it is sufficient to prove that gcd(w, g) = 1. To do this, suppose that gcd(w, g) > 1, and seek a contradiction. Then there is a monomial

x α i i in R such that x i ∈ {x 1 , . . . , x n }, α i ∈ N, g = x α i i v 1 and w = x α i i v 2 for some monomials v 1 and v 2 in R.
Choose two arbitrary generators such as u i and u j in G(I). Then u i = wh 1 and u j = wh 2 for some monomials h 1 and h 2 in R. This gives that u i u j = w 2 h 1 h 2 , and so x α i i w divides gcd(u i u j , u 1 ), this is the required contradiction.

In the next theorems, we prove that any unisplit (or separable) monomial ideal satisfies condition (♯).

Theorem 2.3.10. Every unisplit monomial ideal of R satisfies condition (♯).

Proof. Let I be a monomial ideal of R with G(I) = {u 1 , . . . , u m }. Since I is a unisplit monomial ideal, without loss of generality, we may assume that each of monomials u 2 , . . . , u m has no common factor with u 1 . By [60, Proposition 1.2.2], the generator of (u i : R u 1 ) is the monomial of the form

u i gcd(u i ,u 1 ) = u i for all i = 2, . . . , m. Again, by [60, Proposition 1.2.2], the generator of (u α 2 2 • • • u αm m u i : R u 1 ), for all i = 2, . . . , m, has the following form u α 2 2 • • • u αm m u i gcd(u α 2 2 • • • u αm m u i , u 1 )
.

By virtue of gcd(u α 2 2 • • • u αm m u i , u 1 ) = 1, for all i = 2, . . . , m, it follows that u α 2 2 • • • u αm m u i gcd(u α 2 2 • • • u αm m u i , u 1 ) = u α 2 2 • • • u αm m u i .
This implies that, for all i = 2, . . . , m, we have

(u α 2 2 • • • u αm m u i : R u 1 ) = u α 2 2 • • • u αm m (u i : R u 1 ).
This yields that I satisfies condition (♯), as desired.

Theorem 2.3.11. Every separable monomial ideal of R satisfies condition (♯).

Proof. Let I be a separable monomial ideal of R with G(I) = {u 1 , . . . , u m } such that

u j := x c j,1 1 • • • x c j,n n
for all j = 1, . . . , m and c j,1 , . . . , c j,n ∈ N 0 . Now, without loss of generality, we may assume that there exist monomials g and w :

= x β 1 1 • • • x βn n with β 1 , . . . , β n ∈ N 0 , in R such that u 1 = wg, gcd(w, g) = 1
, and gcd(u j , u 1 ) = w for all j = 2 . . . , m. Due to [60, Proposition 1.2.2], the generator of (u j : R u 1 ) has the form u j gcd(u j ,u 1 ) for all j = 2, . . . , m. Furthermore,

u j gcd(u j ,u 1 ) = u j
w for all j = 2, . . . , m. Again, according to [60, Proposition 1.2.2], the generator of (u α 2 2 • • • u αm m u j : R u 1 ), for all j = 2, . . . , m and α 2 , . . . , α m ∈ N 0 , is the monomial of the following form

u α 2 2 • • • u αm m u j gcd(u α 2 2 • • • u αm m u j , u 1 )
.

For completing the proof, it is enough to verify that gcd(u

α 2 2 • • • u αm m u j , u 1 )
= w for all j = 2, . . . , m. In fact, for all j = 2, . . . , m, one has

u α 2 2 • • • u αm m u j = n t=1 x c 2,t α 2 +•••+c j-1,t α j-1 +c j,t (α j +1)+c j+1,t α j+1 +•••+cm,tαm t .
Since gcd(w, g) = 1 and gcd(u j , u 1 ) = w for all j = 2 . . . , m, one can conclude that

c 1,t ≤ c 2,t α 2 + • • • + c j-1,t α j-1 + c j,t (α j + 1) + c j+1,t α j+1 + • • • + c m,t α m ,
for all t = 1, . . . , n with β t = 0. This implies that gcd(u

α 2 2 • • • u αm m u j , u 1 )
= w for all j = 2, . . . , m. Therefore, we get

(u α 2 2 • • • u αm m u j : R u 1 ) = u α 2 2 • • • u αm m (u j : R u 1
), for all j = 2, . . . , m, and so I satisfies condition (♯), as claimed.

Theorems 2.3.9, 2.3.10, and 2.3.11 characterize any monomial ideal that satisfying condition (♯).

In the following, we give some combinatorial aspects. In fact, we study at the intersection of graph theory and commutative algebra by relating powers of squarefree monomial ideals in polynomial rings to paths of finite simple graphs. We then show that certain classes of these path ideals satisfy condition (♯).

Let G be a finite simple graph on the vertex set [n] := {1, . . . , n}. Then the sequence of distinct vertices i 1 , . . . , i r is called a maximal path when i 1 , . . . , i r is a path and furthermore, for all j ∈ {1, . . . , n} \ {i 1 , . . . , i r }, none of j, i 1 , . . . , i r and i 1 , . . . , i r , j is a path. Proof. Let L T be the set of vertices of degree one in T . It is routine to check that the path i 1 , . . . , i t is a maximal path in T if and only if i 1 , i t ∈ L T . This implies that

P I(G) := (x i 1 • • • x it : i 1 , . . . , i t is a maximal path in G), in R = K[x 1 , . . . , x n ] is called a max-path ideal of G.
P I(T ) = (x i 1 • • • x i j x z x i j+1 • • • x it : i 1 , i t ∈ L T ).
Thus, one can conclude that

P I(T ) ∨ = i 1 ,it∈L T (x i 1 , . . . , x i j , x z , x i j+1 , . . . , x it ).
One can easily observe that this is a minimal primary decomposition of P I(T ) ∨ in the polynomial ring R ′ = K[x z , x 1 , . . . , x n ] over the field K. By Lemma 2.2.5, it follows that x z ∈ G(P I(T ) ∨ ). Assume that G(P I(T ) ∨ ) = {x z , u 1 , . . . , u s }. Since G(P I(T ) ∨ ) is a minimal set, this implies that gcd(x z , u j ) = 1 for all j = 1, . . . , s. Due to Definition 2.3.1, P I(T ) ∨ is a unisplit monomial ideal, and so has the persistence property, as desired.

In the next example we illustrate the major goal of Theorem 2.3.13.

Example 2.3.14. Let T be a tree on the vertex set V (T ) := {z} ∪ [10] and the edge set E(T ) := {{1, z}, {2, z}, {2, 3}, {4, z}, {4, 5}, {5, 6}, {7, z}, {7, 8}, {8, 9}, {9, 10}}.

By figure below, it is easy to verify that T is starlike with the center z. z This implies that

P I(T ) =(x 4 x 5 x 6 x z x 7 x 8 x 9 x 10 , x 1 x z x 4 x 5 x 6 , x 1 x z x 7 x 8 x 9 x 10 , x 2 x 3 x z x 7 x 8 x 9 x 10 , x 2 x 3 x z x 4 x 5 x 6 , x 1 x z x 2 x 3 ),
and so

P I(T ) ∨ =(x 4 , x 5 , x 6 , x z , x 7 , x 8 , x 9 , x 10 ) ∩ (x 1 , x z , x 4 , x 5 , x 6 ) ∩ (x 1 , x z , x 7 , x 8 , x 9 , x 10 ) ∩(x 2 , x 3 , x z , x 7 , x 8 , x 9 , x 10 ) ∩ (x 2 , x 3 , x z , x 4 , x 5 , x 6 ) ∩ (x 1 , x z , x 2 , x 3 ).
According to Theorem 2.3.13, it follows that P I(T ) ∨ has the persistence property.

The following theorem is concerned with the separable monomial ideals. Theorem 2.3.15. Suppose that T is a starlike tree on the vertex set {z} ∪ [n] with the center z and L T = {j 1 , . . . , j r } is the set of vertices of degree one in T . Suppose also that z, i 1,k , i 2,k , . . . , i t,k , j k is the path between z and j k for all k = 1, . . . , r. Then, for all s = 1, . . . , r, the ideal

J s := (x js x i 1,s x i 2,s • • • x it,s x z x i 1,k x i 2,k • • • x i t,k x j k : k ∈ {1, . . . , r} \ {s}),
has the persistence property.

Proof. Without loss of generality, we need only prove the assertion for the case s = 1. To do this, set w :=

x j 1 x i 1,1 x i 2,1 • • • x i t,1 x z and u k := x i 1,k x i 2,k • • • x i t,k x j k for all k = 2, . . . , r.
This implies that G(J 1 ) = {wu 2 , wu 3 , . . . , wu r }. It is routine to verify that gcd(w, u 2 ) = 1 and gcd(wu 2 , wu j ) = w for all j = 3, . . . , r. Then J 1 is a separable monomial ideal, and so has the persistence property.

We conclude our argument with the following example which clarifies the main aim of Theorem 2.3.15.

Example 2.3.16. Let T be the starlike tree which has stated in Example 2.3.14. By putting j 1 := 10, we deduce that the following ideal has the persistence property

J 1 := (x 10 x 9 x 8 x 7 x z x 1 , x 10 x 9 x 8 x 7 x z x 2 x 3 , x 10 x 9 x 8 x 7 x z x 4 x 5 x 6 ).

Strongly superficial elements and Ratliff-Rush closed ideals

Our main goal in this subsection is to show that any monomial ideal that satisfying condition (♯) has the persistence property. To do this, we begin with the following theorem. Recall that an ideal I in a commutative Noetherian ring S has the strong persistence property if (I k+1 : S I) = I k for all k ∈ N. It is important to note that if I has the strong persistence property, then I has the persistence property. On the other hand, there exist some monomial ideals which have the persistence property, but do not necessary have the strong persistence property. As an example, consider the ideal I generated by monomials x 1 x 2 x 3 , x 1 x 2 x 4 , x 1 x 3 x 5 , x 1 x 4 x 6 , x 1 x 5 x 6 , x 2 x 3 x 6 , x 2 x 4 x 5 , x 2 x 5 x 6 , x 3 x 4 x 5 , and x 3 x 4 x 6 in the polynomial ring [START_REF] Martinez-Bernal | Associated primes of powers of edge ideals[END_REF]Example 2.18] shows that I has the persistence property. On the other hand, we obtain (I 3 : R I) = I 2 , and so I has no the strong persistence property. Theorem 2.3.17. Let J be a monomial ideal satisfies condition (♯). We then have (J k+1 : R J) = J k for all k ∈ N, i.e, J has the strong persistence property.

R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ]. Using
Proof. Without loss of generality, suppose that G(J) = {u 1 , . . . , u m } such that

(u α 2 2 • • • u αm m u j : R u 1 ) = u α 2 2 • • • u αm m (u j : R u 1 ),
for all j = 2, . . . , m and α 2 , . . . , α m ∈ N 0 . Since J k ⊆ (J k+1 : R J) for all k ∈ N, we need only show that (J k+1 : R J)

⊆ J k for all k ∈ N. Fix k ∈ N. As J = m j=1 u j R,
this implies that

(J k+1 : R J) = m j=1
(J k+1 : R u j ).

Note also that

J k+1 = J k ( m i=1 u i R) = m i=1 J k u i . It follows from Remark 2.3.5 that (J k+1 : R J) = m j=1 (J k+1 : R u j ) = m j=1 m i=1 (J k u i : R u j ).
On the other hand, according to Proposition 2.3.6, one can conclude that

m i=1 (J k u i : R u 1 ) = (J k u 1 : R u 1 ) + m i=2 (J k u i : R u 1 ) = J k + m i=2 (J k u i : R u 1 ).
However, by Remark 2.3.5, for i ∈ N with 2 ≤ i ≤ m, we have the following equalities

(J k u i : R u 1 ) = ( α 1 +•••+αm=k u α 1 1 • • • u αm m u i : R u 1 ) = α 1 +•••+αm=k (u α 1 1 • • • u αm m u i : R u 1 ) = α 1 +•••+αm=k,α 1 >0 (u α 1 1 u α 2 2 • • • u αm m u i : R u 1 ) + α 2 +•••+αm=k (u α 2 2 • • • u αm m u i : R u 1 ).
Due to Proposition 2.3.6, for i ∈ N with 2 ≤ i ≤ m, one can deduce that

(J k u i : R u 1 ) = α 1 +•••+αm=k,α 1 >0 u α 1 -1 1 u α 2 2 • • • u αm m u i R + α 2 +•••+αm=k u α 2 2 • • • u αm m (u i : R u 1 ). It is routine to check that (J k u i : R u 1 ) ⊆ J k for i ∈ N with 2 ≤ i ≤ m, and hence m i=2 (J k u i : R u 1 ) ⊆ J k . This implies that m i=1 (J k u i : R u 1 ) = J k . Since m j=1 m i=1 (J k u i : R u j ) ⊆ m i=1
(J k u i : R u 1 ), it follows that (J k+1 : R J) ⊆ J k . Therefore, (J k+1 : R J) = J k , as required.

Remarks 2.3.18.

(1) Note that Theorem 2.3.17 can fail for a monomial ideal that is not a unisplit monomial ideal. As an example, consider the following monomial ideal

I = (x 1 x 2 2 x 3 , x 2 x 2 3 x 4 , x 3 x 2 4 x 5 , x 1 x 4 x 2 5 , x 2 1 x 2 x 5 ), in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5
]. Then I is not a unisplit monomial ideal. On the other hand, one can conclude that

(I 2 : R I) = (x 1 x 2 2 x 3 , x 2 x 2 3 x 4 , x 3 x 2 4 x 5 , x 1 x 4 x 2 5 , x 2 1 x 2 x 5 , x 1 x 2 x 3 x 4 x 5 ).
Therefore, we get (I 2 : R I) = I.

(2) Observe that Theorem 2.3.17 can fail for a monomial ideal that is not a separable monomial ideal. As an example, consider the ideal I generated by monomials x 1 x 2 x 3 , x 1 x 2 x 4 , x 1 x 3 x 5 , x 1 x 4 x 6 , x 1 x 5 x 6 , x 2 x 3 x 6 , x 2 x 4 x 5 , x 2 x 5 x 6 , x 3 x 4 x 5 and

x 3 x 4 x 6 in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x
: I) ⊆ (I 3 : I 2 ) ⊆ • • • .
(1) If J is an ideal containing I, then J ⊆ I * if and only if there is an n ≥ 1 with

J n = I n .
(2) If J is an ideal with I ⊆ J ⊆ I * , then A * (J) = A * (I). (5) For all large n, I n * = I n . (6) For n ≥ 1, (I * ) n ⊆ I n * . (7) For n ≥ 1, I n * is the eventual stable value of the increasing chain

(I n+1 : I) ⊆ (I n+2 : I 2 ) ⊆ (I n+3 : I 3 ) ⊆ • • • . (8) If p ∈ Ass(R/I * ), then p ∈ Ass(R/(I * ) n ) for all n ≥ 1. Also, Ass(R/I * ) ⊆ A * (I).
Lemma 2.3.20. Let J be a monomial ideal satisfies condition (♯) and let J * denote the eventual stable value of the ascending chain

(J 2 : R J) ⊆ (J 3 : R J 2 ) ⊆ • • • ⊆ (J i+1 : R J i ) ⊆ • • • . Then J k * = J k for all k ∈ N. Moreover, J * = J.
Proof. Fix k ∈ N. Due to Lemma 2.3.19, J k * is the eventual stable value of the following increasing chain

(J k+1 : R J) ⊆ (J k+2 : R J 2 ) ⊆ • • • ⊆ (J k+i : R J i ) ⊆ • • • .
Thus, it is enough to show that (J k+i : R J i ) = J k for all i ∈ N. To achieve this, we proceed by induction on i. By Theorem 2.3.17, we have (J k+1 : R J) = J k . Now, suppose inductively that i > 1 and that the assertion is proved for i. Then, according to Theorem 2.3.17 and the inductive hypothesis, we have the following equalities

(J k+i+1 : R J i+1 ) = ((J k+i+1 : R J) : R J i ) = (J k+i : R J i ) = J k .
This completes the inductive step and the proof.

To show Theorem 2.3.22, one has to repeatedly apply the next proposition. Moreover, recall that an ideal is called a regular ideal if it contains a non-zerodivisor element.

Proposition 2.3.21. [START_REF] Mcadam | Sporadic and irrelevant prime divisors[END_REF]Proposition 4.15]) Suppose that 0 ≤ m < n. Then (1)

(I n * : I m * ) = (I n * : I m ) = I n-m * . (2) For n ≥ 1, let J be an ideal with J ⊆ I n * . If p ∈ Ass(R/J), then p ∈ Ass(I n-1 * /J). (3) Ass(R/I * ) ⊆ Ass(R/I 2 * ) ⊆ Ass(R/I 3 * ) ⊆ • • • . (4) For n ≥ 1, S n (I) ⊆ Ass(I n * /I n ).
Theorem 2.3.22. Let J be a monomial ideal satisfies condition (♯). We then have the following statements.

(i) For positive integers r and s with r < s, (J s : R J r ) = J s-r .

(ii) Let J ′ be an ideal of R such that J ′ ⊆ J n for some n ∈ N. Then Ass R (R/J ′ ) ⊆ Ass R (J n-1 /J ′ ). (iii) Ass R (R/J) ⊆ Ass R (R/J 2 ) ⊆ • • • ⊆ Ass R (R/J i ) ⊆ • • • . (iv) Ass R (R/J) ⊆ Ass ∞ (J).
Proof. As every nonzero ideal of R is regular, the statements (i), (ii), and (iii) follow from Lemma 2.3.20 and Proposition 2.3.21.

(iv) It follows immediately from Proposition 2.3.21(4) that Ass R (R/J n )\Ass ∞ (J) is a subset of Ass R (J n * /J n ) for all n ∈ N, we can derive from Lemma 2.3.20 that Ass R (R/J n ) ⊆ Ass ∞ (J), and so Ass R (R/J) ⊆ Ass ∞ (J).

Corollary 2.3.23. Every ideal satisfies condition (♯) has the persistence property.

Remark 2.3.24. It is necessary to note that the inclusion Ass R (R/I) ⊆ Ass ∞ (I) in part (iv) of Theorem 2.3.22, can be strict. As an example, let I be the following square-free monomial ideal [START_REF] Hà | Embedded associated primes of powers of square-free monomial ideals[END_REF]Example 3.14], one can deduce that

I = (x 1 x 2 x 3 , x 2 x 3 x 4 , x 3 x 4 x 5 , x 4 x 5 x 1 , x 5 x 1 x 2 ), in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 ]. Then, by
Ass R (R/I) = {(x 1 , x 3 ), (x 2 , x 5 ), (x 3 , x 5 ), (x 2 , x 4 ), (x 1 , x 4 )}, while Ass ∞ (I) = {(x 1 , x 3 ), (x 2 , x 5 ), (x 3 , x 5 ), (x 2 , x 4 ), (x 1 , x 4 ), (x 1 , x 2 , x 3 , x 4 , x 5 )}.
This implies that Ass R (R/I) ⊊ Ass ∞ (I).

According to Corollary 2.3.23, Theorem 2.3.9, Example 2.3.2, and Example 2.3.4, we presented two classes of monomial ideals such that are none of edge ideals, cover ideals, and polymatroidal ideals, but satisfy the persistence property.

Here, we explore the height of any monomial ideal that satisfying condition (♯) in Remarks 2.3.25 and 2.3.26.

Remark 2.3.25.

Let I be a separable monomial ideal of R. Then htI = 1.

Proof. First proof. In the first proof we describe the set of associated prime ideals of I. To achieve this, without loss of generality, one may assume that G(I) = {u 1 , . . . , u m } such that u 1 = wg for some monomials w, g ∈ R, gcd(w, g) = 1 and gcd(u 1 , u j ) = w for all j = 2, . . . , m. Put J := (u 1 /w, . . . , u m /w) and suppose that w = x b 1 j 1

Ass R (R/I) = Ass R (R/J) ∪ {(x j 1 ), . . . , (x js )}.
The claim is therefore an immediate consequence of definition of height of an ideal. Second proof. With the same assumptions and notations in the first proof, one can conclude that I = wJ. Since w = 1, the assertion follows from the Krull's Principal Ideal Theorem.

Remark 2.3.26. Let I with G(I) = {u 1 , . . . , u m-1 , x α 1 j 1 • • • x αr jr } be a unisplit mono- mial ideal of R such that x jt /
∈ supp(u i ) for all i = 1, . . . , m -1 and t = 1, . . . , r. By setting J := (u 1 , . . . , u m-1 ), we then have htI = htJ + 1.

Proof. According to Theorem 2.1.30, one can conclude that Ass R (R/I) = {p | p = q + x jt R for all t = 1, . . . , r and q ∈ Ass R (R/J)}.

The claim follows from this fact that each q ∈ Ass R (R/J) is a monomial prime ideal.

The concept of superficial element was first introduced by P. Samuel in [START_REF] Samuel | La notion de multiplicité en algèbre et en géométrie algébrique[END_REF] for the study of multiplicities of primary ideals in local rings. In fact, it is well-known that superficial elements are not only well-behaved with respect to multiplicities but also with respect to Hilbert polynomials. Strongly superficial elements have important role in commutative algebra (see [START_REF] Smith | Strongly superficial elements[END_REF]), refer to Definition 2.2.16.

Theorem 2.3.27. Suppose that I satisfies condition (♯). Then I has a strongly superficial element of degree 1.

Proof. Without loss of generality, we may assume that G(I) is the unique minimal set of monomial generators of I such that

(u α 2 2 • • • u αm m u j : R u 1 ) = u α 2 2 • • • u αm m (u j : R u 1 ),
for all j = 2, . . . , m and α 2 , . . . , α m ∈ N 0 . As we proved in Theorem 2.3.17, it follows that (I k+1 : R u 1 ) = I k for all k ∈ N. This implies that u 1 is a strongly superficial element of degree 1 for I, as claimed.

We now introduce a class of Ratliff-Rush closed ideals. Let I be an arbitrary ideal in a commutative Noetherian ring S. From the maximal condition on ideals of S, it follows that there exist ideals I * in S which are maximal with respect to the condition

I * n = I n for all large n.
Ratliff and Rush, in [START_REF] Ratliff | Two notes on reductions of ideals[END_REF]Theorem 2.1], proved that if I is a regular ideal, then there exists a unique such I * , which can be presented in terms of I as follows:

I * := n∈N (I n+1 : S I n ).
In fact, the eventual stable value of the ascending chain Proof. According to Theorem 2.3.22(i), it follows that I * = k∈N (I k+1 : R I k ) = I. This implies that I is a Ratliff-Rush closed ideal, as claimed.

(I 2 : S I) ⊆ (I 3 : S I 2 ) ⊆ • • • ⊆ (I i+1 : S I i ) ⊆ • • • is I * .

Towards constructing the other classes

In this subsection, our goal is to introduce a method for constructing new monomial ideals which have the persistence property based on the monomial ideals which we know have the persistence property. For this purpose, we provide a criterion for the persistence property. To do this, we refer to the notion of the expansion operator on monomial ideals which has been stated in Definition 2.2.9, and employ it as a criterion for the persistence property of monomial ideals.

Ass R (R/(uI) n ) = Ass R (R/I n ) ∪ {(x i 1 ), . . . , (x ir )}.
By our hypothesis, it follows that Ass R (R/I n ) ⊆ Ass R (R/I n+1 ). Again, by Theorem 2.1.27, we obtain

Ass R (R/(uI) n+1 ) = Ass R (R/I n+1 ) ∪ {(x i 1 ), . . . , (x ir )}.
Hence, Ass R (R/(uI) n ) ⊆ Ass R (R/(uI) n+1 ), and so uI has the persistence property, as desired.

The following example illuminates our method for constructing new monomial ideals which have the persistence property based on the monomial ideals which we know have the persistence property.

Now, consider the following monomial ideal

I := (x 1 x 2 2 x 2 3 x 4 x 5 , x 2 2 x 3 3 x 5 , x 1 x 2 x 3 3 , x 4 x 5 , x 2 x 2 3 x 2 5 x 6 x 7 ), in the polynomial ring R 1 = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 ].
It is easy to see that J is the expansion of I with respect to the 7-tuple (2, 1, 1, 2, 2, 3, 1). In order to prove our claim, set I 1 := (x 1 x 2 x 4 , x 2 x 3 , x 1 x 3 x 4 , x 5 x 6 x 7 ) and u := x 2 x 2 3 x 5 . Since I 1 is a unisplit monomial ideal, it follows that has the persistence property and, by Proposition 2.3.31, one can conclude that I = uI 1 has the persistence property. Now, Lemma 2.3.30 implies that J has the persistence property, as required.

Persistence property for associated primes of a family of ideals

Throughout this section, R is a commutative Noetherian ring. Additionally, the results of this section can be found in [START_REF] Nasernejad | Persistence property for associated primes of a family of ideals[END_REF]. We start with the following definition. For an ideal c of R, consider the family Φ c = {c n } n≥1 of ideals of R. Clearly, Brodmann's Theorem shows that, for all c m ∈ Φ c , the set

( a∈Φc Ass R (R/a)) ∩ V (c m ) = n≥1 Ass R (R/c n ) is finite.
It is necessary to note that in the above definition the partially order ≤ is not reverse inclusion in general. In this section, we provide two families of ideals (with partially order relations) which have the persistence property. We now state the first example constructed on R-regular sequences.

Theorem 2.4.2. Let n be a positive integer with n ≤ depth(R). Put

Φ = {(x 1 , . . . , x n ) : x 1 , . . . , x n is an R-regular sequence}.
Then (Φ, ⊇) satisfies the persistence property.

Proof. To prove the first part of Definition 2.4.1, suppose that p is an arbitrary element in ( a∈Φ Ass R (R/a)) ∩ V (b), for some b ∈ Φ. Then there exists a ∈ Φ such that p ∈ Ass R (R/a) ∩ V (b). Let z 1 , . . . , z n be an R-sequence generating the ideal b and y 1 , . . . , y n be an R-sequence generating the ideal a. In the local ring R p , we have pR p ∈ Ass Rp (R p /aR p ), and so y 1 /1, . . . , y n /1 is a maximal R p -sequence contained in pR p . Thus, depthR p = n. Now, z 1 /1, . . . , z n /1 is an R p -sequence contained in pR p , and so it must be a maximal such sequence, and we must have pR p ∈ Ass Rp (R p /bR p ). Therefore, p ∈ Ass R (R/b), which is a finite set. Now, consider the relation ≤ on Φ such that, for all a, b ∈ Φ, a ≤ b if and only if a ⊇ b. Then (Φ, ≤) is a partially ordered set.

Assume that a and b are two ideals in Φ such that a ≤ b. Then there exist R-regular sequences x 1 , . . . , x n and y 1 , . . . , y n such that

a := (x 1 , . . . , x n ) ⊇ b := (y 1 , . . . , y n ).
Hence, there exists an n × n matrix A = (a ij ) such that y j = n i=1 a ij x i for all j with 1 ≤ j ≤ n. Let α be the determinant of A. Then, by [START_REF] Gibson | Modules of generalized fractions, matrices and determinantal maps[END_REF]Theorem 3.4], the determinantal map f a b : R/a → R/b given by multiplication in α is an R-monomorphism. Hence, Ass R (R/a) ⊆ Ass R (R/b). This means that Φ satisfies the second part of Definition 2.4.1.

Here, we present the second example constructed on d-sequences. The theory of d-sequences was introduced by Huneke in [START_REF] Huneke | The theory of d-sequences and power of ideals[END_REF].

Definition 2.4.3. ([72, Definition 1.1])

A sequence x 1 , . . . , x n of elements of R is called a d-sequence on R if (i) x i / ∈ Rx 1 + • • • + Rx i-1 + Rx i+1 + • • • + Rx n for all i = 1, . .

. , n; and,

(ii) for all k ≥ i + 1 and all i ∈ N 0 which x 0 = 0, ((x 0 , . . . , x i ) : R x i+1 x k ) = ((x 0 , . . . , x i ) : R x k ).

It is an unconditioned strong d-sequence (u.s.d-sequence) on R if x a 1 1 , . . . , x an n is a d-sequence in any order for all positive integers a 1 , . . . , a n . Now, let x = x 1 , . . . , x n be a sequence of elements of R and consider the family

Φ x := {(x a 1
1 , . . . , x an n ) : a 1 , . . . , a n ∈ N with a 1 , . . . , a n ≥ 2}, of ideals of R. In the following, for two arbitrary elements a = (x a 1 1 , . . . , x an n ) and b = (x b 1 1 , . . . , x bn n ) of Φ x , we say that a ≤ b if and only if b i ≤ a i for all i = 1, . . . , n. Clearly, (Φ x , ≤) is a partially ordered set. Note that if a ≤ b, then a ⊇ b, but the converse is not true. So, the relation ≤ in (Φ x , ≤) is not the reverse inclusion. Theorem 2.4.4. Let n be a positive integer and x = x 1 , . . . , x n be a u.s.d-sequence on R. Then, by using the above notation, (Φ x , ≤) has the persistence property.

Proof. Suppose that a = (x a 1 1 , . . . , x an n ) is an arbitrary element of Φ x . For i ∈ N with 1 ≤ i ≤ n, consider the following homomorphism f :

x a i -1 i R + n j=1 x a j j R n j=1 x a j j R -→ R n j=1,j̸ =i x a j j R + x a i +1 i R ,
given by f (α

+ n j=1 x a j j R) := αx i + n j=1,j̸ =i x a j j R + x a i +1 i R. We claim that f is monomorphism. To do this, suppose that αx i ∈ n j=1,j̸ =i x a j j R + x a i +1 i R for some α ∈ x a i -1 i R.
Then we have the following equality

αx i = x a 1 1 r 1 + • • • + x a i-1 i-1 r i-1 + x a i +1 i r i + x a i+1 i+1 r i+1 + • • • + x an n r n , for some r 1 , . . . , r n ∈ R. Since α ∈ x a i -1 i R, there exists r ∈ R such that α = x a i -1 i r. Thus, (r -r i x i )x a i i ∈ n j=1,j̸ =i x a j j R. Since x = x 1 , . . . , x n is a u.s.d-sequence
on R, we can conclude that

( n j=1,j̸ =i x a j j R : R x a i i ) = ( n j=1,j̸ =i x a j j R : R x i ). Hence, rx i ∈ n j=1,j̸ =i x a j j R + x 2 i R, and so α ∈ n j=1
x a j j R. This implies that f is an R-monomorphism. Now, as mentioned in [START_REF] Khashyarmanesh | On the associated primes of local cohomology modules[END_REF], consider the following exact sequence 0 -→

x a i -1 i R + n j=1 x a j j R n j=1 x a j j R f -→ R n j=1,j̸ =i x a j j R + x a i +1 i R g -→ R n j=1 x a j j R -→ 0,
where g is the natural homomorphism. Thus, one obtains that

Ass R R n j=1,j̸ =i x a j j R + x a i +1 i R ⊆ Ass R x a i -1 i R + n j=1 x a j j R n j=1 x a j j R ∪ Ass R R n j=1 x a j j R .
In the light of

Ass R x a i -1 i R + n j=1 x a j j R n j=1 x a j j R ⊆ Ass R R n j=1 x a j j R , we can derive that Ass R ( R n j=1,j̸ =i x a j j R + x a i +1 i R ) ⊆ Ass R ( R n j=1 x a j j R ), for all (x a 1 1 , . . . , x an n ) ∈ Φ x and i ∈ N with 1 ≤ i ≤ n. This implies that for two arbitrary elements a = (x a 1 1 , . . . , x an n ) and b = (x b 1 1 , . . . , x bn n ) of Φ x with a ≤ b, we have the inclusion Ass R (R/a) ⊆ Ass R (R/b), and also the set c∈Φx Ass R (R/c) is contained in the finite set Ass R (R/(x 2 1 , . . . , x 2 n )).
We thus get (Φ x , ≤) has the persistence property. This completes the proof. Remark 2.4.5. Let x 1 , . . . , x n be a permutable R-sequence. Then x α 1 1 , . . . , x αn n is also an R-sequence for all α 1 , . . . , α n ∈ N. By virtue of (x 1 , . . . , x n ) ⊇ (x

) → R/(x 2 1 , . . . , x 2 n ),
given by

f (r + (x 2 1 , . . . , x 2 t , x t+1 , . . . , x n )) = rx t+1 • • • x n + (x 2 1 , . . . , x 2 n ).
We therefore obtain

Ass R (R/(x α 1 1 , . . . , x αn n )) = Ass R (R/(x 1 , . . . , x n )).

Chapter 3

On the strong persistence property of monomial ideals

Associated primes of cover ideals under graph operations

In this chapter, we focus on the strong persistence property of monomial ideals.

In particular, we argue on the cover ideals of graphs. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Associated primes of powers of cover ideals under graph operations[END_REF]. Throughout this section, G is an undirected finite simple graph, that is to say, G has no loops and no multiple edges. The symbol V (G) (respectively, E(G)) is the set of vertices (respectively, the set of edges) of a graph G. Furthermore, I(G) (respectively, J(G)) stands for the edge ideal (respectively, the cover ideal) of a graph G. In addition, [n] := {1, . . . , n}.

Next lemma is necessary for us which will be used in proving Theorem 3.1.2. Also, recall that an ideal I in a commutative Noetherian ring R has the strong persistence property if (I k+1 : R I) = I k for all k ≥ 1. Lemma 3.1.1. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K with G(I) = {u 1 , . . . , u m }, v and h be two monomials in R such that gcd(v, h) = 1, h ∈ I, and gcd(v, u i ) = 1 for all i = 1, . . . , m. If I has the strong persistence property, then L := vI + hR has so.

Proof. Assume that I has the strong persistence property. Due to the definition, it is sufficient for us to show that (L t+1 : R L) = L t for all positive integers t. Based on the binomial expansion, it follows that L t+1 = t+1 i=0 (hR) t+1-i (vI) i , and according to gcd(v, h) = 1, we get the following equalities

(L t+1 : R L) = ( t+1 i=0 (hR) t+1-i (vI) i : R vI) ∩ ( t+1 i=0 (hR) t+1-i (vI) i : R hR) = (h t+1 R + t+1 i=1 (hR) t+1-i v i-1 I i : R I) ∩ ((v t+1 I t+1 : R hR) + t i=0 (hR) t-i (vI) i ).
Since h ∈ I, one can conclude that (hR) t+1-i I i ⊆ I t+1 for all i = 0, . . . , t + 1, and thus (h t+1 R + t+1 i=1 (hR) t+1-i v i-1 I i : R I) ⊆ (I t+1 : R I). Because I has the strong persistence property, we have (I t+1 : R I) = I t . Hence, one can deduce that

(h t+1 R + t+1 i=1 (hR) t+1-i v i-1 I i : R I) ⊆ I t .
By virtue of t i=0 (hR) t-i (vI) i = L t , this implies the containment below

(L t+1 : R L) ⊆ I t ∩ (v t+1 I t+1 : R hR) + I t ∩ L t .
In the light of gcd(v, h) = 1, we can assert that

I t ∩ (v t+1 I t+1 : R hR) = I t ∩ v t+1 (I t+1 : R hR),
and by gcd(v, u i ) = 1 for all i = 1, . . . , m, one has I t ∩ (v t+1 I t+1 : R hR) ⊆ (vI) t . Accordingly, I t ∩ (v t+1 I t+1 : R hR) ⊆ L t , and so (L t+1 : R L) ⊆ L t . On account of L t ⊆ (L t+1 : R L), one derives (L t+1 : R L) = L t , as required.

We now give one of the main results in the subsequent theorem. Indeed, we show that if we start with any graph G whose cover ideal has the strong persistence property, and if we then add a new vertex and join it to every vertex of G, then the cover ideal of the new graph will also have the strong persistence property.

Theorem 3.1.2. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple graphs such that V (H) = V (G) ∪ {w} with w / ∈ V (G), and
E(H) = E(G) ∪ {{v, w} : for all v ∈ V (G)}.
If J(G) has the strong persistence property, then J(H) has so.

Proof. Suppose that J(G) has the strong persistence property. Without loss of generality, one may assume that V (G) = [n] and V (H) = V (G) ∪ {n + 1}. Also, let R = K[x 1 , . . . , x n+1 ] be a polynomial ring over a field K. Let us simplify the notation by L := J(H), I := J(G), and g := n i=1 x i . 53

In the following, our goal is to apply Lemma 3.1.1. Now, one can easily derive

L = I ∩ n i=1 (x i , x n+1 ). Due to n i=1 (x i , x n+1 ) = ( n i=1 x i , x n+1 ), we deduce that L = I ∩ ( n i=1 x i , x n+1 )
. By virtue of g = n i=1 x i and G(I) ⊆ K[x 1 , . . . , x n ], we obtain g ∈ I. It is routine to check that L = x n+1 I + gR. Assume that G(I) = {u 1 , . . . , u m }. This leads to G(L) = {x n+1 u 1 , . . . , x n+1 u m , g}. In the light of L = x n+1 I + gR, gcd(x n+1 , g) = 1, g ∈ I, gcd(x n+1 , u i ) = 1 for all i = 1, . . . , m, and because I has the strong persistence property, Lemma 3.1.1 yields that L has the strong persistence property, as claimed.

In order to prove Corollary 3.1.7 and Theorem 3.1.8, we need the following theorem. To do this, we recall some definitions that will be necessary for understanding Theorem 3.1.6. Definition 4.2]) Let H = (V (H), E(H)) be a hypergraph with V (H) = {x 1 , . . . , x n }. For each s, the s-th expansion of H is defined to be the hypergraph obtained by replacing each vertex x i ∈ V (H) by a collection {x ij | j = 1, . . . , s}, and replacing E(H) by the edge set that consists of edges {x i 1 l 1 , . . . , x irlr } whenever {x i 1 , . . . , x ir } ∈ E(H) and edges {x il , x ik } for l = k. We denote this hypergraph by H s . The new variables x ij are called the shadows of x i . The process of setting x il to equal to x i for all i and l is called the depolarization.

Definition 3.1.3. ([45, Definition 2.7]) Let H = (V, E) be a hypergraph. A d- coloring of H is any partition of V = C 1 ∪ • • • ∪ C d into d disjoint
Theorem 3.1.6. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple graphs such that V (H) = V (G) ∪ {w} with w / ∈ V (G), and
E(H) = E(G) ∪ {{v, w}} for some vertex v ∈ V (G). Then Ass R ′ (R ′ /J(H) s ) = Ass R (R/J(G) s ) ∪ {(x v , x w )} for all s, where R = K[x α : α ∈ V (G)] and R ′ = K[x α : α ∈ V (H)].
Proof. To simplify the notation, set I := J(G) and J := J(H). We first show that

Ass R (R/I s ) ∪ {(x v , x w )} ⊆ Ass R ′ (R ′ /J s )
for all s. To do this, fix s ∈ N, and assume that p = (x i 1 , . . . , x ir ) is an arbitrary element of Ass R (R/I s ). It follows from [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11]

that p ∈ Ass(K[p]/J(G p ) s ), where K[p] = K[x i 1 , . . . , x ir ] and G p is the induced subgraph of G on the vertex set {i 1 , . . . , i r } ⊆ V (G). As G p = H p , we obtain p ∈ Ass(K[p]/J(H p ) s ). This implies that p ∈ Ass R ′ (R ′ /J s ). By virtue of (x v , x w ) ∈ Ass R ′ (R ′ /J s ), one derives Ass R (R/I s ) ∪ {(x v , x w )} ⊆ Ass R ′ (R ′ /J s ).
For completing the proof, it is sufficient to verify the reverse inclusion. To do this, assume that p = (x i 1 , . . . , x ir ) is an arbitrary element of Ass R ′ (R ′ /J s ) with [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11] yields that p ∈ Ass R (R/I s ), and the proof is complete. We thus let w ∈ {i 1 , . . . , i r }. Without loss of generality, one may assume that i 1 = v and i 2 = w. According to [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Corollary 4.5], the associated primes of J(H) s will correspond to critical chromatic subgraphs of size s + 1 in the s-th expansion of H. This means that one can take the induced subgraph on the vertex set {i 1 , . . . , i r }, and then form the s-th expansion on this induced subgraph, and within this new graph find a critical (s + 1)-chromatic graph. Due to w is only connected to v in the graph H, and since this induced subgraph is critical, if we remove the vertex w, we can color the resulting graph with s colors. This gives that w has to be adjacent to s vertices. But the only thing w is adjacent to is the shadows of w and the shadows of v, and hence one has a clique among these vertices. Consequently, w and its neighbors will form a clique of size s + 1. By virtue of a clique is a critical graph, it follows that we do not need any element of {i 3 , . . . , i r } or their shadows when making the critical (s + 1)-chromatic graph. This implies that p = (x v , x w ), as required.

{i 1 , . . . , i r } ⊆ V (H). If {i 1 , . . . , i r } ⊆ V (G), then
In the following, we present the other main results in this section in Corollary 3.1.7 and Theorem 3.1.8. In fact, we prove that if we take any graph G which its cover ideal has the (strong) persistence property, and if we then add a leaf to G, then the cover ideal of the new graph satisfies the (strong) persistence property as well.

Corollary 3.1.7. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple graphs such that V (H) = V (G) ∪ {w} with w / ∈ V (G), and E(H) = E(G) ∪ {{v, w}} for some vertex v ∈ V (G). If J(G) has the persistence property, then J(H) has so.

Proof. Let J(G) have the persistence property, and fix a positive integer s. Based on Theorem 3.1.6, we gain

Ass R ′ (R ′ /J(H) s ) = Ass R (R/J(G) s ) ∪ {(x v , x w )}, and
Ass R ′ (R ′ /J(H) s+1 ) = Ass R (R/J(G) s+1 ) ∪ {(x v , x w )}, where R = K[x α : α ∈ V (G)] and R ′ = K[x α : α ∈ V (H)]. Due to Ass R (R/J(G) s ) ⊆ Ass R (R/J(G) s+1 ), one can conclude that Ass R ′ (R ′ /J(H) s ) ⊆ Ass R ′ (R ′ /J(H) s+1
). This means that J(H) has the persistence property, and the desired result has been proved.

Theorem 3.1.8. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple graphs such that V (H) = V (G) ∪ {w} with w / ∈ V (G)
, and E(H) = E(G) ∪ {{v, w}} for some vertex v ∈ V (G). If J(G) has the strong persistence property, then J(H) has so.

Proof. Suppose that J(G) has the strong persistence property. Without loss of generality, one may assume that V (G) = [n], V (H) = V (G) ∪ {n + 1}, and E(H) = E(G) ∪ {{1, n + 1}}. In addition, let R = K[x 1 , . . . , x n+1 ] be a polynomial ring over a field K. In order to simplify our notation, put I := J(G), L := J(H), and Q := (x 1 , x n+1 ). By virtue of L = I ∩ Q, we need only prove that I ∩ Q has the strong persistence property. To achieve this, one has to demonstrate

((I ∩ Q) k+1 : R I ∩ Q) = (I ∩ Q) k , for all k. Fix k ∈ N.
In the light of [START_REF] Matsumura | Commutative Ring Theory[END_REF]Exercise 6.4], it is enough to verify that

(L k+1 a : Ra L a ) = L k a for every a ∈ Ass R (R/L k ).
Let a be an arbitrary element in Ass R (R/L k ). It follows also from Theorem 3.1.6 that Ass R (R/L k ) = Ass(I k ) ∪ {Q}. We thus can consider the following two cases: Case 1. a ∈ Min(L k ). Since Min(L k ) = Min(L), one derives a ∈ Min(L). In the following, we show a / ∈ Supp R (R/b) for every b ∈ Ass R (R/L) \ {a}. Suppose, on the contrary, that there exists b

∈ Ass R (R/L) \ {a} such that a ∈ Supp R (R/b). Due to Supp R (R/b) = V (Ann R (R/b)), one has Ann R (R/b) ⊆ a,
and hence b ⊆ a. On account of a is a minimal prime ideal of L, we get b = a, which is a contradiction. This implies that b a = R a for every b ∈ Ass R (R/L) \ {a}. Therefore, L a = a a , and hence (L k ) a = (a k ) a . It also arises from direct computation that (a k+1 : R a) = a k . We thus gain (a k+1 a : Ra a a ) = a k a , and so (L

k+1 a : Ra L a ) = L k a , as required. Case 2. a ∈ Ass R (R/L k ) \ Min(L k ). By Min(L k ) = Min(L) = Min(I) ∪ {Q},
we deduce that a = Q, and thus a ∈ Ass(I k ) \ Min(I). This implies that x n+1 / ∈ a. We next prove that (I ∩ Q) a = I a . By virtue of (I ∩ Q) a ⊆ I a , it is sufficient to show that I a ⊆ (I ∩ Q) a . For this purpose, assume that r/s ∈ I a . This gives that r/s = α/β for some α ∈ I and β / ∈ a. It is routine to check that αx n+1 ∈ I ∩ Q and βx n+1 / ∈ a. On the other hand, on account of the following equality From now onwards, we give some applications of the results which appeared in this section. Indeed, we show that the cover ideals of some imperfect graphs satisfy the strong persistence property. To achieve this, one should remember the definition of imperfect graphs, see Section 1.2.

α β = αx n+1 βx n+1 , one derives r/s ∈ (I ∩ Q) a ,

Cover ideals of odd cycle graphs

In the first subsection, we investigate the cover ideals of cycle graphs of odd orders. In other words, Theorem 3.1.11 states that the cover ideal of every cycle graph of odd order satisfies the strong persistence property. For this purpose, we require Remark 3.1.9 and Lemma 3.1.10. We start with the following comment.

Let R = K[x 1 , . . . , x 2n+1 ] be a polynomial ring over a field K. Let I = J(C 2n+1 ) ⊂ R be the cover ideal associated to the cycle graph C 2n+1 with vertex set

[2n + 1].
That is,

I = 2n+1 i=1 (x i , x i+1 ) = (x 1 , x 2 ) ∩ • • • ∩ (x 2n , x 2n+1 ) ∩ (x 2n+1 , x 1 )
,

where 2n + 2 = 1. Note that a monomial m := x l 1 1 • • • x l 2n+1
2n+1 belongs to I if and only if m ∈ (x i , x i+1 ), if and only if l i + l i+1 ≥ 1 for every i ∈ [2n + 1]. As R has an odd number of indeterminates, then l j + l j+1 ≥ 2 for some j ∈ [2n + 1]. Therefore, we conclude the following remark.

Remark 3.1.9. A monomial m = x l 1 1 • • • x l 2n+1 2n+1 ∈ I if and only if l i + l i+1 ≥ 1 for all i ∈ [2n + 1] and 2n+1 i=1 [(l i + l i+1 ) -1] ≥ 1.
According to the remark above, consider a special subset {µ 1 , . . . , µ 2n+1 } of generators of I with (emphasizing that 2n + 2 = 1) such that

µ j := x j x j+1 n-1 k=1 x j+1+2k . Lemma 3.1.10. A monomial f = x l 1 1 • • • x l 2n+1 2n+1 ∈ I t if and only if l i + l i+1 ≥ t for every i ∈ [2n + 1] and 2n+1 i=1 [(l i + l i+1 ) -t] ≥ t. Proof.
The necessity is direct from the above remark. To prove the sufficiency, we proceed by induction on t. Based on Remark 3.1.9, the claim can be proved for the case t = 1. Assume that the statement is true for some t > 1.

Let f = x l 1 1 • • • x l 2n+1 2n+1
be with l i +l i+1 ≥ t +1 for every i ∈ [2n+1] and 2n+1 i=1 [(l i + l i+1 ) -(t + 1)] ≥ t+1. Want to show f ∈ I t+1 . First, we claim that if µ j | f for some j and if l j +l j+1 ≥ t+2, then f /µ j ∈ I t , and hence f ∈ I t+1 . To prove the claim, write f /µ j = x

l ′ 1 1 • • • x l ′ 2n+1 2n+1 and note that l ′ j + l ′ j+1 = l j + l j+1 -2 and l ′ i + l ′ i+1 = l i + l i+1 -1 for i = j. Consider, 2n+1 i=1 (l ′ i + l ′ i+1 ) -t = (l j + l j+1 -2) -t + 2n+1 i=1,i̸ =j [(l i + l i+1 -1) -t] = (l j + l j+1 ) -(t + 1) -1 + 2n+1 i=1,i̸ =j [(l i + l i+1 ) -(t + 1)] = 2n+1 i=1 [(l i + l i+1 ) -(t + 1)] -1 ≥ t;
thus, f /µ j ∈ I t by induction. Define the set A = {j | l j = 0} and rewrite A = {j 1 , . . . , j s } with j i < j i+1 . Define a monomial N , and for every two consecutive elements j i , j i+1 in A with 1 ≤ i ≤ s -1 define a monomial M i,i+1 , as follows (with addition of indices is taken modulo 2n + 1 as of the cyclic behavior of the graph)

N = x σ j 1 -1 (j 1 +2n-1-σ-js)/2 k=0 x js+1+2k and M i,i+1 = x ε j i +1 (j i+1 -j i -2-ε)/2 k=0 x j i +1+ε+2k ,
where σ = 1 or 0 according as j s -j 1 is even or odd, and ε = 0 or 1 according as j i+1 -j i is even or odd. Roughly speaking, whenever j i and j i+1 are both odd or both even, then M i,i+1 is the product of every second variable between x j i and x j i+1 beginning with x j i +1 and ending with x j i+1 -1 . Also, whenever one of j i and j i+1 is odd and the other is even, then M i,i+1 is x j i +1 multiplied by the product of every second variable beginning with x j i +2 and ending with x j i+1 -1 . Similarly, whenever one of j 1 and j s is odd and the other is even, then N is the product of every second variable beginning with x js+1 and ending with x j 1 -1 taken in a cyclic order. Also, whenever j i and j i+1 are both odd or both even, then N is x j 1 -1 multiplied by the product of every second variable beginning with x js+1 and ending with x j 1 -2 , taken in a cyclic order.

Define

g = N s-1 i=1 M i,i+1 = x h 1 1 • • • x h 2n+1 2n+1 . It is clear that h i + h i+1 = 1
or 2 for all i. Precisely, h i + h i+1 = 2 if and only if i = j 1 -2 with j s -j 1 is even or i = j r + 1 with j r+1 -j r is odd. Hence, g ∈ I. Recall that l i-2 + l i-1 ≥ t + 2 for any i ∈ A with i -2 / ∈ A, and also

l i+1 + l i+2 ≥ t + 2 for any i ∈ A with i + 2 / ∈ A. Therefore, if f /g = x l ′ 1 1 • • • x l ′ 2n+1 2n+1 with l ′ i = l i -h i , then l ′ i + l ′ i+1 ≥ t for all i.
If A consists (or can be represented) entirely of odd (or even) integers, then g = µ j 1 -2 ; thus, the proof is over by the above claim (note if j 1 = 1, then j 1 -2 equals 2n because computations are taken modulo 2n + 1). Pick j r , j r+1 ∈ A such that j r+1 -j r is odd. Note that l jr+1 ≥ t + 1 since j r ∈ A, and l jr+2 = a ≥ 1 since j r + 2 / ∈ A. Also, since l i + l i+1 ≥ t + 1 for all i, then l jr+3 ≥ t + 1 -a, which in turn forces l jr+4 ≥ a. Continuing in this manner, and since j r+1 -j r is odd, we end up with l j r+1 -2 ≥ t + 1 -a. But l j r+1 -1 ≥ t + 1 since j r+1 ∈ A. Therefore, we have

(l jr+1 + l jr+2 ) + l j r+1 -2 + l j r+1 -1 ≥ 3(t + 1). Thus, l ′ jr+1 + l ′ jr+2 + l ′ j r+1 -2 + l ′ j r+1 -1 = (l jr+1 + l jr+2 ) -(h jr+1 + h jr+2 ) + l j r+1 -2 + l j r+1 -1 -h j r+1 -2 + h j r+1 -1 ≥ 3(t + 1) -2 -1 = 3t; therefore, 2n+1 i=1 l ′ i + l ′ i+1 -t ≥ t; thus, f /g ∈ I t
by the inductive hypothesis. This completes the inductive step, and therefore the claim has been proved by induction.

We are now in a position to show Theorem 3.1.11. Let I = J(C 2n+1 ) be the cover ideal. Our aim is to show that (I t+1 : I) = I t for all t ≥ 1. Fix t ≥ 1. We only need to show that (I t+1 : I) ⊆ I t . Suppose

f ′ = x l ′ 1 1 • • • x l ′ 2n+1 2n+1 ∈ (I t+1 : I). For any i ∈ [2n + 1] choose i 0 = i and consider f ′ µ i 0 := x l 1 1 • • • x l 2n+1 2n+1 ∈ I t+1 since µ i 0 ∈ I; hence, l i +l i+1 ≥ t+1 for every i ∈ [2n+1]; and therefore, l ′ i + l ′ i+1 ≥ t for every i ∈ [2n + 1]. Now, fix j 0 ∈ [2n + 1] and consider f ′ µ j 0 := x l 1 1 • • • x l 2n+1 2n+1 ∈ I t+1 ; thus, 2n+1 i=1 [(l i + l i+1 ) -(t + 1)] ≥ t+1. Note l j 0 +l j 0 +1 = l ′ j 0 +l ′ j 0 +1 +2 and l i +l i+1 = l ′ i +l ′ i+1 +1 for i = j 0 .
Therefore, by the same process as the first part of the above proof, we obtain 2n+1 i=1

(l ′ i + l ′ i+1 ) -t ≥ t; hence, f ′ ∈ I t , which finishes the proof.
In order to establish Proposition 3.1.13, we require the following proposition. This proposition is a well-known result, but we re-prove it by a new proof. Proposition 3.1.12. Let I be an ideal in a commutative Noetherian ring S such that satisfies the strong persistence property. Then I has the persistence property.

Proof. Fix k ≥ 1, and choose an arbitrary element p ∈ Ass S (S/I k ). This implies that p = (I k : S h) for some h ∈ S. Since I satisfies the strong persistence property, we have (I k+1 : S I) = I k , and so p = ((I k+1 : S I) : S h). Let G(I) = {u 1 , . . . , u m }. Hence, one obtains p = (I k+1 : S h m i=1 u i S) = m i=1 (I k+1 : S hu i ). Accordingly, we get p = (I k+1 : S hu i ) for some 1 ≤ i ≤ m. Therefore, p ∈ Ass S (S/I k+1 ). This means that I has the persistence property, as claimed.

In the next proposition, our goal is to detect when the unique homogeneous maximal ideal appears. This result can also be deduced directly from [ x k . To achieve this, we need to show that x i 2n+1 k=1 x k ∈ I 2 for all i = 1, . . . , 2n + 1. To see this, let i be an arbitrary positive integer with 1 ≤ i ≤ 2n + 1, and set g i := x i 2n+1 k=1 x k . Now, write

g i = x i 2n+1 k=1 x k = 2n+1 k=1 x ℓ k k .
It is routine to check that ℓ j + ℓ j+1 ≥ 2 for all j ∈ [2n + 1], and also

2n+1 k=1 [(ℓ k + ℓ k+1 ) -2] = (3 + 3) -2(2) = 2.
Now, Lemma 3.1.10 yields that g i ∈ I 2 , as required. Moreover, since the ideal I is a square-free monomial ideal and 2n + 1 ≥ 3, one can see that m / ∈ Ass R (R/I). This gives m ∈ Ass R (R/(J(C 2n+1 )) s ) for all s ≥ 2, and the desired conclusion has been obtained.

We now explore the associated prime ideals of powers of the cover ideals of cycle graphs of odd orders. To do this, we first recall the definition of the monomial localization of a monomial ideal with respect to a monomial prime ideal which has been stated in [START_REF] Herzog | Persistence and stability properties of powers of ideals[END_REF], and then use it in the proof of Proposition 3.1.15. Definition 3.1.14. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. We also denote by V * (I) the set of monomial prime ideals containing I. Let p = (x i 1 , . . . , x ir ) be a monomial prime ideal. The monomial localization of I with respect to p, denoted by I(p), is the ideal in the polynomial ring It should be noted that a special case of Proposition 3.1.15, when s = 2, has already appeared in [START_REF] Francisco | Associated primes of monomial ideals and odd holes in graphs[END_REF].

R(p) = K[x i 1 , . . . ,
R = K[x 1 , . . . , x 2n+1
] is a polynomial ring over a field K, and m is the unique homogeneous maximal ideal of R. Then

Ass R (R/(J(C 2n+1 )) s ) = Ass R (R/J(C 2n+1 )) ∪ {m},
for all s ≥ 2. In particular,

Ass ∞ (J(C 2n+1 )) = {(x i , x i+1 ) : i = 1, . . . 2n} ∪ {(x 2n+1 , x 1 )} ∪ {m}.
Proof. Label the vertices of C 2n+1 in counterclockwise order with 1, . . . , 2n + 1. To simplify our notation, we take I := J(C 2n+1 ). Fix s ∈ N with s ≥ 2. It follows from Proposition 3.1.13 that Ass R (R/I) ∪ {m} ⊆ Ass R (R/I s ). Therefore, one has to demonstrate the reverse inclusion. For this purpose, let p = (x i 1 , . . . , x ir ) be an arbitrary element in Ass R (R/I s )\{m}. This implies that (x i 1 , . . . , x ir ) ∈ Ass(I s (p)). On account of above argument, one can assume that the induced subgraph of C 2n+1 on the vertex set {i 1 , . . . , i r } is a union of some disjoint path subgraphs. On the other hand, union of these disjoint path subgraphs can be viewed as a bipartite graph, say H. Accordingly, I(p) is the cover ideal of H. Due to [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6], it follows that I(p) is normally torsion-free, i.e., Ass((I(p)) t ) = Ass(I(p)) for all t. Since I t (p) = (I(p)) t for all t, we get (x i 1 , . . . , x ir ) ∈ Ass(I(p)). This leads to p ∈ Ass R (R/I), and the proof is complete. The last assertion follows readily from the fact that Ass R (R/J(C 2n+1 )) = {(x i , x i+1 ) : i = 1, . . . 2n} ∪ {(x 2n+1 , x 1 )}.

Cover ideals of wheel graphs

In the second subsection, we probe the cover ideals of wheel graphs. To do this, we first express the definition of wheel graphs. Definition 3.1.16. A wheel graph W n of order n is a graph that contains a cycle of order n -1, and for which every vertex in the cycle is connected to one other vertex which is known as the hub. The edges of a wheel which include the hub are called spokes.

In the following, we show that every cover ideal of wheel graphs of even order has the strong persistence property. Theorem 3.1.17. Suppose that W 2n is a wheel graph of order 2n on the vertex set [2n]. Then J(W 2n ) satisfies the strong persistence property.

Proof. Label the vertices of the cycle graph C 2n-1 of order 2n-1 in counterclockwise order with 1, . . . , 2n -1, and the hub with 2n, as shown in the figure below. Now, the desired conclusion follows readily from Theorems 3.1.11 and 3.1.2. This finishes the proof.

Proposition below illuminates when the unique homogeneous maximal ideal appears. Proposition 3.1.18. Suppose that W 2n is a wheel graph of order 2n on the vertex set [2n], R = K[x 1 , . . . , x 2n ] is a polynomial ring over a field K, and m is the unique homogeneous maximal ideal of R. Then m ∈ Ass R (R/(J(W 2n )) s ) for all s ≥ 3, and

m / ∈ Ass R (R/(J(W 2n )) s ) for s = 1, 2.
Proof. Label the vertices of C 2n-1 in counterclockwise order with 1, . . . , 2n -1, and the hub with 2n. Set I := J(W 2n ). By virtue of Theorem 3.1.17 and Proposition 3.1.12, one can conclude that the ideal I has the persistence property. On the other hand, [78, Theorem 2.2] yields that m ∈ Ass R (R/I 3 ) \ Ass R (R/I 2 ).

According to 2n ≥ 4, this gives rise to m / ∈ Ass R (R/I). Therefore, we gain m ∈ Ass R (R/(J(W 2n )) s ) for all s ≥ 3, and m / ∈ Ass R (R/(J(W 2n )) s ) for s = 1, 2, as claimed.

Cover ideals of helm graphs

The third subsection is devoted to the cover ideals of helm graphs. For this purpose, we begin with the definition of helm graphs. Definition 3.1.19. The helm graph H n , where has 2n + 1 vertices, is the graph obtained from a wheel graph W n+1 of order n + 1 by adjoining a pendant edge at each node of the outer circuit of the wheel graph W n+1 .

The following theorem illustrates every cover ideal of helm graphs of odd order with greater than or equal to 5, has the strong persistence property. We can now combine together Theorem 3.1.17 and the iteration of Theorem 3.1.8 to obtain the claim.

In the following proposition, one can see when appears the unique associated prime ideal which is maximal with respect to inclusion. It follows from Proposition 3.1.21 that for the unique associated prime ideal m which is maximal with respect to inclusion, one derives m ∈ Ass R (R/(J(H 5 )) s ) for all s ≥ 3, and m / ∈ Ass R (R/(J(H 5 )) s ) for s = 1, 2.

Superficial ideals in commutative Noetherian rings

The concept of superficial elements was first introduced by P. Samuel in [START_REF] Samuel | La notion de multiplicité en algèbre et en géométrie algébrique[END_REF]. Indeed, let I be an ideal in a commutative ring S, and let k ∈ N. An element x in S is called a superficial element of degree k for I if x ∈ I k and there exists c ∈ N such that (I n+k : S x) ∩ I c = I n for all n ≥ c. If (I n+k : S x) = I n for all n ∈ N, we say that x is a strongly superficial element of degree k for I (see [90, 4.1.2]). The motivation of this section comes back to the notion of strong persistence property. In fact, exploring the proof of Proposition 2.2.17 gives us that if an ideal I has a strongly superficial element of degree 1, then I has the strong persistence property. This result gives rise to a natural question in the context of the notion of strong persistence property: May one consider a part of a minimal set of generators of I instead of all elements of the minimal set of generators in the definition of the strong persistence property? To answer this question, we provide the definition of superficial ideals. Definition 3.2.1. let I and J be two ideals in a commutative Noetherian ring S. We say that J is a superficial ideal for I if the following conditions are satisfied: (i) G(J) ⊆ G(I); and, (ii) (I k+1 : S J) = I k for all positive integers k.

On the other hand, it is easy to see that an ideal I has the strong persistence property if and only if I has a superficial ideal. Therefore, one can replace the concept of the strong persistence property with superficial ideals. It should be noted that the results of this section can be found in [START_REF] Rajaee | Superficial ideals for monomial ideals[END_REF].

We begin with the following proposition which will be used in the proof of Proposition 3.2.3.

Proposition 3.2.2. Let I and J be two ideals of a commutative Noetherian ring S with G(J) ⊆ G(I).

Then J is a superficial ideal for I if and only if (I r : S J s ) = I r-s for all r, s ∈ N with r ≥ s.

Proof. First, assume that J is a superficial ideal for I. We use induction on s.

Our assumption gives that the assertion is true in the case s = 1. Now, suppose, inductively, that s > 1 and that the result has been proved for s -1. Due to J is a superficial ideal for I, and on account of (I r : S J s ) = ((I r : S J) : S J s-1 ), one derives (I r : S J s ) = (I r-1 : S J s-1 ). Now, the inductive hypothesis implies that (I r-1 : S J s-1 ) = I r-s . Thus, we have (I r : S J s ) = I r-s . This completes the inductive step, and so the claim has been proved by induction. The converse implication is straightforward.

The next result describes the relation between powers of ideals and powers of their superficial ideals. Proposition 3.2.3. Let I and J be two ideals of a commutative Noetherian ring S such that J is a superficial ideal for I. Then J s is a superficial ideal for I s for all s ∈ N.

Proof. Fix s ∈ N. By virtue of J is a superficial ideal for I, Proposition 3.2.2 yields that (I sk+s : S J s ) = I sk for all positive integers k. Thus, one can deduce that ((I s ) k+1 : S J s ) = (I s ) k for all positive integers k. Moreover, one can easily verify that G(J s ) ⊆ G(I s ). Accordingly, J s is a superficial ideal for I s , as required.

Along with these consequences, the following question arises: Question 3.2.4. Does every commutative ring have the property that any of its ideals has a superficial ideal? Proposition 3.2.5. Every commutative Noetherian ring does not have this property that any of its ideals has a superficial ideal.

Proof. We give a counterexample. To achieve this goal, one should use a graph which has been stated in [START_REF] Kaiser | Replication in critical graphs and the persistence of monomial ideals[END_REF]. Indeed, for a positive integer n, let [n] denote the set {0, . . . , n -1}. Assume that P n is a path with vertex set [n], with vertices in the increasing order along P n . Let also K 3 be the complete graph whose vertex set is the group Z 3 . For n ≥ 4, we define H n as the graph obtained from the Cartesian product P n 2K 3 by adding the three edges joining (0, j) to (n -1, -j) Assume that I = J(H 4 ) denotes the cover ideal of H 4 in the polynomial ring R = K[x 1 , . . . , x 12 ] over a field K. Suppose, on the contrary, that there exists an ideal J with G(J) ⊆ G(I) such that (I k+1 : R J) = I k for all k. This implies that (I k+1 : R I) = I k for all k, i.e., the ideal I has the strong persistence property, and so the ideal I has the persistence property, i.e., Ass R (R/I k ) ⊆ Ass R (R/I k+1 ) for all k. By computing the set of associated primes of I 3 and I 4 , one derives m = (x 1 , . . . , x 12 ) ∈ Ass R (R/I 3 ) \ Ass R (R/I 4 ). This leads to a contradiction. In other words, there exists a commutative Noetherian ring R and an ideal I of R such that I has no superficial ideal.

for j ∈ Z 3 . Figure below is the graph of H 4 . v 0,2 v 0,0 v 0,1 v 1,2 v 1,0 v 1,1 v 2,2 v 2,0 v 2,1 v 3,2 v 3,0 v 3,1
The following proposition states that under a certain condition, one can always find a superficial ideal for some power of an ideal. Proposition 3.2.6. Let S be a commutative Noetherian ring and I be an ideal of S such that I has a non-zerodivisor element. Then there exists a positive integer s such that I s is a superficial ideal for I s .

Proof. By the assumption, I has a non-zerodivisor element, say z. It follows from the Artin-Rees lemma that there exists a positive integer ℓ such that for all n > ℓ, we have I n ∩ zS = I n-ℓ (I ℓ ∩ zS). In the light of z(I n : S z) = I n ∩ zS and I n-ℓ (I ℓ ∩ zS) ⊆ zI n-ℓ , one can conclude that (I n : S z) ⊆ I n-ℓ for all n > ℓ. Now, assume that R = gr I (S) = S/I ⊕ I/I 2 ⊕ I 2 /I 3 ⊕ • • • . This implies that R is a Noetherian N-graded ring. The ideal (0 : R I/I 2 ) is finitely generated, in degrees at most some positive integer r. Accordingly, one derives (0 : R I/I 2 ) ∩ I n /I n+1 = 0 for all n > r. This yields that (I n+1 : S I) ∩ I r = I n for all n > r. Therefore, for all n > ℓ+r, we get (I n+1 : S I) ⊆ (I n+1 : S z) ⊆ I n+1-ℓ . This gives that (I n+1 : S I) ⊆ I r for all n > ℓ + r, and so (I n+1 : S I) = I n for all n > ℓ + r. Here put s := ℓ + r + 1 and fix k ∈ N. It is routine to check that ((I s ) k+1 : R I s ) = (I s ) k . This means that I s is a superficial ideal for I s , as claimed.

Some recursive theorems on the superficiality

Here, we give two recursive theorems which allow us to build new ideals that have superficial ideals. In order to demonstrate Theorem 3.2.8, one needs to prove the following lemma.

Lemma 3.2.7. Let I be a monomial ideal of R = K[x 1 , . . . , x n ] such that I = I 1 R + I 2 R, where G(I 1 ) ⊂ R 1 = K[x 1 , . . . , x m ] and G(I 2 ) ⊂ R 2 = K[x m+1 , . . . , x n ] for some positive integer 1 ≤ m < n. Then I s = s i=1 (I i 1 + I s+1-i 2 ) for all s ∈ N. Proof. Fix s ∈ N and set L r := I r 1 + r-1 i=0 I i 1 ∩ I s-i 2 with 1 ≤ r ≤ s. Our aim is to prove that L r = r i=1 (I i 1 + I s+1-i 2
). To do this, we proceed by induction on r. One can easily see that the assertion is true for the case in which r = 1. Now, suppose, inductively, that r > 1 and that the result has been proved for all r less than t with t ≤ s. It follows from the inductive hypothesis that

L t-1 = t-1 i=1 (I i 1 + I s+1-i 2 ). On account of I t 1 ∩ t-2 i=0 I i 1 ∩ I s-i 2 ⊆ t-2 i=0 I i 1 ∩ I s-i 2 and t-2 i=0 I i 1 ∩ I s-i 2 ⊆ I s+1-t 2
, one can derive the following equalities

L t = I t 1 + t-1 i=0 I i 1 ∩ I s-i 2 = I t 1 + I t-1 1 ∩ I s-t+1 2 + t-2 i=0 I i 1 ∩ I s-i 2 = I t 1 ∩ I t-1 1 + I t 1 ∩ t-2 i=0 I i 1 ∩ I s-i 2 + I t-1 1 ∩ I s-t+1 2 + t-2 i=0 I i 1 ∩ I s-i 2 = (I t 1 + I s+1-t 2 ) ∩ (I t-1 1 + t-2 i=0 I i 1 ∩ I s-i 2 ) = (I t 1 + I s+1-t 2 ) ∩ L t-1 = (I t 1 + I s+1-t 2 ) ∩ t-1 i=1 (I i 1 + I s+1-i 2 ) = t i=1 (I i 1 + I s+1-i 2 ).
This completes the inductive step, and so the claim has been proved by induction.

In particular, this implies that

L s = s i=1 (I i 1 + I s+1-i 2
), and hence s i=1

(I i 1 + I s+1-i 2 ) = I s 1 + s-1 i=0 I i 1 ∩ I s-i 2 = s i=0 I i 1 ∩ I s-i 2 .
The assumption yields that

I i 1 ∩ I s-i 2 = I i 1 I s-i 2
for each i = 0, . . . , s. This gives that

s i=1 (I i 1 + I s+1-i 2 ) = s i=0 I i 1 I s-i 2 .
Finally, it follows from the binomial expansion that s i=1 (I i 1 + I s+1-i 2 ) = (I 1 + I 2 ) s = I s , as required.

We are now ready to express Theorem 3.2.8, which is fundamental for proving Theorem 3.2.26.

Theorem 3.2.8. Let I be a monomial ideal of R = K[x 1 , . . . , x n ] such that I = I 1 R + I 2 R, where G(I 1 ) ⊆ R 1 = K[x 1 , . . . , x m ] and G(I 2 ) ⊆ R 2 = K[x m+1 , . . . , x n ]
for some positive integer m. If I 1 or I 2 has a superficial ideal, then I has a superficial ideal.

Proof. Without loss of generality, assume that J 1 is a superficial ideal for I 1 with G(J 1 ) = {v 1 , . . . , v t }. Because G(J 1 ) ⊆ G(I), it is sufficient for us to verify that (I k+1 : R J 1 ) = I k for all k. To do this, fix k ∈ N. By virtue of the assumption, one has (I i 1 : R J 1 ) = I i-1

1
for each i = 1, . . . , k + 1. In view of Lemma 3.2.7, we obtain the following equalities

(I k+1 : R J 1 ) = ( k+1 i=1 (I i 1 + I k+2-i 2 ) : R J 1 ) = k+1 i=1 t j=1 ((I i 1 : R v j ) + (I k+2-i 2 : R v j )) = k+1 i=1 t j=1 ((I i 1 : R v j ) + I k+2-i 2 ) = k+1 i=1 (I k+2-i 2 + ∩ t j=1 (I i 1 : R v j )) = k+1 i=1 (I k+2-i 2 + (I i 1 : R J 1 )) = k+1 i=2 (I k+2-i 2 + I i-1 1 ) = k ℓ=1 (I k+1-ℓ 2 + I ℓ 1 ) = I k .
This yields that J 1 is a superficial ideal for I. A similar argument shows that if J 2 is a superficial ideal for I 2 , then one obtains (I k+1 : R J 2 ) = I k , i.e., J 2 is a superficial ideal for I. This completes the proof.

We state the second main result in the following theorem, which will be used in the proof of Theorem 3.2.29. Theorem 3.2.9. Let I be a monomial ideal in R = K[x 1 , . . . , x n ] with G(I) = {u 1 , . . . , u m }, v and h be two monomials in R such that gcd(v, h) = 1, h ∈ I, and gcd(v, u i ) = 1 for all i = 1, . . . , m. If J is a superficial ideal for I, then vJ + hR is a superficial ideal for vI + hR.

Proof. In order to simplify our notation, set L := vI + hR. Let J be a superficial ideal for I. This implies that (I t+1 : R J) = I t for all positive integers t. Fix t ∈ N. Due to the binomial expansion, it follows that L t+1 = t+1 i=0 (hR) t+1-i (vI) i . In the light of gcd(v, h) = 1, one obtains the following equalities

(L t+1 : R vJ + hR) = ( t+1 i=0 (hR) t+1-i (vI) i : R vJ) ∩ ( t+1 i=0 (hR) t+1-i (vI) i : R hR) = (h t+1 R + t+1 i=1 (hR) t+1-i v i-1 I i : R J) ∩ ((v t+1 I t+1 : R hR) + t i=0 (hR) t-i (vI) i ).
Because h ∈ I, one can conclude that (hR) t+1-i I i ⊆ I t+1 for each i = 0, . . . , t + 1, and so (h t+1 R + t+1 i=1 (hR) t+1-i v i-1 I i : R J) ⊆ (I t+1 : R J). Accordingly, one has

(h t+1 R + t+1 i=1 (hR) t+1-i v i-1 I i : R J) ⊆ I t .
On account of t i=0 (hR) t-i (vI) i = L t , one can deduce the containment below

(L t+1 : R vJ + hR) ⊆ I t ∩ (v t+1 I t+1 : R hR) + I t ∩ L t .
By virtue of gcd(v, h) = 1, one can yield the following equality

I t ∩ (v t+1 I t+1 : R hR) = I t ∩ v t+1 (I t+1 : R hR).
As gcd(v, u i ) = 1 for each i = 1, . . . , m, we get I t ∩ (v t+1 I t+1 : R hR) ⊆ (vI) t . Consequently, I t ∩ (v t+1 I t+1 : R hR) ⊆ L t , and so (L t+1 : R vJ + hR) ⊆ L t . Based on L t ⊆ (L t+1 : R vJ + hR), we gain (L t+1 : R vJ + hR) = L t . On the other hand, it follows from G(J) ⊆ G(I) that G(vJ + hR) ⊆ G(vI + hR). Therefore, vJ + hR is a superficial ideal for vI + hR, and the proof is complete.

Superficiality under some monomial operators

In the following, we introduce several methods for constructing new classes of monomial ideals which have superficial ideals. To accomplish this purpose, we start with the first method. Thus, we need to consider the expansion operator on monomial ideals which has been stated in Definition 2.2.9, and utilize it as a criterion for the existence of superficial ideals.

The following theorem provides a powerful tool for producing new superficial ideals.

Theorem 3.2.10. Let I and J be two monomial ideals of R. Then J is a superficial ideal for I if and only if J * is a superficial ideal for I * . Proof. For necessity, assume that J is a superficial ideal for I. Fix k ∈ N. This implies that (I k+1 : R J) = I k . In the light of [13, Lemma 1.1(v)], we obtain ((I k+1 ) * : R * J * ) = (I k ) * . Furthermore, [13, Lemma 1.1(iii)] yields that (I k ) * = (I * ) k and (I k+1 ) * = (I * ) k+1 . Accordingly, ((I * ) k+1 : R * J * ) = (I * ) k . It is clear that G(J * ) ⊆ G(I * ). Therefore, one can deduce that J * is a superficial ideal for I * . Sufficiency follows similarly, and the proof is over.

We now state the second method for constructing new superficial ideals. To achieve this, one requires to recall the next definition. Definition 3.2.11. A weight over a polynomial ring R = K[x 1 , . . . , x n ] is a function W : {x 1 , . . . , x n } → N, and w i = W (x i ) is called the weight of the variable x i . Given a monomial ideal I and a weight W , we define the weighted ideal, denoted by I W , to be the ideal generated by {h(u) : u ∈ G(I)}, where h is the unique homomorphism

h : R → R given by h(x i ) = x w i i . For a monomial u ∈ R, we denote h(u) = u W .
For instance, consider the monomial ideal

I = (x 2 1 x 2 x 6 3 , x 3 2 x 4 x 4 5 ) in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 ]. Furthermore, let W : {x 1 , x 2 , x 3 , x 4 , x 5 } → N be a weight over R with W (x 1 ) = 2, W (x 2 ) = 4, W (x 3 ) = 2, W (x 4 ) = 3, and W (x 5 ) = 1.
Therefore, the weighted ideal I W is given by

I W = (x 4 1 x 4 2 x 12 3 , x 12 2 x 3 4 x 4 5
). The next lemma is essential for us to prove Theorem 3.2.13. Lemma 3.2.12. Let I and J be two monomial ideals of R, and W a weight over R. Then the following statements hold. andG((I +

(i) (I + J) W = I W + J W ; (ii) (IJ) W = I W J W ; (iii) (I ∩ J) W = I W ∩ J W ; (iv) (I : R J) W = (I W : R J W ). Proof. (i) It is routine to check that if u | v (respectively, v | u) for some u ∈ G(I) and v ∈ G(J), then u W | v W (respectively, v W | u W ). Now, the desired conclusion follows from the facts that G(I W ) = {u W : u ∈ G(I)}, G(J W ) = {v W : v ∈ G(J)},
J) W ) ⊆ G(I W ) ∪ G(J W ).
(ii) One can easily see that (uv) W = u W v W for all u ∈ G(I) and v ∈ G(J). Here, the assertion follows directly from the fact that G((IJ) W ) ⊆ G(I W )G(J W ).

(iii) It is straightforward to show that (lcm(u, v)) W = lcm(u W , v W ), and thus one can conclude the claim from the fact that I W ∩ J W is generated by the following set

{lcm(u W , v W ) : u W ∈ G(I W ), v W ∈ G(J W )}.
(iv) In the light of (I : R J) = v∈G(J) (I : R v), and by part (iii), one can derive (I : R J) W = v∈G(J) (I : R v) W . Our aim is to show that (I : R v) W = (I W : R v W ) for all v ∈ G(J). On account of I W = u∈G(I) u W R, it follows that

(I W : R v W ) = ( u∈G(I) u W R : R v W ) = u∈G(I) (u W : R v W ).
For completing the proof, one has to demonstrate (u W : R v W ) = (u : R v) W , and then apply part (i). This claim follows readily form the fact that (gcd(u, v)) W = gcd(u W , v W ).

We are now ready to prove Theorem 3.2.13 by using Lemma 3.2.12. Theorem 3.2.13. Let I and J be two monomial ideals of R, and W a weight over R. Then J is a superficial ideal for I if and only if J W is a superficial ideal for I W .

Proof. For the forward implication, assume that J is a superficial ideal for I. Fix k ∈ N. This gives that (I k+1 : R J) = I k . On account of Lemma 3.2.12(iv), one has ((I k+1 ) W : R J W ) = (I k ) W . It follows from Lemma 3.2.12(ii) that (I k+1 ) W = (I W ) k+1 and (I k ) W = (I W ) k . Hence, we get ((I W ) k+1 : R J W ) = (I W ) k . Therefore, J W is a superficial ideal for I W . A similar argument shows the converse implication, and the proof is complete.

The subsequent lemma says that under certain conditions the monomial multiple of a monomial ideal which has a superficial ideal, always has a superficial ideal. Lemma 3.2.14. Let I and J be two monomial ideals of R with G(J) = {v 1 , . . . , v t }. Let also h be a monomial in R such that gcd(h, v i ) = 1 for each i = 1, . . . , t. Then J is a superficial ideal for I if and only if hJ is a superficial ideal for hI.

Proof. To simplify our notation, we set L := hI. First, let J be a superficial ideal for I. Fix k ∈ N. Thanks to G(hJ) ⊆ G(L), for completing the proof, one has to demonstrate (L k+1 : R hJ) = L k . Due to (h k+1 I k+1 : R h) = h k I k+1 , one derives (L k+1 : R hJ) = (h k I k+1 : R J). By virtue of gcd(h, v i ) = 1 for each i = 1, . . . , t, this gives that (h k I k+1 : R J) = h k (I k+1 : R J). It follows from (I k+1 : R J) = I k that (L k+1 : R hJ) = L k , as required. The converse implication is straightforward. This completes the proof.

In order to prove Theorem 3.2.16, we need to verify Lemma 3.2.15. To do this, we should recall the definition of the monomial localization of a monomial ideal with respect to a monomial prime ideal as has been stated in Definition 3.1.14. Lemma 3.2.15. Let I and J be two monomial ideals in R = K[x 1 , . . . , x n ], and p be a monomial prime ideal of R. Then the following statements hold. (vi) If Q is a q-primary monomial ideal in R with q ⊆ p, then Q(p) is a q-primary monomial ideal in R(p);

(vii) If p ∈ V * (I), and

I = Q 1 ∩ • • • ∩ Q s is
a minimal primary decomposition of I, then To conclude the argument, one has to demonstrate u(p) : R(p) v(p) = (u : R v)(p) for all u ∈ G(I) and v ∈ G(J). Here, the assertion follows directly from the fact that gcd(u, v)(p) = gcd(u(p), v(p)) for all u ∈ G(I) and v ∈ G(J).

I(p) = i=1,...,s, √ Q i ⊆p Q i , is a minimal primary decomposition of I(p) in R(p); (viii) If p ∈ V * (I), then Ass R(p) (R(p)/I(p)) = {q : q ∈ Ass R (R/I) and q ⊆ p}; (ix) If p ∈ V * (I), then Ass ∞ (I(p)) = {q : q ∈ Ass ∞ (I) and q ⊆ p}
(v) Suppose that G(I) = {u 1 , . . . , u m }. Applying statement (i) implies the subsequent equalities

I(p) = ( m i=1 u i R)(p) = m i=1 u i (p)R(p) = m i=1 u i (p)R(p) = m i=1 u i (p)R(p) = m i=1 √ u i (p)R(p).
On the other hand, since √ I is generated by { √ u i : i = 1, . . . , m}, using statement (i) yields the following equalities

√ I(p) = ( m i=1 √ u i R)(p) = m i=1 √ u i (p)R(p).
Therefore, one can conclude that I(p) = √ I(p). (vi) Let Q be a q-primary monomial ideal in R with q ⊆ p. Assume that q = (x i 1 , . . . , x it ). We thus get Q = (x α 1 i 1 , . . . , x αt it ) with α 1 , . . . , α t are some positive integers. On account of q ⊆ p, one has Q(p) = (x α 1 i 1 , . . . , x αt it ), Q(p) = R(p), and q(p) = (x i 1 , . . . , x it ). Accordingly, one derives Q(p) is a q-primary monomial ideal in R(p).

(vii) Let p ∈ V * (I), and

I = Q 1 ∩ • • • ∩ Q s be a minimal primary decomposition of I. Since p ∈ V * (I), one has s i=1 Q i ⊆ p.
Hence, there exists a positive integer

1 ≤ i ≤ s such that Q i ⊆ p,
I(p) = Q 1 (p) ∩ • • • ∩ Q t (p) is a minimal primary decomposition of I(p).
(viii) This claim follows directly from statement (vii).

(ix) Let p ∈ V * (I). Thanks to statement (ii), one has I ℓ (p) = (I(p)) ℓ for all positive integers ℓ. Furthermore, statement (viii) gives the following equality Ass R(p) (R(p)/I ℓ (p)) = {q : q ∈ Ass R (R/I ℓ ) and q ⊆ p}.

We thus gain

Ass R(p) (R(p)/(I(p)) ℓ ) = {q : q ∈ Ass R (R/I ℓ ) and q ⊆ p}.

Therefore, one derives Ass ∞ (I(p)) = {q : q ∈ Ass ∞ (I) and q ⊆ p}.

(x) This assertion follows readily from statement (ix). (ii) Let I have the persistence property, and k be an arbitrary positive integer. Consider a ∈ Ass R(p) (R(p)/(I(p)) k ). Since (I(p)) k = I k (p), it follows from part (viii) of Lemma 3.2.15 that a = q(p) with q ∈ Ass R (R/I k ) and q ⊆ p. Due to I has the persistence property, one can conclude that q ∈ Ass R (R/I k+1 ). Once again, by (I(p)) k+1 = I k+1 (p) and part (viii) of Lemma 3.2.15, we obtain q ∈ Ass R(p) (R(p)/(I(p)) k+1 ). Accordingly, one derives I(p) has the persistence property, as desired.

It should be noted that Theorem 3.2.16 allows us to refute the (strong) persistence property for some monomial ideals, and helps us to construct new counterexamples inspired by some well-known counterexamples. Corollary 3.2.17. Let I be a monomial ideal in R 1 = K[x 1 , . . . , x r ] with G(I) = {u 1 , . . . , u m }, and J 1 , . . . , J m be some monomial ideals in R 2 = K[x r+1 , . . . , x n ]. If I does not satisfy the persistence (respectively, strong persistence) property, then

L := u 1 J 1 R + • • • + u m J m R does not satisfy the persistence (respectively, strong persistence) property, where R = K[x 1 , . . . , x n ].
Proof. Suppose, on the contrary, that L has the persistence (respectively, strong persistence) property. Put p := ( m i=1 supp(u i )). It is clear that p is a monomial prime ideal in R, and L ⊆ p. Thanks to Theorem 3.2.16, one derives L(p) has the persistence (respectively, strong persistence) property. Since L(p) = I, we get a contradition. Therefore, L does not satisfy the persistence (respectively, strong persistence) property, as claimed.

Example 3.2.18. (1) Consider the monomial ideal I generated by x 1 x 2 x 3 , x 1 x 2 x 4 , x 1 x 3 x 5 , x 1 x 4 x 6 , x 1 x 5 x 6 , x 2 x 3 x 6 , x 2 x 4 x 5 , x 2 x 5 x 6 , x 3 x 4 x 5 , x 3 x 4 x 6 in the polynomial ring R 1 = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ] over a field K, and the monomial ideal L generated by x 1 x 2 x 3 x 8 , x 1 x 2 x 4 , x 1 x 3 x 5 x 7 , x 1 x 4 x 6 , x 1 x 5 x 6 , x 2 x 3 x 6 x 7 , x 2 x 4 x 5 , x 2 x 5 x 6 x 8 , x 3 x 4 x 5 , x 3 x 4 x 6 x 7 in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 , x 8 ]. Now, set p := (x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ). Certainly, p ∈ V * (L) and L(p) = I. By virtue of (I 3 : R I) = I 2 , one can conclude that the monomial ideal I does not satisfy the strong persistence property. Finally, based on Corollary 3.2.17, we deduce that the monomial ideal L does not satisfy the strong persistence property. On the other hand, direct computations show that (L 3 : R L) = L 2 which confirm our result.

(2) Suppose that

I := (x 1 x 2 2 x 3 , x 2 x 2 3 x 4 , x 3 x 2 4 x 5 , x 4 x 2 5 x 1 , x 5 x 2 1 x 2 )
and

L := (x 1 x 2 2 x 3 x 7 , x 2 x 2 3 x 4 x 6 , x 3 x 2 4 x 5 x 6 , x 4 x 2 5 x 1 x 7 , x 5 x 2 1 x 2 x 8 ), be two monomial ideals in R = K[x 1 , x 2 ,
(x 1 , x 2 , x 3 , x 4 , x 5 ) ∈ Ass R (R/L) \ Ass R (R/L 2 ), as required.
To state the subsequent results, we need the definitions of the contraction and deletion operators as given in [50, P. 303]. Definition 3.2.19. Let I be a monomial ideal in R = K[x 1 , . . . , x n ] with G(I) = {u 1 , . . . , u m }. For some 1 ≤ j ≤ n, the contraction I/x j (respectively, deletion I \ x j ) is obtained by setting x j = 1 in u i (respectively, x j = 0 in u i ) for each i = 1, . . . , m. It should be observed that the contraction I/x j is exactly the monomial localization of I with respect to p = m \ {x j }, where m = (x 1 , . . . ,

x n ) is the graded maximal ideal of R = K[x 1 , . . . , x n ].
Inspired by Lemma 3.2.15 and Theorem 3.2.16, one can conclude Lemma 3.2.20 and Theorem 3.2.21. Lemma 3.2.20. Let I and J be two monomial ideals in R = K[x 1 , . . . , x n ] over a field K, m = (x 1 , . . . , x n ), and 1 ≤ j ≤ n. Then the following statements hold.

(i) (I + J)/x j = I/x j + J/x j ;

(ii) (IJ)/x j = (I/x j )(J/x j );

(iii) (I ∩ J)/x j = (I/x j ) ∩ (J/x j );

(iv) (I : R J)/x j = (I/x j : R/x j J/x j );

(v) I/x j = √ I/x j ;
(vi) If Q is a q-primary monomial ideal in R with x j / ∈ q, then Q/x j is a q-primary monomial ideal in R/x j ;

(vii) If I ⊆ m \ {x j }, and

I = Q 1 ∩ • • • ∩ Q s is a minimal primary decomposition of I, then I/x j = i=1,...,s,x j / ∈ √ Q i Q i , is a minimal primary decomposition of I/x j in R/x j ; (viii) If I ⊆ m \ {x j }, then
Ass R/x j ((R/x j )/(I/x j )) = {q : q ∈ Ass R (R/I) and x j / ∈ q};

(ix) If I ⊆ m \ {x j }, then Ass ∞ (I/x j ) = {q : q ∈ Ass ∞ (I) and x j / ∈ q}, where Ass ∞ (I) denotes the stable set of associated prime ideals of I;

(x) If I ⊆ m\{x j }, then astab(I) ≥ astab(I/x j ), where astab(I) denotes the index of stability for the associated prime ideals of I.

Theorem 3.2.21. Let I be a monomial ideal in R = K[x 1 , . . . , x n ], and 1 ≤ j ≤ n.

Then the following statements hold.

(i) If I has the strong persistence property, then I/x j has the strong persistence property. In particular, I/x j has a superficial ideal.

(ii) If I has the persistence property, then I/x j has the persistence property.

To express Lemma 3.2.22 and Question 3.2.23, we must employ the definition of the deletion operator as given in Definition 3.2.19. Lemma 3.2.22. Let I and J be two monomial ideals in R = K[x 1 , . . . , x n ] over a field K, and 1 ≤ j ≤ n. Then the following statements hold.

(i) (I + J) \ x j = I \ x j + J \ x j ; (ii) (IJ) \ x j = (I \ x j )(J \ x j ); (iii) (I ∩ J) \ x j = (I \ x j ) ∩ (J \ x j ); (iv) I \ x j = √ I \ x j ; (v) If Q is a q-primary monomial ideal in R, then Q \ x j is a (q \ x j )-primary monomial ideal in R \ x j ; (vi) If I = Q 1 ∩ • • • ∩ Q s is a minimal primary decomposition of I in R, then I \ x j = (Q 1 \ x j ) ∩ • • • ∩ (Q s \ x j ) is a primary decomposition of I \ x j in R \ x j ; (vii) Ass R\x j ((R \ x j )/(I \ x j )) ⊆ {q \ x j : q ∈ Ass R (R/I)}.
Proof. (i) It is routine to ckeck that if x j |u (respectively, x j |v) for some u ∈ G(I)

(respectively, v ∈ G(J)), then u / ∈ G(I \ x j ) (respectively, v / ∈ G(J \ x j )) and u / ∈ G((I + J) \ x j ) (respectively, v / ∈ G((I + J) \ x j ))
. Also, if x j ∤ uv, then u \ x j = u and v \ x j = v. Since I \ x j (respectively, J \ x j ) is generated by {u : u ∈ G(I) and x j ∤ u} (respectively, {v : v ∈ G(J) and x j ∤ v}), and because

G((I + J) \ x j ) ⊆ G(I \ x j ) ∪ G(J \ x j ), we get (I + J) \ x j = I \ x j + J \ x j .
(ii) One can easily see that (uv)\x j = (u\x j )(v\x j ) for all u ∈ G(I) and v ∈ G(J).

Since G((IJ) \ x j ) ⊆ G(I \ x j )G(J \ x j ), we obtain (IJ) \ x j = (I \ x j )(J \ x j ).
(iii) By using the convention that if x j |u or x j |v, then lcm(u \ x j , v \ x j ) = 0, one can conclude that (I \ x j ) ∩ (J \ x j ) is generated by

{lcm(u \ x j , v \ x j ) : for all u ∈ G(I) and v ∈ G(J)}.
Furthermore, we know that (I ∩ J) \ x j is generated by

{lcm(u, v) \ x j : for all u ∈ G(I) and v ∈ G(J)}.
Based on lcm(u, v) \ x j = lcm(u \ x j , v \ x j ) for all u ∈ G(I) and v ∈ G(J), we gain

(I ∩ J) \ x j = (I \ x j ) ∩ (J \ x j ).
(iv) Assume that G(I) = {u 1 , . . . , u m }. Using statement (i) yields the subsequent equalities

I \ x j = ( m i=1 u i R) \ x j = m i=1 (u i \ x j )(R \ x j ) = m i=1 u i \ x j (R \ x j ) = m i=1 u i \ x j (R \ x j ) = m i=1 ( √ u i \ x j )(R \ x j ).
In addition, by virtue of √ I is generated by { √ u i : i = 1, . . . , m}, applying statement (i) gives the following equalities

√ I \ x j = ( m i=1 √ u i R) \ x j = m i=1 ( √ u i \ x j )(R \ x j ).
Consequently, we get

I \ x j = √ I \ x j . (v) Suppose that Q is a q-primary monomial ideal in R. Let q = (x i 1 , . . . , x it ). This implies that Q = (x α 1 i 1 , . . . , x αt it ) with α 1 , . . . , α t are some positive integers. If x j / ∈ q, then Q \ x j = Q and q \ x j = q. Accordingly, Q \ x j is a (q \ x j )-primary monomial ideal in R \ x j . If x j ∈ q, then there exists a positive integer 1 ≤ s ≤ t such that i s = j. This yields that Q \ x j = (x α 1 i 1 , . . . , x α s-1 i s-1 , x α s+1 i s+1 , . . . , x αt it ) and q \ x j = (x i 1 , . . . , x i s-1 , x i s+1 , . . . , x it ). Therefore, one can conclude that Q \ x j is a (q \ x j )-primary monomial ideal in R \ x j . (vi) Suppose that I = Q 1 ∩ • • • ∩ Q s is a minimal primary decomposition of I. It follows from statements (iii) and (v) that I \ x j = (Q 1 \ x j ) ∩ • • • ∩ (Q s \ x j ) is a primary decomposition of I \ x j in R \ x j .
(vii) In the light of statement (vi), one can immediately deduce this assertion.

Here, we continue with the following question. We will argue on this question in Question 10.2.15.

Question 3.2.23. Suppose that

I ⊂ R = K[x 1 , . . . , x n ] is a square-free monomial ideal with G(I) = {u 1 , . . . , u m }, m i=1 supp(u i ) = {x 1 , . . . , x n }, and m = (x 1 , . . . , x n ) is the graded maximal ideal of R. If there exists a positive integer 1 ≤ j ≤ n such that m \ x j ∈ Ass R\x j ((R \ x j )/(I \ x j ) k ) for some positive integer k, then can we deduce that m ∈ Ass R (R/I k )?
Remarks 3.2.24. (1) We note that the converse of Question 3.2.23 may be false. For example, suppose that C 5 denotes the cycle graph on the vertex set V (C 5 ) = {1, 2, 3, 4, 5}, and the edge set

E(C 5 ) = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 1}}. Let a := J(C 5 ) = (x 1 , x 2 ) ∩ (x 2 , x 3 ) ∩ (x 3 , x 4 ) ∩ (x 4 , x 5 ) ∩ (x 5 , x 1 ),
be the cover ideal associated to C 5 in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 ] over a field K. Hence, we get a = (x 2 x 4 x 5 , x 2 x 3 x 5 , x 1 x 3 x 5 , x 1 x 3 x 4 , x 1 x 2 x 4 ). Now, one can easily see that m = (x 1 , x 2 , x 3 , x 4 , x 5 ) ∈ Ass R (R/a 2 ). Here, direct computations yield the subsequent table:

a \ x 1 = (x 2 x 4 x 5 , x 2 x 3 x 5 ) Ass R\x 1 ((R \ x 1 )/(a \ x 1 ) 2 ) = {(x 2 ), (x 5 ), (x 3 , x 4 )} a \ x 2 = (x 1 x 3 x 5 , x 1 x 3 x 4 ) Ass R\x 2 ((R \ x 2 )/(a \ x 2 ) 2 ) = {(x 3 ), (x 1 ), (x 5 , x 4 )} a \ x 3 = (x 2 x 4 x 5 , x 1 x 2 x 4 ) Ass R\x 3 ((R \ x 3 )/(a \ x 3 ) 2 ) = {(x 4 ), (x 2 ), (x 5 , x 1 )} a \ x 4 = (x 2 x 3 x 5 , x 1 x 3 x 5 ) Ass R\x 4 ((R \ x 4 )/(a \ x 4 ) 2 ) = {(x 3 ), (x 5 ), (x 2 , x 1 )} a \ x 5 = (x 1 x 3 x 4 , x 1 x 2 x 4 ) Ass R\x 5 ((R \ x 5 )/(a \ x 5 ) 2 ) = {(x 4 ), (x 1 ), (x 3 , x 2 )} For example, for a \ x 1 = (x 2 x 4 x 5 , x 2 x 3 x 5 ) in R \ x 1 = K[x 2 , x 3 , x 4 , x 5 ], one obtains a \ x 1 = (x 2 ) ∩ (x 5 ) ∩ (x 4 , x 3 ) = (x 2 )(x 5 )(x 4 , x 3 )

, and inspired by Proposition 2.1.8, we gain

Ass R\x 1 ((R \ x 1 )/(a \ x 1 ) t ) = Ass R\x 1 ((R \ x 1 )/(a \ x 1 )),
for all positive integers t. A similar argument shows that

Ass R\x j ((R \ x j )/(a \ x j ) t ) = Ass R\x j ((R \ x j )/(a \ x j )),
for all positive integers t and for all j = 2, 3, 4, 5. This implies that

m \ x j / ∈ Ass R\x j ((R \ x j )/(a \ x j ) t ),
for all positive integers t and for all j = 1, 2, 3, 4, 5.

(2) Question 3.2.23 can fail if we drop the word square-free. As an example, consider the monomial ideal

I = (x 1 x 2 2 x 3 , x 2 x 2 3 x 4 , x 3 x 2 4 x 5 , x 4 x 2 5 x 1 , x 5 x 2 1 x 2 , x 1 x 2 3 x 6 ) in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ] over a field K. It is routine to check that m \ x 6 = (x 1 , x 2 , x 3 , x 4 , x 5 ) ∈ Ass R\x 6 ((R \ x 6 )/(I 4 \ x 6 )), while m = (x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) / ∈ Ass R (R/I 4 ).

Examples of monomial superficial ideals

In the following, we introduce some classes of monomial ideals in which every ideal has a superficial ideal. To do this, one has to recall the definition of centipede graphs from Definition 2.2.18.

In view of [START_REF] Nasernejad | Asymptotic behaviour of associated primes of monomial ideals with combinatorial applications[END_REF], one can provide the following result which gives that the path ideals generated by paths of length two in centipede graphs have superficial ideals. Theorem 3.2.25. Let W n , for some n ∈ N with n ≥ 2, be a centipede graph with corresponding path ideal I 2 (W n ), i.e., the path ideal generated by paths of length two. Then I 2 (W n ) has a superficial ideal. Proof. To simplify our notation, one may consider the following centipede graph on the vertex set V (W n ) = {1, 2, 3, . . . , 2n -2, 2n -1, 2n}, and also set

L := I 2 (W n ). 2 4 6 2n -2 2n 1 3 5 2n -3 2n -1 Put g := x 1 x 2 x 3 . Certainly, g ∈ G(L). On account of the proof of Theorem 2.2.19, one derives (L k+1 : R g) = L k for all k ∈ N, where R = K[x 1 , . . . , x 2n ]. It follows directly that gR is a superficial ideal for L, as required.
Here, we focus on two classes of monomial ideals which were introduced in Section 2.3, see Definitions 2.3.1 and 2.3.3. In fact, our aim is to verify that every ideal in these two classes has a superficial ideal. We are now going to show that every unisplit monomial ideal has a superficial ideal.

Theorem 3.2.26. Every unisplit monomial ideal has a superficial ideal.

Proof. Let I be a unisplit monomial ideal of R with G(I) = {u 1 , . . . , u m }. Without loss of generality, assume that gcd(u 1 , u i ) = 1 for each i = 2, . . . , m. By virtue of (u k+1 1 : R u 1 ) = u k 1 for all k, one can conclude that J := u 1 R is a superficial ideal for 79 ideal for I, i.e., (I s+1 : R u 1 ) = I s for all s, as desired.

The next theorem illustrates every separable monomial ideal has a superficial ideal. In this theorem, we see an application of Theorem 3.2.26. Theorem 3.2.27. Every separable monomial ideal has a superficial ideal.

Proof. Assume that I is a separable monomial ideal of R with G(I) = {u 1 , . . . , u m }. Without loss of generality, assume there are monomials f and w in R such that w = 1, u 1 = wf , gcd(w, f ) = 1, and gcd(u 1 , u i ) = w for each i = 2, . . . , m. This gives that there exist monomials h 2 , . . . , h m in R such that u i = wh i for each i = 2, . . . , m. Accordingly, one can deduce that

I = w(f R + m i=2 h i R). Put L := f R + m i=2 h i R.
This implies that L is a unisplit monomial ideal, and Theorem 3.2.26 yields that

(L k+1 : R f ) = L k for all k.
Here, our goal is to demonstrate (I k+1 : R wf ) = I k for all k. We thus have the following equalities

(I k+1 : R wf ) = (w k+1 L k+1 : R wf ) = (w k L k+1 : R f ) = w k (L k+1 : R f ) = w k L k = I k .
Therefore, J := u 1 R is a superficial ideal for I, as required.

As another application of Theorems 3.2.9 and 3.2.26, we provide the following theorem. To see this, one needs the proposition below. Proposition 3.2.28. Let R = K[x 1 , . . . , x n ] be a polynomial ring over a field K, v be a square-free monomial in R, and u 1 , . . . , u m be some monomials in R. Then, for all positive integers α 1 , . . . , α m , the following statement holds

(u α 1 1 • • • u αm m : R v) = u α 1 -1 1 • • • u αm-1 m (u 1 • • • u m : R v).
Proof. Let α 1 , . . . , α m be some arbitrary positive integers. By virtue of v is a squarefree monomial in R, one can easily see that

gcd(u α 1 1 • • • u αm m , v) = gcd(u 1 • • • u m , v).
We thus gain the following equalities

(u α 1 1 • • • u αm m : R v) = u α 1 1 • • • u αm m gcd(u α 1 1 • • • u αm m , v) = u α 1 1 • • • u αm m gcd(u 1 • • • u m , v) = u α 1 -1 1 • • • u αm-1 m u 1 • • • u m gcd(u 1 • • • u m , v) = u α 1 -1 1 • • • u αm-1 m (u 1 • • • u m : R v).
This completes the proof.

Theorem 3.2.29. Suppose that I is a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K with G(I) = {u 1 , . . . , u m } such that, for each i = 2, . . . , m -1, gcd(u 1 , u 2 , u i+1 ) = 1 and (u i : R u 1 ) divides (u i+1 : R u 1 ). Then I has a superficial ideal. Proof. Fix k ∈ N. If m = 1, then G(I) = {u 1 }. Since u 1 is a non-zerodivisor on R, one has (u k+1 1 R : R u 1 R) = u k 1 R, and so (I k+1 : R I) = I k . Hence, the claim is true for the case m = 1. If m = 2, then G(I) = {u 1 , u 2 }. It follows from the binomial expansion that I k+1 = k+1 ℓ=0 u k+1-ℓ 1 u ℓ 2 R
. This implies the subsequent equalities

(I k+1 : R u 1 ) = ( k+1 ℓ=0 u k+1-ℓ 1 u ℓ 2 R : R u 1 ) = k+1 ℓ=0 (u k+1-ℓ 1 u ℓ 2 R : R u 1 ) = k ℓ=0 (u k+1-ℓ 1 u ℓ 2 R : R u 1 ) + (u k+1 2 : R u 1 )R = k ℓ=0 u k-ℓ 1 u ℓ 2 R + (u k+1 2 : R u 1 )R.
Furthermore, Proposition 3.2.28 yields that (u k+1

2 : R u 1 ) = u k 2 (u 2 : R u 1 ). Due to u k-ℓ 1 u ℓ 2 ∈ I k for all ℓ = 0, . . . , k
, and by virtue of u k 2 (u 2 : R u 1 ) ∈ I k , we deduce that (I k+1 : R u 1 ) ⊆ I k . As I k ⊆ (I k+1 : R u 1 ), one has (I k+1 : R u 1 ) = I k . Thus, the assertion holds for the case m = 2. So let m ≥ 3. To simplify our notation, we take M i := gcd(u i+1 , u 1 ) and N i := (u i+1 : R u 1 ) for all i = 1, . . . , m -1. This gives that u i+1 = M i N i for all i = 1, . . . , m -1. On the other hand, the hypothesis implies that N i |N i+1 for all i = 1, . . . , m -2, and thus N 1 |N j for all j = 2, . . . , m -1. Hence,

N j = N 1 H j for some monomial H j in R for all j = 2, . . . , m -1. Accordingly, one can write I = (u 1 , N 1 M 1 , N 1 H 2 M 2 , . . . , N 1 H m-1 M m-1
). This gives rise to the equation below

I = N 1 (M 1 , H 2 M 2 , . . . , H m-1 M m-1 ) + u 1 R. Here, set L := (M 1 , H 2 M 2 , . . . , H m-1 M m-1 ). Therefore, one has I = N 1 L + u 1 R.
Because M 1 = gcd(u 2 , u 1 ), we have M 1 |u 1 , and so u 1 ∈ L. Also, by virtue of u i , for all i = 1, . . . , m, is a square-free monomial, one can conclude that gcd(N 1 , u

1 ) = 1, gcd(N 1 , M 1 ) = 1, and gcd(N 1 , H i M i ) = 1 for all i = 2, . . . , m -1. On account of gcd(u 1 , u 2 , u i+1 ) = gcd(gcd(u 1 , u 2 ), u i+1 ), for all i = 2, . . . , m -1, we get gcd(M 1 , u i+1 ) = 1 for all i = 2, . . . , m -1. This gives that gcd(M 1 , M i N i ) = 1 for all i = 2, . . . , m -1, and hence gcd(M 1 , M i H i ) = 1 for all i = 2, . . . , m -1.
This yields that L is a unisplit monomial ideal. Thanks to Theorem 3.2.26, one derives that the monomial ideal M 1 R is a superficial ideal for L. It follows now from Theorem 3.2.9 that

N 1 M 1 R + u 1 R = (u 1 , u 2 )R is a superficial ideal for I, as required. Corollary 3.2.30. Suppose that I is a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K with G(I) = {u 1 , . . . , u m } such that, for each i = 2, . . . , m -1, gcd(u 1 , u 2 , u i+1 ) = 1 and lcm(u 1 , u i ) divides lcm(u 1 , u i+1 ). Then I has a superficial ideal. Proof. We know that (u i : R u 1 ) = u i /gcd(u i , u 1 ) and u i u 1 = lcm(u i , u 1 )gcd(u i , u 1 )
for each i = 1, . . . , m. The assertion follows readily from Theorem 3.2.29. We next give an example which illustrates Theorem 3.2.29.

Example 3.2.31. Consider the following monomial ideal

I = (x 1 x 2 x 3 x 7 x 8 , x 1 x 2 x 4 , x 3 x 4 x 5 x 8 , x 3 x 4 x 5 x 6 x 7 , x 4 x 5 x 6 x 7 x 8 ), in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 , x 8 ] over a field K. Now, set u 1 := x 1 x 2 x 3 x 7 x 8 , u 2 := x 1 x 2 x 4 , u 3 := x 3 x 4 x 5 x 8 , u 4 := x 3 x 4 x 5 x 6 x 7
, and u 5 := x 4 x 5 x 6 x 7 x 8 . It is routine to check that, for each i = 2, 3, 4, gcd(u 1 , u 2 , u i+1 ) = 1 and (u i : R u 1 ) divides (u i+1 : R u 1 ). Now, Theorem 3.2.29 yields that the monomial ideal

(u 1 , u 2 ) = (x 1 x 2 x 3 x 7 x 8 , x 1 x 2 x 4 ) is a superficial ideal for I.
We close this argument by giving two results which show that for an arbitrary monomial ideal I in a polynomial ring R = K[x 1 , . . . , x n ], several superficial ideals can be constructed.

Proposition 3.2.32. Suppose that G = (V (G), E(G)) is a graph, I = I(G) the edge ideal associated to G, v a pendant of G, and f is the unique neighbor of v. Then vf R is a superficial ideal for I, where R = K[x t : t ∈ V (G)].
Proof. This assertion follows readily from [94, Lemma 2.10].

Proposition 3.2.33. Suppose G = (V (G), E(G)) is a bipartite graph, J = J(G) the cover ideal associated to G, and V (G) = U ∪ W is a bipartition for the V (G).

Put u := α∈U x α and w := β∈W x β . Then vR and wR are superficial ideals for

J, where R = K[x t : t ∈ V (G)].
Proof. In the light of [START_REF] Seyed Fakhari | Depth, Stanley depth, and regularity of ideals associated to graphs[END_REF]Lemma 3.2], one can immediately deduce this claim.

A criterion for detecting the non-normality

In this subsection, we present an elegant and useful criterion for detecting the nonnormality of monomial ideals. We express it in the following theorem.

Theorem 3.2.34. Every normal monomial ideal has the strong persistence property.

Proof. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K with G(I) = {u 1 , . . . , u m }. Without loss of generality, one may assume that x 1 ∈ m i=1 supp(u i ). After relabeling, we have

u i = x 1 α i g i with α i ∈ N for all i = 1, .
. . , t, and gcd(u i , x 1 ) = 1 for all i = t + 1, . . . , m. To complete the proof, one has to prove that (I k+1 : R I) = I k for all k ∈ N. To achieve this, fix k ∈ N, and set L := (I : R x 1 ). We thus gain the following consequences The following corollary allows us to refute the normality of a monomial ideal based on the persistence property.

(I k+1 : R I) = (I k+1 : R (u 1 , . . . , u m )) = (I k+1 : R x 1 (x α 1 -1 1 g 1 , . . . , x αt-1 1 g t ) + (u t+1 , . . . , u m )) ⊆ (I k+1 : R x 1 (x α 1 -1 1 g 1 , . . . , x αt-1 1 g t , u t+1 , . . . , u m )) = ((I k+1 : R x 1 ) : R (x α 1 -1 1 g 1 , . . . , x αt-1 1 g t , u t+1 , . . . , u m )) = (I k (I : R x 1 ) : R (x α 1 -1 1 g 1 , . . . , x αt-1 1 g t , u t+1 , . . . , u m )) = (I k (I :

Corollary 3.2.35.

Let I be a monomial ideal in R = K[x 1 , . . . , x n ] over a field K which does not satisfy the persistence property. Then I is non-normal.

Corollary 3.2.36. Every normal monomial ideal has a superficial ideal.

We now present the following question which is devoted to the superficiality of monomial ideals under polarization operator. To do this, we recall the definition of polarization. Definition 3.2.37. [START_REF] Hà | Embedded associated primes of powers of square-free monomial ideals[END_REF]Definition 4.1]) The process of polarization replaces a power x t i by a product of t new variables x (i,1) • • • x (i,t) . We call x (i,j) a shadow of x i . We will use I t to denote the polarization of I t , will use S t for the new polynomial ring in this polarization, and will use w to denote the polarization in S t of a monomial

w in R = K[x 1 , . . . , x n ].
The depolarization of an ideal in S t is formed by setting x (i,j) = x i for all i, j.

Question 3.2.38. Let I and J be two monomial ideals in

R = K[x 1 , . . . , x n ]. If J is a superficial ideal for I, then J is a superficial ideal for I?
As an application of Corollary 3.2.36, we show that Question 3.2.38 can fail in general. To do this, we need Lemma 3.2.39 and Theorem 3.2.40 to provide a counterexample. Lemma 3.2.39. Let R = K[x 1 , . . . , x 2n ] be a polynomial ring over a field K, and I be a square-free monomial ideal in R with

G(I) = {x 2i-3 x 2i-1 x 2i x 2i+1 : i = 1, . . . , n}, where 2n + 1 = 1. Then, for all positive integers t, x ℓ 1 1 • • • x ℓ 2n 2n ∈ G(I t ) if and only if (1) n i=1 ℓ 2i = t and (2) ℓ 2i+1 = ℓ 2i + ℓ 2i+2 + ℓ 2i+4 for each i = 1, . . . , n.
Proof. (=⇒) Set u 2i := x 2i-3 x 2i-1 x 2i x 2i+1 for each i = 1, . . . , n. Clearly, our claim is true for the case t = 1, because of u 2i satisfies the statement for each i = 1, . . . , n. For the same reason, the proof follows directly by induction.

(⇐=) Any square-free monomial in R that satisfies the sufficient condition, must clearly be of the form u 2i for some 1 ≤ i ≤ n. Hence, the statement is true for the case t = 1. Follow by induction. Assume that

n i=1 ℓ 2i = t + 1 and ℓ 2i+1 = ℓ 2i + ℓ 2i+2 + ℓ 2i+4 for each i = 1, . . . , n, where 2n + 1 = 1. Choose i 0 such that x 2i 0 +2 |x ℓ 1 1 • • • x ℓ 2n 2n for some nonnegative integers ℓ 1 , . . . , ℓ 2n . Put f := x ℓ 1 1 • • • x ℓ 2n 2n . Then, by (2), we must have each of x 2i 0 -1 , x 2i 0 +1 , x 2i 0 +3 divides f ; that is, u 2i 0 +2 |f . Write f = u 2i 0 +2 f ′ . Let f ′ = x ℓ ′ 1 1 • • • x ℓ ′ 2n 2n for some nonnegative integers ℓ ′ 1 , . . . , ℓ ′ 2n .
Then it must be clear that (1) and (2) hold for the ℓ ′ 's, and so f ′ ∈ G(I t ). Therefore, f ∈ G(I t+1 ). This completes the inductive step, and so the claim has been proved by induction. Theorem 3.2.40. Let R = K[x 1 , . . . , x 2n ] be a polynomial ring over a field K, and I be a square-free monomial ideal in R with

G(I) = {x 2i-3 x 2i-1 x 2i x 2i+1 : i = 1, . . . , n}, where 2n + 1 = 1. Then I is a normal ideal.
Proof. Our strategy is to use [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.2]. For this purpose, assume that m ∈ I t for some positive integer t. Let α ∈ Z 2n be the tuple of exponents of m. Also, let {µ i : i = 1, . . . , r} be the tuples of exponents of the minimal generators of [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.4.3]). On account of {µ i : i = 1, . . . , r} is the set of tuples of exponents of the minimal generators of I t , Lemma 3.2.39 implies the following equalities ( * )

I t . Suppose that µ ij is the j-th component of µ i . Since m ∈ I t , one can conclude that there exist λ 1 , . . . , λ r ∈ Q ≥0 with r i=1 λ i = 1 such that α = λ 1 µ 1 + • • • + λ r µ r (see
n i=1 α 2i = n i=1 r j=1 λ j µ j,2i = r j=1 λ j t = t.
Furthermore, for each i = 1, . . . , n, applying Lemma 3.2.39 derives that

α 2i+1 = r j=1 λ j µ j,2i+1 = r j=1 λ j (µ j,2i + µ j,2i+2 + µ j,2i+4 ) = α 2i + α 2i+2 + α 2i+4 . ( * * )
We can now combine ( * ) and ( * * ) to obtain m ∈ I t , as claimed.

We are now in a position to state the following counterexample.

Example 3.2.41. Let

I = (x 1 x 2 2 x 3 , x 2 x 2 3 x 4 , x 3 x 2 4 x 5 , x 4 x 2 5 x 1 , x 5 x 2 1 x 2 ) be a monomial ideal in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5
] over a field K. Suppose, on the contrary, that I has a superficial ideal. This implies that I satisfies the persistence property, while one can easily check that

m = (x 1 , x 2 , x 3 , x 4 , x 5 ) ∈ Ass R (R/I) \ Ass R (R/I 2 ).
This gives rise to a contradiction. Consequently, the monomial ideal I has no superficial ideal. We now explore the superficiality of the polarization of I. It is routine to compute that

I = (x

K.

In order to simplify the notation, we set L := I, z 2k-1 := x k1 and z 2k := x k2 for all k = 1, . . . , 5. Accordingly, one obtains

L = (z 1 z 3 z 4 z 5 , z 3 z 5 z 6 z 7 , z 5 z 7 z 8 z 9 , z 1 z 7 z 9 z 10 , z 1 z 2 z 3 z 9 ), in the polynomial ring S = K[z 1 , z 2 , z 3 , z 4 , z 5 , z 6 , z 7 , z 8 , z 9 , z 10 ].
Our aim is to verify that L has a superficial ideal. To establish this claim, we use Theorem 3.2.40 and Corollary 3.2.36. Indeed, it follows from Theorem 3.2.40 that L is a normal ideal, and Corollary 3.2.36 yields that L has a superficial ideal. Therefore, I has a superficial ideal, while I has no superficial ideal, as desired.

In the next theorem, we probe the relation between normally torsion-freeness and superficiality. Theorem 3.2.42. Every normally torsion-free square-free monomial ideal has the strong persistence property.

Proof. Assume that I is a normally torsion-free square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. It follows from the assumption that Min(I) = Ass R (R/I) = Ass R (R/I k ) for all k. Since I t ⊆ (I t+1 : R I) for all t, we need to show that (I t+1 : R I) ⊆ I t for all t. Fix t ∈ N. Due to [START_REF] Matsumura | Commutative Ring Theory[END_REF]Exercise 6.4], it is sufficient to demonstrate that (I t+1 a : Ra I a ) = I t a for every a ∈ Ass R (R/I t ). To do this, consider an arbitrary element a ∈ Ass R (R/I t ). Hence, a ∈ Min(I). We next prove that, for every b ∈ Ass R (R/I) \ {a}, one has a / ∈ Supp R (R/b). On the contrary, assume that there exists b

∈ Ass R (R/I) \ {a} such that a ∈ Supp R (R/b). As Supp R (R/b) = V (Ann R (R/b)), this implies that Ann R (R/b) ⊆ a, and so b ⊆ a.
By virtue of a ∈ Min(I), one derives b = a. This leads to a contradiction. This yields that b a = R a for every b ∈ Ass R (R/I) \ {a}. Thus, I a = a a , and so (I t ) a = (a t ) a . In addition, we know that (a t+1 : R a) = a t . This implies that (a t+1 a : Ra a a ) = a t a , and hence (I t+1 a : Ra I a ) = I t a , as required.

Corollary 3.2.43. Every normally torsion-free square-free monomial ideal has a superficial ideal.

We conclude this subsection with showing that if I is a square-free monomial ideal of R, then we always have (I 2 : R I) = I. Proof. Since K is a field, one derives R is an integral domain. Furthermore, it is well-known that prime ideals are integrally closed, and also intersections of integrally closed ideals are integrally closed. This gives that I is integrally closed. Thus, the desired conclusion follows from Proposition 3.2.44. The last assertion is clear.

More results on the strong persistence property

In the following, we study the strong persistence property of monomial ideals with some suitable assumption on its support. It should be noted that the results of this section can be found in [START_REF] Nasernejad | The strong persistence property and symbolic strong persistence property[END_REF].

We begin with the following lemma which allows us to discuss the strong persistence property of the intersection and product of two monomial ideals which are generated by two disjoint sets of variables. Moreover, recall that for a monomial ideal

I of R with G(I) = {u 1 , . . . , u m }, we define supp(I) := m i=1 supp(u i ).
Lemma 3.3.1. Suppose that I 1 and I 2 are two monomial ideals in a polynomial ring To end the proof, it is enough to consider the following equalities,

R = K[x 1 , . . . , x n ] over a field K such that supp(I 1 ) ∩ supp(I 2 ) = ∅. If
((I 1 ∩ I 2 ) k+1 : R I 1 ∩ I 2 ) = ((I 1 I 2 ) k+1 : R I 1 I 2 ) = ((I k+1 1 I k+1 2 : R I 1 ) : R I 2 ) = (I k+1 2 (I k+1 1 : R I 1 ) : R I 2 ) = (I k 1 I k+1 2 : R I 2 ) = I k 1 (I k+1 2 : R I 2 ) = I k 1 I k 2 = (I 1 ∩ I 2 ) k .
Lemma 3.3.2. Suppose that I 1 and I 2 are two monomial ideals in a polynomial ring

R = K[x 1 , . . . , x n ] over a field K such that supp(I 1 ) ∩ supp(I 2 ) = ∅. Then, I 1 + I 2
has the strong persistence property if and only if I 1 or I 2 has the strong persistence property.

Proof. The backward implication can be immediately deduced from Theorem 3.2.8.

To establish the forward implication, suppose, on the contrary, that I 1 and I 2 do not satisfy the strong persistence property. This implies that there exist a positive integer k 1 (respectively, k 2 ) and a monomial m 1 (respectively, m 2 ) such that

m 1 ∈ G(I k 1 +1 1 : R I 1 ) \ G(I k 1 1 ) (respectively, m 2 ∈ G(I k 2 +1 2 : R I 2 ) \ G(I k 2 2 )). Take the nonnegative integer a 1 (respectively, a 2 ) such that m 1 ∈ I a 1 1 \ I a 1 +1 1 (respectively, m 2 ∈ I a 2 2 \ I a 2 +1
2

). This gives that a 1 ≤ k 1 -1 (respectively, a 2 ≤ k 2 -1). Put

I := I 1 + I 2 , m := m 1 m 2 , and b := a 1 + a 2 . Thus, one has m ∈ I b . Note that m / ∈ I i 1 ∩ I j 2 for either i > a 1 or j > a 2 , so, by [70, Lemma 1.1], m / ∈ I b+1 .
To conclude the proof, it is enough to show that m ∈ (I b+2 : R I), this contradicts the assumption that I has the strong persistence property. So, take a monomial

u ∈ I = I 1 + I 2 . Without loss of generality, we can assume that u ∈ I 1 . Thus, um = (um 1 )m 2 ∈ I k 1 +1 1 I a 2 2 ⊆ (I 1 + I 2 ) k 1 +1+a
= (x α 1 i 1 , . . . , x αt it ) in a polynomial ring R = K[x 1 , . . . ,
x n ] over a field K with α 1 , . . . , α t are positive integers and {x i 1 , . . . , x it } ⊆ {x 1 , . . . , x n }. Set I 1 := (x α 1 i 1 ) and I 2 = (x α 2 i 2 , . . . , x αt it ). Since I 1 has the strong persistence property, Lemma 3.3.2 implies that Q has the strong persistence property, as desired.

By a repeated application of Lemma 3.1.1, we have the following result. andh a monomial in R such that for each i = 1, . . . , m and j = 1, . . . , s, gcd(h, v j ) = 1, gcd(v j , u i ) = 1, and h ∈ I. If I has the strong persistence property, then L := JI+hR has the strong persistence property.

Lemma 3.3.4. Let I and J be two monomial ideals in a polynomial

ring R = K[x 1 , . . . , x n ] over a field K with G(I) = {u 1 , . . . , u m }, G(J) = {v 1 , . . . , v s },

Remark 3.3.5. It should be noted that Lemma 3.3.4 may be false if we consider the ideal L as L = JI + H with H not a principal monomial ideal. To see this, assume that L is the Stanley-Reisner ideal that corresponds to the natural triangulation of the projective plane, that is, L

⊂ R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ], one has L = (x 1 x 2 x 3 , x 1 x 2 x 4 , x 1 x 3 x 5 , x 1 x 4 x 6 , x 1 x 5 x 6 , x 2 x 3 x 6 , x 2 x 4 x 5 , x 2 x 5 x 6 , x 3 x 4 x 5 , x 3 x 4 x 6 ).
In the following, put J := (x 3 ), I := (x 1 x 2 , x 1 x 5 , x 2 x 6 , x 4 x 5 , x 4 x 6 ), and

H := (x 1 x 2 x 4 , x 1 x 4 x 6 , x 1 x 5 x 6 , x 2 x 4 x 5 , x 2 x 5 x 6 ).

It can be rapidly checked that

L = JI + H, H ⊆ I, gcd(x 3 , h) = 1 for all h ∈ G(H), gcd(x 3 , u) = 1 for all u ∈ G(I).
We prove that I has the strong persistence property. To do this, assume that G is the cycle graph with V (G) = {1, 2, 4, 5, 6} and

E(G) = {{1, 2}, {2, 6}, {6, 4}, {4, 5}, {5, 1}}. It is easy to detect that the edge ideal of G is I(G) = (x 1 x 2 , x 1 x 5 , x 2 x 6 , x 4 x 5 , x 4 x 6 )
. This implies that I(G) = I. In addition, [START_REF] Villarreal | Monomial algebras[END_REF]Theorem 7.7.14] yields that I(G) has the strong persistence property, and thus the monomial ideal I has the strong persistence property. On the other hand, by using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], one can detect that (L 3 : R L) = L 2 , that is, L does not satisfy the strong persistence property.

As an application of Lemma 3.1.1, we present Theorem 3.3.7. To understand the importance of this theorem, we first review some background.

Definition 3.3.6. ([133, Definition 10.5.4]) The cone C(G), over the graph G, is obtained by adding a new vertex t to G and joining every vertex of G to t.

More generally, we can take any graph G and form a new graph H by adding new vertices, joining each to every vertex of G. Then J(G) has the strong persistence property if and only if J(H) has the strong persistence property. Explicitly,

Theorem 3.3.7. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple graphs such that V (H) = V (G) ∪ {w 1 , . . . , w r } with w i / ∈ V (G) for all i = 1, . . . ,

r, and

E(H) = E(G) ∪ {{v, w i } : for all v ∈ V (G)
and for all i = 1, . . . , r}.

Then J(G) has the strong persistence property if and only if J(H) has the strong persistence property.

Proof. We give a sketch of the proof. Assume that

V (G) = [n] and V (H) = V (G) ∪ {n + 1, . . . , n + r}. In addition, let R = K[x 1 , . . . , x n+r ] be the polynomial ring over a field K. Put L := J(H), I := J(G), h := n i=1 x i , and g := n+r i=n+1 x i . It follows from [133, Exercise 6.1.23] that n i=1 n+r j=n+1 (x i , x j ) = ( n i=1 x i , n+r j=n+1 x j ).
Also, it is easy to see that L = gIR + hR. Suppose that I has the strong persistence property. Now, Lemma 3.1.1 yields that L has the strong persistence property, as claimed. Conversely, let J(H) have the strong persistence property. Put p := (x 1 , . . . , x n ). Since I = L(p), where L(p) denotes the monomial localization of L with respect to p, the claim follows at once from Theorem 3.2.16.

Here, we concentrate on the notion of a whisker graph. Our aim is to explore the strong persistence property of the cover ideal of a whisker graph. To do this, we recollect the subsequent definition. Definition 3.3.8. [START_REF] Villarreal | Monomial algebras[END_REF]Definition 7.3.10]) Let G 0 be a graph on the vertex set Y = {y 1 , . . . , y n } and take a new set of variables X = {x 1 , . . . , x n }. The whisker graph or suspension of G 0 , denoted by G 0 ∪ W (Y ), is the graph obtained from G 0 by attaching to each vertex y i a new vertex x i and the edge {x i , y i }. The edge {x i , y i } is called a whisker. Theorem 3.1.2 says that if we take any graph G whose cover ideal has the strong persistence property, and if we then add a leaf to G, then the cover ideal of the new graph satisfies the strong persistence property as well.

Corollary 3.3.9. With the notation of Definition 3.3.8, if the cover ideal of a finite simple graph G 0 has the strong persistence property, then the cover ideal of the whisker graph of G 0 has the strong persistence property.

Proof. Proceed by using Theorem 3.1.2 repeatedly, n times.

Strong persistence property of primary monomial ideals

Several questions may be asked along our argument. In this subsection, we investigate the strong persistence property of primary monomial ideals as a case study. To accomplish this, we start with the following main question. In addition, the symbol Q will always denote the set of rational numbers.

Question 3.3.10. Which classes of monomial ideals have the strong persistence property?

We present a class of primary monomial ideals which do not satisfy the strong persistence property. To do this, consider the monomial ideals

I n = (x n , x n-1 y, xy n-1 , y n ), in the polynomial ring R = K[x, y], n ≥ 4 over a field K.
The idea is to describe the monomials in (I d n ) nd+i for i ≥ 0, where subscripts denote the degree in R. We list all the monomials in (I d n ) nd+i in lexicographic order with respect to x < y (i.e., x nd+i , x nd+i-1 y, . . . , y nd+i ). In order to simplify the notation, we will often describe a monomial in (I d n ) nd+i with its power of y because we can identify such a monomial with its power of y.

It is clear that (I d n ) j = 0 for j < nd and that the monomials in (I d n ) nd are exactly the products of d generators of I n (with repetition). First of all, we consider the monomials {(x n ) α (y n ) β } α+β=d,α≥0,β≥0 which we will refer to as "subdivision points". Start with the subdivision point (x n ) α (y n ) β . If α ≥ 1, then some of the x n can be replaced by x n-1 y, and if β ≥ 1, then some of the y n can be replaced by xy n-1 . This leads to an "interval" of consecutive monomials {x nα-j y nβ+j } -β≤j≤α about the subdivision point (x n ) α (y n ) β . The first monomial in this interval is {x nα+β y nβ-β } and the last is {x nα-α y nβ+α }. Describing this interval with the powers of y, we have an increasing sequence of consecutive powers of y, {y j } nβ-β≤j≤nβ+α . The number of monomials in this interval is 1) and so to avoid overlap of intervals we require that n(β -1)

(nβ + α) -(nβ -β) + 1 = α + β + 1 = d + 1. The previous subdivision point (if β ≥ 1) is x n(α+1) y n(β-
+ (α + 1) < nβ -β or equivalently d + 1 < n, so that we have 1 ≤ d ≤ n -2. If d = n-2, then there is no overlap of intervals, but also no gap. If 1 ≤ d ≤ n-3, between adjacent intervals {x n(α+1)-j y n(β-1)+j } -(β-1)≤j≤(α+1) and {x nα-j y nβ+j } -β≤j≤α ,
there is a gap. For the left-hand side interval the largest exponent of y is n(β -1) + (α + 1) and the smallest exponent of y in the right-hand side interval is nβ -β, so the powers of y in the gap are {y j } n(β-1)+(α+1)<j<nβ-β , there being

(nβ -β) -(n(β -1) + (α + 1)) -1 = n -d -2,
monomials in the gap. As a simple check for consistency, we have d + 1 intervals each containing d + 1 monomials, and d gaps each containing n -d + 2 monomials, and

(d + 1) 2 + d(n -d -2) = nd + 1, the total number of monomials in R nd . Now, consider the monomials in (I d n ) nd+i , i ≥ 0. Then the interval {x n(α+1)-j y n(β-1)+j } -(β-1)≤j≤(α+1) , (of cardinality α + β + 1 = d + 1) expands to include {x n(α+1)-j+r y n(β-1)+j+s } -(β-1)≤j≤(α+1),r≥0,s≥0,r+s=i , (of cardinality α + β + 1 + i = d + i + 1
) and {x nα-j y nβ+j } -β≤j≤α expands to include {x nα-j+r y nβ+j+s } -β≤j≤α,r≥0,s≥0,r+s=i (again respectively of cardinalities d + 1 and d + i + 1). As above we will describe the gap between these intervals by giving the exponents of y in the monomials in this gap.

Lemma 3.3.11. [Gap Lemma]

(1

) If i ≥ n -d -2, then (I d n ) nd+i = R nd+i . 91 (2) If 0 ≤ i ≤ n -d -2, then (I d n ) nd+i has d + 1 subdivision intervals each of whose cardinality is d + i + 1, which is i more than that for (I d n ) nd . (3) If 0 ≤ i < n -d -2,

then the list of y-exponents of monomials in the gap between consecutive intervals

{x n(α+1)-j+r y n(β-1)+j+s } -(β-1)≤j≤(α+1),r≥0,s≥0,r+s=i , and

{x nα-j+r y nβ+j+s } -β≤j≤α,r≥0,s≥0,r+s=i , of (I d n ) nd+i , i ≥ 0 contains n -d -2 -i consecutive values (that is, i less than for (I d n ) nd ). Also, the largest y-exponent in the gap is nβ -β -1.
Proof. Part (2) was observed in the paragraph before the statement of the lemma.

The largest exponent of y of a monomial in the first interval is n(β -1) + α + 1 + i (coming from j = α + 1 and s = i) and the smallest exponent of y in the right most interval is nβ -β (coming from j = -β and s = 0). If

(nβ -β) -(n(β -1) + α + 1 + i) -1 = n -d -2 -i ≤ 0,
there are no gaps and part (1) follows. Otherwise we have (3).

Thus, we start with gap size n -d -2 previously obtained for i = 0, and drop one exponent from the beginning of the list of exponents for each increase of 1 in the value of i, until

(I d n ) nd+i consists of all R nd+i for i = n -d -2.
To make the notation more manageable we will refer to {x nα-j+r y nβ+j+s } -β≤j≤α,r≥0,s≥0,r+s=i ,

as the (α, β) subdivision interval of (I d n ) nd+i (0 ≤ β ≤ d, α + β = d).
With this notation, Lemma 3.3.11(3) takes on the simpler appearance: Lemma 3.3.12. [simplified Gap Lemma] For 1 ≤ β ≤ d, the list of y-exponents of monomials in the gap between consecutive subdivision intervals (α + 1, β -1) and

(α, β) of (I d n ) nd+i with 0 ≤ i < n -d -2 contains n -d -2 -i consecutive values, with the largest equal to nβ -β -1.
: I n ) = I d n for d ≥ n -2. Now, consider 1 ≤ d ≤ n -3. Let i = n -d -
: I n )\I d n . These monomials are {x nd+n-d-3-nβ+β+1 y nβ-β-1 } 1≤β≤d = {x nd-d-1 y n-2 , x nd-n-d y 2n-3 , . . . , x n-2 y nd-d-1 }. If 0 ≤ i < n -d -3, then (I d n )
nd+i has gaps of size t > 1, which are again mapped into R n(d+1)+i when multiplied by any minimal generator of I n , in which

(I d+1 n ) n(d+1)+i has gaps of size t -1 > 0.
Suppose that one of our gaps of (I d n ) nd+i lies between subdivision intervals (α + 1, β -1) and (α, β). Then, by Lemma 3.3.12, the monomials in the gap (described by giving only their exponent of y) consist of t consecutive powers of y with the largest being nβ -β -1. There is not a one-to-one correspondence between the gaps of (I d+1 n ) n(d+1)+i and (I d n ) nd+i , the former having d + 1 gaps, and the latter having d gaps so we have to make a choice. Choose the gap of (I d+1 n ) n(d+1)+i between subdivision intervals (α+2, β -1) and (α+1, β). Then the monomials in the gap between these intervals consist of t -1 consecutive powers of y with the largest again being nβ -β -1. The monomials in our gap for (I d n ) nd+i are then represented by {y nβ-β-t , y nβ-β-(t-1) , . . . , y nβ-β-1 } and the monomials in our gap for (I d+1 n ) n(d+1)+i are represented by {y nβ-β-(t-1) , . . . , y nβ-β-1 }. Recall that x n and x n-1 y are minimal generators of I n and multiplication by them raises the degree by n. Thus, if we multiply a monomial represented by y a of the gap sequence for (I d n ) nd+i by x n we get a monomial of the gap sequence for (I d+1 n ) n(d+1)+i that is also represented by y a , except for the first power y nβ-β-t where we can multiply instead by x n-1 y and get the monomial represented by y nβ-β-(t-1) which is in the gap sequence for (I d+1 n ) n(d+1)+i . This shows that if 0 ≤ i < n -d -3, then the monomials in the gaps of (I d n ) nd+i do not lie in I d+1 n : I n . Therefore, (I d+1 n : I n )\I d+1 n consists only of the monomials {x nd-d-1 y n-2 , x nd-n-d y 2n-3 , . . . , x n-2 y nd-d-1 } of degree nd+i where i = n -d -3 found in the previous paragraph.

In the expression {x nd+n-d-3-nβ+β+1 y nβ-β-1 } 1≤β≤d there is an annoying apparent lack of symmetry between x and y and between α and β. This can be removed by setting i = β -1, j = α, a = d -1 so that this expression becomes {x n-2 y n-2 (x n-1 ) j (y n-1 ) i } and our result becomes:

Proposition 3.3.13. Let I n = (x n , x n-1 y, xy n-1 , y n ) in the polynomial ring R = Q[x, y], n ≥ 4. Then I a+2 n : I n ⊋ I a+1 n for 0 ≤ a ≤ n -4 and I a+2 n : I n = I a+1 n for a ≥ n -3. More precisely, if 0 ≤ a ≤ n -4, then (I a+2 n : I n )/I a+1 n is an (a + 1)-dimensional vector space with basis {x n-2 y n-2 (x n-1 ) j (y n-1 ) i } i≥0,j≥0,i+j=a .
It should be noted that, in particular, if a = 0, then Proposition 3.3.13 yields the result that x n-2 y n-2 is the only monomial in I 2 n : I n but not in I n . To clarify our discussion, we provide the following example. ) j = 0 for j < 12. Furthermore, the set of all monomials in (I 2 6 ) 13 is {x 13 , x 12 y, x 11 y 2 , x 10 y 3 , x 8 y 5 , x 7 y 6 , x 6 y 7 , x 5 y 8 , x 3 y 10 , x 2 y 11 , xy 12 , y 13 }, and the set of all monomials in (I 2 6 ) 14 is {x 14 , x 13 y, x 12 y 2 , x 11 y 3 , x 10 y 4 , x 9 y 5 , x 8 y 6 , x 7 y 7 , x 6 y 8 , x 5 y 9 ,

x 4 y 10 , x 3 y 11 , x 2 y 12 , xy 13 , y 14 }, that is, all monomials of degree 14. Obviously, (I 2 6 ) j = R j for j > 14. In the notation of Lemma 3.3.11, we have n = 6 and d = 2. Part (1) of this Lemma says that Lemma 3.3.11 part (3). We are interested in I 3 6 : I 6 . This contains I 2 6 and potentially the gaps of (I 6 by all the minimal generators of I 6 . The gaps of (I 3 6 ) 18 are {x 14 y 4 }, {x 9 y 9 } and {x 4 y 14 }, and for example x 6 (x 8 y 4 ) = x 14 y 4 / ∈ (I 6 ) 3 so x 8 y 4 / ∈ I 3 6 : I 6 . Similarly, x 5 y(x 9 y 3 ) = x 14 y 4 / ∈ (I 6 ) 3 so x 9 y 3 / ∈ I 3 6 : I 6 either. Instead of the gap {x 14 y 4 } we can use {x 9 y 9 }. Thus, y 6 (x 9 y 3 ) = x 9 y 9 / ∈ (I 6 ) 3 , again yielding that x 9 y 3 / ∈ I 3 6 : I 6 . We thus conclude that (I 3 6 : I 6 )\(I 6 ) 2 = {x 9 y 4 , x 4 y 9 }. (In the language of Proposition 3.3.13, a = 1.)

(I 2 6 ) nd+i = (I 2 6 ) 12+i = R 12+i for i ≥ 6 -2 -2 = 2, or (I 2 6 ) j = R j for j ≥
We are ready to present the second question. Let R = K[x 1 , . . . , x n ] be a polynomial ring over a field K and let I be a monomial ideal in R that has the strong persistence property. Also, let Q be an irreducible primary monomial ideal of R. Does the monomial ideal I + Q satisfy the strong persistence property?

We provide a counterexample. For this purpose, assume that R = K[x, y] is the polynomial ring over a field K, and put I := (x 4 , x 3 y, xy 3 , y 5 ), and Q = (y 4 ). We thus have I + Q = (x 4 , x 3 y, xy 3 , y 4 ). It follows from Proposition 3.3.13 that I + Q does not satisfy the strong persistence property. To conclude our argument, it remains to verify that the ideal I has the strong persistence property. To see this, fix k ≥ 1. Since I k ⊆ (I k+1 : R I) ⊆ (I k+1 : R xy 3 ), it is enough for us to show that (I k+1 : R xy 3 ) ⊆ I k . It is well-known that

I k+1 = λ 1 +λ 2 +λ 3 +λ 4 =k+1 (x 4 ) λ 1 (x 3 y) λ 2 (y 5 ) λ 3 (xy 3 ) λ 4 R. If λ 4 ≥ 1, then we have ((x 4 ) λ 1 (x 3 y) λ 2 (y 5 ) λ 3 (xy 3 ) λ 4 : R xy 3 ) = ((x 4 ) λ 1 (x 3 y) λ 2 (y 5 ) λ 3 (xy 3 ) λ 4 -1 ) ⊆ I k . Thus, let λ 4 = 0. If λ 2 ≥ 3, then one obtains ((x 4 ) λ 1 (x 3 y) λ 2 (y 5 ) λ 3 : R xy 3 ) = ((x 4 ) λ 1 (x 3 y) λ 2 -3 (y 5 ) λ 3 x 8 ) = ((x 4 ) λ 1 +2 (x 3 y) λ 2 -3 (y 5 ) λ 3 ) ⊆ I k .
Hence, let λ 4 = 0 and 0 ≤ λ 2 ≤ 2. Accordingly, one may consider the following cases:

Case 1. λ 4 = 0 and λ 2 = 0. If λ 1 = 0, then λ 3 = k + 1. This leads to ((y 5 ) λ 3 : R xy 3 ) = ((y 5 ) λ 3 -1 y 2 ) ⊆ I k . Let λ 1 ≥ 1. If λ 3 = 0, then λ 1 = k + 1, and thus ((x 4 ) λ 1 : R xy 3 ) = ((x 4 ) λ 1 -1 x 3 ) ⊆ I k . Hence, let λ 3 ≥ 1. Due to λ 1 ≥ 1 and λ 3 ≥ 1, we have ((x 4 ) λ 1 (y 5 ) λ 3 : R xy 3 ) = ((x 4 ) λ 1 -1 (y 5 ) λ 3 -1 (x 3 y)y) ⊆ I k .
Case 2. λ 4 = 0 and λ 2 = 1. Let λ 1 = 0. Hence, λ 3 = k ≥ 1, and so

((x 3 y)(y 5 ) λ 3 : R xy 3 ) = (x(xy 3 )(y 5 ) λ 3 -1 ) ⊆ I k . Therefore, let λ 1 ≥ 1. If λ 3 = 0, then λ 1 = k, and thus ((x 4 ) λ 1 (x 3 y) : R xy 3 ) = ((x 4 ) λ 1 x 2 ) ⊆ I k . Hence, let λ 3 ≥ 1. Thanks to λ 1 ≥ 1 and λ 3 ≥ 1, one derives that ((x 4 ) λ 1 (x 3 y)(y 5 ) λ 3 : R xy 3 ) = ((x 4 ) λ 1 -1 (x 3 y) 2 (y 5 ) λ 3 -1 y) ⊆ I k . Case 3. λ 4 = 0 and λ 2 = 2. If λ 3 ≥ 1, then ((x 4 ) λ 1 (x 3 y) 2 (y 5 ) λ 3 : R xy 3 ) = ((x 4 ) λ 1 (x 5 y 2 )(y 5 ) λ 3 -1 y 2 ) = ((x 4 ) λ 1 (x 3 y)(y 5 ) λ 3 -1 (xy 3 )x) ⊆ I k .
Therefore, let λ 3 = 0. This gives that λ 1 = k -1. Hence, we gain

((x 4 ) λ 1 (x 3 y) 2 : R xy 3 ) = ((x 4 ) λ 1 x 5 ) = ((x 4 ) λ 1 +1 x) ⊆ I k .
This terminates our argument.

Strong persistence property of the cover ideals

In this subsection, we focus on the strong persistence property of the cover ideals of the union of simple finite graphs. For this purpose, we consider the following lemma which examines the relation between associated primes of powers of the cover ideal of the union of a finite simple connected graph and a tree with the assoicated primes of powers of the cover ideals of each of them, under the condition that they have only one common vertex.

It should be noted that throughout this subsection, all trees are non-trivial, that is, they have at least two vertices.

A repeated application of Theorem 3.1.6 yields the following lemma:

Lemma 3.3.15. Let G = (V (G), E(G)) be a finite simple connected graph and T be a tree such that |V (G) ∩ V (T )| = 1. Let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (T ) and E(L) := E(G) ∪ E(T ). Then Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(T ) s ), for all s, where R 1 = K[x α : α ∈ V (G)], R 2 = K[x α : α ∈ V (T )], and R = K[x α : α ∈ V (L)] over a field K.
The next lemma explores the relation between associated primes of powers of the cover ideal of the union of a finite simple connected graph and a tree with the associated primes of powers of the cover ideals of each of them, under the condition that they have only a path in common. In fact, a repeated application of Lemma 3.3.15 gives the following lemma:

Lemma 3.3.16. Let G = (V (G), E(G)) be a finite simple connected graph, T 1 , . . . , T r be some trees with V (G) ∩ V (T i ) = {v i } for each i = 1, . . . , r, V (T i ) ∩ V (T j ) = ∅ for i = j, and P = (V (P ), E(P )) be a path of G with V (P ) = {v 1 , . . . , v r , v r+1 , . . . , v m } ⊆ V (G), and
E(P ) = {{v i , v i+1 } : for i = 1, . . . , m -1} ⊆ E(G).
Let T = (V (T ), E(T )) be the tree with

V (T ) = r i=1 V (T i ) ∪ V (P ) and E(T ) = r i=1 E(T i ) ∪ E(P ).
Also, let L = (V (L), E(L)) be the finite simple graph such that

V (L) := V (G) ∪ V (T ) and E(L) := E(G) ∪ E(T ). Then Ass R (R/J(L) s ) = Ass R ′ (R ′ /J(G) s ) ∪ Ass R ′′ (R ′′ /J(T ) s ), for all s, where R ′ = K[x α : α ∈ V (G)], R ′′ = K[x α : α ∈ V (T )], and R = K[x α : α ∈ V (L)] over a field K.
To establish Theorem 3.3.19, one needs to know the following auxiliary propositions. Indeed, by considering the fact that localizing at a minimal prime ideal inverts everything outside of it, one can deduce the following proposition: Proposition 3.3.17. Let I be an ideal in a commutative Noetherian ring R. Also, let

I = Q 1 ∩ • • • ∩ Q t ∩ Q t+1 ∩ • • • ∩ Q r be a minimal primary decomposition of I with p i = √ Q i for i = 1, . . . ,

r, and

Min(I) = {p 1 , . . . , p t }. Then I p i = (Q i ) p i for i = 1, . . . , t. Proposition 3.3.18. Let I be a monomial ideal in R = K[x 1 , . . . , x n ] over a field K with G(I) = {u 1 , . . . , u m } and Ass R (R/I) = {p 1 , . . . , p s }.
Then, the following statements hold. (i) If x i |u t for some i with 1 ≤ i ≤ n, and for some t with 1 ≤ t ≤ m, then there exists j with 1 ≤ j ≤ s, such that x i ∈ p j .

(ii) If x i ∈ p j for some i with 1 ≤ i ≤ n, and for some j with 1 ≤ j ≤ s, then there exists t with 1 ≤ t ≤ m, such that x i |u t .

In particular, s j=1 supp(p j ) = m t=1 supp(u t ).

Proof. (i) Suppose that

I = Q 1 ∩ • • • ∩ Q s is a minimal primary decomposition of I such that √ Q z = p z for all z = 1, . . . , s. Also, let x i |u t for some 1 ≤ i ≤ n and 1 ≤ t ≤ m. Since u t ∈ G(I), one has u t ∈ Q z for all z = 1, . . . ,
s, and so u t ∈ p z for all z = 1, . . . , s. It follows also from x i |u t that there exists some monomial v in R such that u t = x i v. Suppose, on the contrary, that

x i / ∈ p z for all z = 1, . . . , s. Then one can conclude that v ∈ Q z for all z = 1, . . . , s because x i v ∈ Q z , x i / ∈ p z , and Q z is primary. Therefore, v ∈ s z=1 Q z , that is, v ∈ I.
This contradicts the minimality of u t . We thus have there exists some 1 ≤ j ≤ s such that x i ∈ p j .

(ii) Let x i ∈ p j for some 1 ≤ i ≤ n and 1 ≤ j ≤ s. Since p j ∈ Ass R (R/I), there exists some monomial v in R such that p j = (I : R v). In addition, the assumption

G(I) = {u 1 , . . . , u m } yields that I = m r=1 u r R. By virtue of p j = (I : R v) = m r=1 (u r R : R v) and x i ∈ p j , one can conclude that there exists some 1 ≤ t ≤ m such that x i ∈ (u t R : R v). We thus have u t h = x i v for some monomial h in R. If x i |h, then v ∈ I,
which is a contradiction. Therefore, one can derive that x i |u t , as claimed.

The last assertion is an immediate consequence from parts (i) and (ii).

In the next theorem, we turn our attention to studying the strong persistence property of the cover ideal of the union of two finite simple connected graphs.

Theorem 3.3.19. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple connected graphs such that J(G) and J(H) have the strong persistence property. Also

, let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G)∪V (H), E(L) := E(G) ∪ E(H). Assume that Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ), for all s, where R 1 = K[x α : α ∈ V (G)], R 2 = K[x α : α ∈ V (H)], and R = K[x α : α ∈ V (L)] over a field K.
Then, under each of the following cases, J(L) has the strong persistence property.

(i) V (G) ∩ V (H) = {v}. (ii) V (G) ∩ V (H) = {v, w} and E(G) ∩ E(H) = {{v, w}}. (iii) V (G) ∩ V (H) = {v, w, z} and E(G) ∩ E(H) = {{v, w}, {w, z}}.
Proof. To simplify our notation, set I 1 := J(G), I 2 := J(H), and I := J(L). Note that I = I 1 ∩ I 2 . We requires to show that (I k+1 : R I) = I k for all k ≥ 1. To achieve this, fix k ≥ 1. In view of [START_REF] Matsumura | Commutative Ring Theory[END_REF]Exercise 6.4], it is sufficient to prove that

(I k+1 p : Rp I p ) = I k p for every p ∈ Ass R (R/I k ).
For this purpose, pick an arbitrary element p ∈ Ass R (R/I k ). We therefore can consider the following two cases:

: Rp I p ) = I k p , as required. Case (2). p ∈ Ass R (R/I k ) \ Min(I k ). By considering the assumption, one can deduce that p ∈ Ass R 1 (R 1 /I k 1 ) or p ∈ Ass R 2 (R 2 /I k 2 ). In view of the fact that Min(I k ) = Min(I) = Min(I 1 ) ∪ Min(I 2 ), we get p ∈ Ass R (R/I k 1 ) \ Min(I 1 ) or p ∈ Ass R (R/I k 2 ) \ Min(I 2 ).
We only demonstrate the case p ∈ Ass R (R/I k 1 )\Min(I 1 ), while another case is proved similarly. To finish the argument, we show that I p = (I 1 ) p . Thanks to I p ⊆ (I 1 ) p , it remains to establish (I 1 ) p ⊆ I p . To do this, take an arbitrary element r/s in (I 1 ) p . This gives that r/s = α/β for some α ∈ I 1 and β / ∈ p. We now have to consider the following cases:

Case (i). V (G) ∩ V (H) = {v}. One can easily see that α ℓ∈V (H)\{v} x ℓ ∈ I 1 ∩ I 2 .
Furthermore, Proposition 3.3.18 implies that x ℓ / ∈ p for any ℓ ∈ V (H) \ {v}, and thus

β ℓ∈V (H)\{v} x ℓ / ∈ p.
Due to the following equality

α β = α ℓ∈V (H)\{v} x ℓ β ℓ∈V (H)\{v} x ℓ , we obtain r/s ∈ I p . Case (ii). V (G) ∩ V (H) = {v, w} and E(G) ∩ E(H) = {{v, w}}. Since α ∈ I 1 , we get α ∈ (x v , x w )
, and so x v |α or x w |α. Hence, one can readily deduce that

x v |α ℓ∈V (H)\{v,w} x ℓ or x w |α ℓ∈V (H)\{v,w} x ℓ .
This leads to α ℓ∈V (H)\{v,w} x ℓ ∈ I 2 , and so α ℓ∈V (H)\{v,w} x ℓ ∈ I 1 ∩ I 2 . Based on Proposition 3.3.18, we derive that x ℓ / ∈ p for any ℓ ∈ V (H) \ {v, w}, and hence β ℓ∈V (H)\{v,w} x ℓ / ∈ p. In the light of the following equality

α β = α ℓ∈V (H)\{v,w} x ℓ β ℓ∈V (H)\{v,w} x ℓ , one can conclude that r/s ∈ I p . Case (iii). V (G) ∩ V (H) = {v, w, z} and E(G) ∩ E(H) = {{v, w}, {w, z}}. On account of α ∈ I 1 , one has α ∈ (x v , x w ) ∩ (x w , x z ). If x w |α, then we obtain
x w |α ℓ∈V (H)\{v,w,z} x ℓ , and so α ℓ∈V (H)\{v,w,z} x ℓ ∈ I 2 . Let x w ∤ α. Because I 1 is the cover ideal of the graph G, one must have x v |α and x z |α. This gives rise to α ℓ∈V (H)\{v,w,z} x ℓ ∈ I 2 . It follows also from Proposition 3.3.18 that x ℓ / ∈ p for any ℓ ∈ V (H) \ {v, w, z}, and thus β ℓ∈V (H)\{v,w,z} x ℓ / ∈ p. Now, by observing the following equality

α β = α ℓ∈V (H)\{v,w,z} x ℓ β ℓ∈V (H)\{v,w,z} x ℓ , one derives that r/s ∈ I p .
As the ideal I 1 has the strong persistence property, one gains (I k+1

i) V (G) ∩ V (T ) = {v}. (ii) V (G) ∩ V (T ) = {v, w} and E(G) ∩ E(T ) = {{v, w}}. (iii) V (G) ∩ V (T ) = {v, w, z} and E(G) ∩ E(T ) = {{v, w}, {w, z}}.
Proof. Since T is a tree, [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6] yields that J(T ) is a normally torsion-free square-free monomial ideal. It follows now from Theorem 3.2.42 that J(T ) has the strong persistence property. Hence, this claim is a direct consequence from Lemmas 3.3.15, 3.3.16, and Theorem 3.3.19.

The following remark says that, with the notation of Theorem 3.3.19, it is possible J(G), J(H), and J(L) have the strong persistence property, while, for some s, we have

Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ).

Remark 3.3.21. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple connected graphs such that J(G) and J(H) have the strong persistence property,

|V (G) ∩ V (H)| > 2, and |E(G) ∩ E(H)| > 1. Also, let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (H) and E(L) := E(G) ∪ E(H).
Then, it is possible that J(L) has the strong persistence property, while, for some s, we have

Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ) ⊊ Ass R (R/J(L) s ), where R 1 = K[x α : α ∈ V (G)], R 2 = K[x α : α ∈ V (H)], and R = K[x α : α ∈ V (L)
] are polynomial rings over a field K. As an example, consider the graph

G = (V (G), E(G)) with V (G) = {1, 2, 3, 4, 5} and
E(G) = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 1}, {5, 2}},
and also the graph

H = (V (H), E(H)) with V (H) = {1, 4, 5, 6} and
E(H) = {{1, 5}, {4, 5}, {1, 6}, {5, 6}, {4, 6}}.
Hence, as shown in the figure below, we have V (L) = {1, 2, 3, 4, 5, 6}, and

V (L) = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 1}, {5, 2}, {1, 6}, {5, 6}, {4, 6}}.

It is easy to compute that

I 1 := J(G) =(x 1 , x 2 ) ∩ (x 2 , x 3 ) ∩ (x 3 , x 4 ) ∩ (x 4 , x 5 ) ∩ (x 5 , x 1 ) ∩ (x 5 , x 2 ) =(x 2 x 4 x 5 , x 2 x 3 x 5 , x 1 x 3 x 5 , x 1 x 2 x 4 ), and
I 2 := J(H) =(x 1 , x 5 ) ∩ (x 4 , x 5 ) ∩ (x 1 , x 6 ) ∩ (x 5 , x 6 ) ∩ (x 4 , x 6 ) =(x 5 x 6 , x 1 x 4 x 6 , x 1 x 4 x 5 ),
and

I := J(L) = (x 1 , x 2 ) ∩ (x 2 , x 3 ) ∩ (x 3 , x 4 ) ∩ (x 5 , x 1 ) ∩ (x 5 , x 2 ) ∩(x 4 , x 5 ) ∩ (x 1 , x 6 ) ∩ (x 5 , x 6 ) ∩ (x 4 , x 6 ) = (x 2 x 4 x 5 x 6 , x 2 x 3 x 5 x 6 , x 1 x 3 x 5 x 6 , x 1 x 2 x 4 x 6 , x 1 x 3 x 4 x 5 , x 1 x 2 x 4 x 5 ). G H L 1 2 3 4 5 6
In the first step, we verify that I 1 and I 2 have the strong persistence property. To accomplish this, one can write

I 1 = x 5 (x 2 x 4 , x 2 x 3 , x 1 x 3 ) + x 1 x 2 x 4 R 1 and I 2 =
x 1 x 4 (x 5 , x 6 ) + x 5 x 6 R 2 . Let P be the path graph with V (P ) = {1, 2, 3, 4} and E(P ) = {{1, 3}, {3, 2}, {2, 4}}. Since the edge ideal of P is I(P ) = (x 2 x 4 , x 2 x 3 , x 1 x 3 ), by virtue of [START_REF] Villarreal | Monomial algebras[END_REF]Theorem 7.7.14], we gain the monomial ideal (x 2 x 4 , x 2 x 3 , x 1 x 3 ) has the strong persistence property. Moreover, it follows from Corollary 3.3.3 that the prime ideal (x 5 , x 6 ) has the strong persistence property. Consequently, Lemma 3.3.4 gives that I 1 and I 2 have the strong persistence property. In the second step, we demonstrate that I has the strong persistence property. For this purpose, set

u 1 := x 1 x 3 x 4 , u 2 := x 1 x 3 x 6 , u 3 := x 2 x 3 x 6 , u 4 := x 2 x 4 x 6 , u 5 := x 1 x 2 x 4 , and F := (u 1 , u 2 , u 3 , u 4 , u 5 )R. This gives rise to I = x 5 F + x 1 x 2 x 4 x 6 R.
Our strategy is to show that F has the strong persistence property. For this purpose, let

G = (V (G), E(G)) be the graph with the vertex set V (G) = {1, 2, 3, 4, 6} and the edge set E(G) = {{1, 2}, {2, 3}, {3, 4}, {4, 6}, {6, 1}}. It is routine to check that G
is the odd cycle graph of order 5. By using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], we can deduce that F is the cover ideal of G. Also, by virtue of Theorem 3.1.11, F has the strong persistence property. It follows now from Lemma 3.3.4 that I has the strong persistence property. Ultimately, by using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], we note that

(x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) ∈ Ass R (R/I 3 ) \ (Ass R 1 (R 1 /I 3 1 ) ∪ Ass R 2 (R 2 /I 3 2 )).
This completes our discussion.

We continue this argument with the following result, which examines the relation between associated primes of powers of cover ideal of a finite simple connected graph with the associated primes of powers of the cover ideals of each connected subgraph of that graph.

Proposition 3.3.22. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple connected graphs. Let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (H) and E(L) := E(G) ∪ E(H). Then Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ) ⊆ Ass R (R/J(L) s ).
Proof. To simplify our notation, put I 1 := J(G), I 2 := J(H), and I := J(L). Take an arbitrary element p

= (x i 1 , . . . , x ir ) in Ass R 1 (R 1 /I s 1 ) ∪ Ass R 2 (R 2 /I s 2 ) with {i 1 , . . . , i r } ⊆ V (L). Let p ∈ Ass R 1 (R 1 /I s 1 )
. In view of Proposition 3.3.18, one has supp(p) ⊆ u∈G(I 1 ) supp(u). This leads to {i 1 , . . . , i r } ⊆ V (G). It follows from [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11]

that p ∈ Ass(K[p]/J(G p ) s ), where K[p] = K[x i 1 , . . . , x ir ] and G p is the induced subgraph of G on the vertex set {i 1 , . . . , i r } ⊆ V (G). Thanks to G p = L p , we get p ∈ Ass(K[p]/J(L p ) s ). Once again, using [45, Lemma 2.11] yields that p ∈ Ass R (R/I s ). A similar discussion shows that if p ∈ Ass R 2 (R 2 /I s 2 ), then p ∈ Ass R (R/I s ). Therefore, one can conclude that Ass R 1 (R 1 /I s 1 ) ∪ Ass R 2 (R 2 /I s 2 ) ⊆ Ass R (R/I s ).
This finishes the proof.

The following theorem examines the relation between assoicated primes of powers of the cover ideal of the union of two finite simple graphs with the assoicated primes of powers of the cover ideals of each of them, under the condition that they have only one common vertex.

Theorem 3.3.23. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple connected graphs such that |V (G) ∩ V (H)| = 1. Let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (H) and E(L) := E(G) ∪ E(H). Then Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ), for all s, where R 1 = K[x α : α ∈ V (G)], R 2 = K[x α : α ∈ V (H)], and R = K[x α : α ∈ V (L)].
In particular, if J(G) and J(H) have the strong persistence property, then J(L) has the strong persistence property. Proof. Fix s ≥ 1, and let V (G) ∩ V (H) = {v}. If G or H is a tree, then the claim follows rapidly from Lemma 3.3.15 and Theorem 3.3.19. Hence, we can assume that neither G nor H is a tree. On account of Proposition 3.3.22, we gain the following containment

Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ) ⊆ Ass R (R/J(L) s ).
In order to complete the proof, it is sufficient for us to verify the reverse inclusion. To achieve this, pick an arbitrary element p = (x i 1 , . . . , x ir ) in Ass R (R/J(L) s ). Based on Proposition 3.3.18, we have {i 1 , . . . , i r } ⊆ V (L). Moreover, it follows from [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11] that p ∈ Ass(K[p]/J(L p ) s ), where K[p] = K[x i 1 , . . . , x ir ] and L p is the induced subgraph of L on the vertex set {i 1 , . . . , i r }. Hereafter, we assume that p is the maximal ideal in the polynomial ring R = K[p]. To simplify the notation, put Γ := {x c : c ∈ V (G) \ {v}} and Λ := {x c : c ∈ V (H) \ {v}}. Take an arbitrary element u in G(J(L)). Because of neither G nor H is a tree, one can conclude that neither G nor H is a star graph or an edge, and so supp(u) ∩ Γ = ∅ and The argument above gives that if u ∈ G(J(L)), then u = Af g, where A|x v and f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ) such that Af ∈ J(G) (respectively, Ag ∈ J(H)). Since p ∈ Ass R (R/J(L) s ), we get there exists some monomial h in R such that p = (J(L) s : R h). Assume that

supp(u) ∩ Λ = ∅. Since J(L) = J(G) ∩ J(H),
h = h 1 h 2 x ρ v with h 1 (respectively, h 2
) is a monomial in the variables Γ (respectively, Λ), and ρ is a nonnegative integer. It follows readily from Lemma 2.1.12 that

(J(G) s : R h 1 h 2 x ρ v ) = (J(G) s : R h 1 x ρ v ) and (J(H) s : R h 1 h 2 x ρ v ) = (J(H) s : R h 2 x ρ v ). Since (J(L) s : R h) ⊆ (J(G) s : R h) (respectively, (J(L) s : R h) ⊆ (J(H) s : R h)), one can conclude that p ⊆ (J(G) s : R h 1 x ρ v ) (respectively, p ⊆ (J(H) s : R h 2 x ρ v )). Our aim is to demonstrate that h 1 x ρ v / ∈ J(G) or h 2 x ρ v / ∈ J(H). Suppose, on the contrary, that h 1 x ρ v ∈ J(G) and h 2 x ρ v ∈ J(H). It follows from h 1 x ρ v ∈ J(G) s (respectively, h 2 x ρ v ∈ J(H) s
) that there exist square-free monomials M 1 , f 1 , . . . , f s (respectively, M 2 , g 1 , . . . , g s ) in the variables Γ (respectively, Λ), and A 1 , . . . , A s (respectively,

A ′ 1 , . . . , A ′ s ) with A i |x v (respectively, A ′ i |x v ) and A i f i ∈ J(G) (respectively, A ′ i g i ∈ 103 J(H)) for each i = 1, . . . , s, such that h 1 x ρ v = ( s i=1 A i f i )M 1 x θ v (respectively, h 2 x ρ v = ( s i=1 A ′ i g i )M 2 x λ v ),
for some nonnegative integer θ (respectively, λ). We thus have

h 1 = ( s i=1 f i )M 1 , h 2 = ( s i=1 g i )M 2 , and ( s i=1 A i )x θ v = x ρ v = ( s i=1 A ′ i )x λ v .
Accordingly, one has

h 1 h 2 x ρ v = ( s i=1 f i )M 1 ( s i=1 g i )M 2 x ρ v = ( s i=1 f i g i )M 1 M 2 x ρ v . Since ( s i=1 A i )x θ v = x ρ v = ( s i=1 A ′ i )x λ v , we get h 1 h 2 x ρ v ∈ J(L) s
, and so h ∈ J(L) s ; this contradicts the fact that p = (J(L) s : R h). We therefore gain

h 1 x ρ v / ∈ J(G) or h 2 x ρ v / ∈ J(H). As p is the maximal ideal, one has p ∈ Ass R 1 (R 1 /J(G) s ) or p ∈ Ass R 2 (R 2 /J(H) s )
, and so we have the following equality

(3.3.1) Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ),
as required. The last claim is an immediate consequence of Theorem 3.

= V (G) ∪ V (H) and E(L) := E(G) ∪ E(H). Then Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ), for all s, where R 1 = K[x α : α ∈ V (G)], R 2 = K[x α : α ∈ V (H)], and R = K[x α : α ∈ V (L)
] over a field K. In particular, if J(G) has the strong persistence property, then J(L) has the strong persistence property.

Proof. Let V (H) = {v, w, z}, V (G)∩V (H) = {v,
Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ) ⊆ Ass R (R/J(L) s ).
For completing the proof, it is enough for us to prove the reverse inclusion. To accomplish this, select an arbitrary element p = (x i 1 , . . . , x ir ) in Ass R (R/J(L) s ). One can conclude from Proposition 3.3.18 that {i 1 , . . . , i r } ⊆ V (L). If z / ∈ {i 1 , . . . , i r }, then {i 1 , . . . , i r } ⊆ V (G), and [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11]

implies that p ∈ Ass R 1 (R 1 /J(G) s ).
We thus let z ∈ {i 1 , . . . , i r }. In view of [17,Lemma 2.4], L p is a connected graph, where L p denotes the induced graph on {i 1 , . . . , i r }. This leads to the following cases:

Case 1. v ∈ {i 1 , . . . , i r } but w / ∈ {i 1 , . . . , i r } (or w ∈ {i 1 , . . . , i r } but v / ∈ {i 1 , . . . , i r }).
We only consider the case v ∈ {i 1 , . . . , i r } but w / ∈ {i 1 , . . . , i r }, while the other case is proved similarly. Because p ∈ Ass R (R/J(L) s ), [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11]

concludes that p ∈ Ass R (R/J(L p ) s ). Since w / ∈ {i 1 , . . . , i r }, one derives that V (L p ) = V (G p ) ∪ {v, z} and E(L p ) = E(G p ) ∪ {{v, z}}. By Theorem 3.1.6, we have Ass(J(L p ) s ) = Ass(J(G p ) s ) ∪ {(x v , x z )}. If p = (x v , x z ), then p ∈ Min(J(H)), and hence p ∈ Ass R 2 (R 2 /J(H) s ). If p ∈ Ass(J(G p ) s ), then [45, Lemma 2.11] yields that p ∈ Ass R 1 (R 1 /J(G) s ), and so p ∈ Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ).
Case 2. v, w ∈ {i 1 , . . . , i r }. Without loss of generality, one may assume that i 1 = v, i 2 = w, and i 3 = z. Based on [45, Corollary 4.5], the associated primes of J(L) s will correspond to critical chromatic subgraphs of size s + 1 in the s-th expansion of L. This means that one can take the induced subgraph on the vertex set {i 1 , . . . , i r }, and then form the s-th expansion on this induced subgraph, and within this new graph find a critical (s + 1)-chromatic graph. Thanks to z is only connected to v and w in the graph L, and since this induced subgraph is critical, if we remove the vertex z, we can color the resulting graph with s colors. Also, by virtue of [18,Theorem 14.6], the vertex z has to be adjacent to at least s vertices. But the only things z is adjacent to are the shadows of z and the shadows of v and w, and hence one has a clique among these vertices. We thus gain that z and its neighbors will form a clique of size s + 1. On account of a clique is a critical graph, this gives that we do not need any element of {i 4 , . . . , i r } or their shadows when making the critical (s + 1)-chromatic graph. This implies that p = (x v , x w , x z ). Due to Proposition 3.1.15, one can conclude that

(x v , x w , x z ) ∈ Ass R 2 (R 2 /J(H) s ).
Accordingly, we obtain the following equality

(3.3.2) Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ).
Since the cover ideal of any triangle has the strong persistence property, the last assertion is a straightforward consequence of Theorem 3. The next theorem probes the relation between assoicated primes of powers of the cover ideal of the union of two finite simple connected graphs with the assoicated primes of powers of the cover ideals of each of them, under the condition that they have only one edge in common.

Theorem 3.3.25. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple connected graphs such that |V (G) ∩ V (H)| = 2 and |E(G) ∩ E(H)| = 1. Let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (H) and E(L) := E(G) ∪ E(H). Then Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ), for all s, where R 1 = K[x α : α ∈ V (G)], R 2 = K[x α : α ∈ V (H)], and R = K[x α : α ∈ V (L)].
In particular, if J(G) and J(H) have the strong persistence property, then J(L) has the strong persistence property.

Proof. Fix s ≥ 1. Let V (G) ∩ V (H) = {v, w} and E(G) ∩ E(H) = {{v, w}}. If G
or H is a tree or triangle, then the assertion is true by virtue of Lemma 3.3.16 and Theorem 3.3.24. We therefore can assume that neither G nor H is a tree or triangle. As a direct consequence of Proposition 3.3.22, we have the following containment

Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ) ⊆ Ass R (R/J(L) s ).
To finish the argument, one has to establish the reverse inclusion. To do this, choose an arbitrary element p = (x i 1 , . . . , x ir ) in Ass R (R/J(L) s ). By Proposition 3.3.18, one has {i 1 , . . . , i r } ⊆ V (L). On account of [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11], one can derive that

p ∈ Ass(K[p]/J(L p ) s ), where K[p] = K[x i 1 , . . . ,
x ir ] and L p is the induced subgraph of L on the vertex set {i 1 , . . . , i r }. We thus assume that p is the maximal ideal in the polynomial ring R = K[p]. To simplify the notation, set

Γ := {x c : c ∈ V (G) \ {v, w}} and Λ := {x c : c ∈ V (H) \ {v, w}}.
Pick an arbitrary element u in G(J(L)). Due to neither G nor H is a tree or triangle, we can deduce that supp(u) ∩ Γ = ∅ and supp(u) ∩ Λ = ∅. It follows also from

J(L) = J(G) ∩ J(H) that u = ab/gcd(a, b) for some a ∈ G(J(G)) and b ∈ G(J(H)).
Consequently, one of the following cases occurs: Case 1. a = x v f and b = x v g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). Because gcd(a, b) = x v , we have u = x v f g. Case 2. a = x v f and b = x w g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). Since gcd(f, g) = 1, this implies that Case 7. a = x v x w f and b = x v g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). Similarly to Case 6, one can conclude that u = x v x w f g. Case 8. a = x v x w f and b = x w g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). On account of gcd(a, b) = x w , this gives that u = x v x w f g. Case 9. a = x v x w f and b = x v x w g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). According to gcd(a, b) = x v x w , we get u = x v x w f g.

u = x v x w f g. Case 3. a = x v f and b = x v x w g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). By gcd(a, b) = x v , this yields that u = x v x w f g.
It follows from the discussion above that if u ∈ G(J(L)), then u = Af g, where

A = 1, A|x v x w , and f (respectively, g) is a square-free monomial in the vari- ables Γ (respectively, Λ) such that Af ∈ J(G) (respectively, Ag ∈ J(H)). Since p ∈ Ass R (R/J(L) s ), we obtain p = (J(L) s : R h) for some monomial h in R. Sup- pose that h = h 1 h 2 x ρ v x δ w with h 1 (respectively, h 2
) is a monomial in the variables Γ (respectively, Λ), and ρ (respectively, δ) is a nonnegative integer. One can concludes from Lemma 2.1.12 that (J(G)

s : R h 1 h 2 x ρ v x δ w ) = (J(G) s : R h 1 x ρ v x δ w ) and also (J(H) s : R h 1 h 2 x ρ v x δ w ) = (J(H) s : R h 2 x ρ v x δ w ). Thanks to (J(L) s : R h) ⊆ (J(G) s : R h) (respectively, (J(L) s : R h) ⊆ (J(H) s : R h)), this implies that p ⊆ (J(G) s : R h 1 x ρ v x δ w ) (respectively, p ⊆ (J(H) s : R h 2 x ρ v x δ w )). Our purpose is to verify h 1 x ρ v x δ w / ∈ J(G) or h 2 x ρ v x δ w / ∈ J(H). Assume to the contrary that h 1 x ρ v x δ w ∈ J(G) and h 2 x ρ v x δ w ∈ J(H). Because of h 1 x ρ v x δ w ∈ J(G) s (respectively, h 2 x ρ v x δ w ∈ J(H) s
), this gives that there exist square-free monomials M 1 , f 1 , . . . , f s (respectively, M 2 , g 1 , . . . , g s ) in the variables Γ (respectively, Λ), and A 1 , . . . , A s (respectively, A ′ 1 , . . . , A ′ s ) with

A i = 1 (respectively, A ′ i = 1), A i |x v x w (respectively, A ′ i |x v x w ), and A i f i ∈ J(G) (respectively, A ′ i g i ∈ J(H)) for each i = 1, . . . , s, such that we have h 1 x ρ v x δ w = ( s i=1 A i f i )M 1 x θ 1 v x θ 2 w (respectively, h 2 x ρ v x δ w = ( s i=1 A ′ i g i )M 2 x λ 1 v x λ 2 w ),
for some nonnegative integers θ 1 and θ 2 (respectively, λ 1 and λ 2 ). We thus get

( s i=1 A i )x θ 1 v x θ 2 w = x ρ v x δ w = ( s i=1 A ′ i )x λ 1 v x λ 2 w , h 1 = ( s i=1 f i )M 1 , and h 2 = ( s i=1 g i )M 2 .
This gives rise to the following equalities

h 1 h 2 x ρ v x δ w = ( s i=1 f i )M 1 ( s i=1 g i )M 2 x ρ v x δ w = ( s i=1 f i g i )M 1 M 2 x ρ v x δ w . Due to ( s i=1 A i )x θ 1 v x θ 2 w = x ρ v x δ w = ( s i=1 A ′ i )x λ 1 v x λ 2 w , we obtain h 1 h 2 x ρ v x δ w ∈ J(L) s . Hence, h ∈ J(L) s ,
which contradicts the fact that p = (J(L) s : R h). Accordingly, 107

one can derive h 1 x ρ v x δ w / ∈ J(G) or h 2 x ρ v x δ w / ∈ J(H). On account of p is the maximal ideal, this implies that p ∈ Ass R 1 (R 1 /J(G) s ) or p ∈ Ass R 2 (R 2 /J(H) s )
, and so we get the following equality

(3.3.3) Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ).
We can now combine together Theorem 3. As an application of Theorem 3.3.25, we express the following proposition. (ii) The assertion can be deduced from Theorem 3.3.25 and part (i).

The next question examines the relation between assoicated primes of powers of the cover ideal of the union of two finite simple connected graphs with the assoicated primes of powers of the cover ideals of each of them, in a general case.

Question 3.3.27. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple connected graphs. Let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (H) and E(L) := E(G) ∪ E(H).
Then can we deduce that, for all s, one of the following statements holds?

(i) Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ). (ii) Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(H) s ) ∪ {m}, where R 1 = K[x α : α ∈ V (G)], R 2 = K[x α : α ∈ V (H)], R = K[x α : α ∈ V (L)] over a field K, and m = (x α : α ∈ V (L)) is the unique homogeneous maximal ideal of R.
The answer is negative. We provide a counterexample. Let G := K 4 be the complete graph with V (G) = {1, 2, 3, 4} and

E(G) = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}},
and H := S 4 be the star graph with V (H) = {1, 2, 3, 4, 5} and

E(H) = {{1, 5}, {2, 5}, {3, 5}, {4, 5}}. Let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (H) and E(L) := E(G) ∪ E(H)
. By using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], we obtain

(x 2 , x 3 , x 4 , x 5 ) ∈ Ass(J(L) 3 ) \ (Ass(J(G) 3 ) ∪ Ass(J(H) 3 )),
and

(x 1 , x 2 , x 4 , x 5 ) ∈ Ass(J(L) 4 ) \ (Ass(J(G) 4 ) ∪ Ass(J(H) 4 ) ∪ m),
where m = (x 1 , x 2 , x 3 , x 4 , x 5 ). In view of [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6] and Theorem 3.2.42, we deduce that the cover ideal of every tree has the strong persistence property, and so J(H) has the strong persistence property. On the other hand, note that the graph L is exactly the complete graph K 5 . It follows also from Corollary 7.1.6 that the cover ideal of every complete graph is normal, and by virtue of Theorem 3.2.34, one can cnclude that J(G) and J(L) have the strong persistence property.

In the subsequent question, our aim is to investigate the relation between assoicated primes of powers of the cover ideal of the union of a finite simple connected graph and a complete graph with the assoicated primes of powers of the cover ideals of each of them, under the condition that they are common in a path with length 2.

(G)∩E(K n )| = 2. Let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (K n ) and E(L) := E(G) ∪ E(K n ). Then can we conclude that Ass R (R/J(L) s ) = Ass R 1 (R 1 /J(G) s ) ∪ Ass R 2 (R 2 /J(K n ) s ), for all s, where R 1 = K[x α : α ∈ V (G)], R 2 = K[x α : α ∈ V (K n )], and R = K[x α : α ∈ V (L)] over a field K? In particular, if J(G)

has the strong persistence property, then does J(L) have the strong persistence property?

By giving a counterexample, we show that the answer is negative. To do this, consider the graph G = (V (G), E(G)), the left graph in the figure below, with

V (G) = {1, 2, 3, 4, 5, 6} and
E(G) = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 1}, {5, 2}, {4, 6}, {5, 6}},
and also the graph L = G ∪ K 3 , the right graph in the figure below, with

V (K 3 ) = {1, 6, 5}, E(K 3 ) = {{5, 6}, {1, 6}, {1, 5}}, V (L) = {1, 2, 3, 4, 5, 6}, and
V (L) = {{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 1}, {5, 2}, {1, 6}, {5, 6}, {4, 6}}.

G

It is routine to check that

J(G) =(x 1 , x 2 ) ∩ (x 2 , x 3 ) ∩ (x 3 , x 4 ) ∩ (x 4 , x 5 ) ∩ (x 5 , x 1 ) ∩ (x 5 , x 2 ) ∩(x 4 , x 6 ) ∩ (x 5 , x 6 ) =(x 2 x 4 x 5 , x 2 x 3 x 5 x 6 , x 1 x 3 x 5 x 6 , x 1 x 2 x 4 x 6 , x 1 x 3 x 4 x 5 ),
and

J(L) = (x 1 , x 2 ) ∩ (x 2 , x 3 ) ∩ (x 3 , x 4 ) ∩ (x 4 , x 5 ) ∩ (x 5 , x 1 ) ∩ (x 5 , x 2 ) ∩(x 1 , x 6 ) ∩ (x 5 , x 6 ) ∩ (x 4 , x 6 ) = (x 2 x 4 x 5 x 6 , x 2 x 3 x 5 x 6 , x 1 x 3 x 5 x 6 , x 1 x 2 x 4 x 6 , x 1 x 3 x 4 x 5 , x 1 x 2 x 4 x 5 ).
Using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF] yields that

(x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) ∈ Ass(J(L) 3 ) \ (Ass(J(G) 3 ) ∪ Ass(J(K 3 ) 3 )).
On the other hand, since the graph G is the union of a cycle graph with one chord and a triangle, Proposition 3.3.26(ii) and Lemma 3.3.24 imply that J(G) has the strong persistence property. Furthermore, one can write

J(L) = x 5 (x 2 x 4 x 6 , x 2 x 3 x 6 , x 1 x 3 x 6 , x 1 x 3 x 4 , x 1 x 2 x 4 ) + (x 1 x 2 x 4 x 6 ). Because F := (x 2 x 4 x 6 , x 2 x 3 x 6 , x 1 x 3 x 6 , x 1 x 3 x 4 , x 1 x 2 x 4 ) is the cover ideal of the odd cycle H with V (H) = {1, 2, 3, 4, 6} and E(H) = {{1, 2}, {2, 3}, {3, 4}, {4, 6}, {6, 1}}, Proposition 3.3.26(i)
gives that F has the strong persistence property, and by virtue of x 1 x 2 x 4 x 6 ∈ F , it follows from Lemma 3.1.1 that J(L) = x 5 F + (x 1 x 2 x 4 x 6 ) has the strong persistence property. This finishes our discussion.

Chapter 4

On the symbolic strong persistence property of monomial ideals

Some results on the symbolic strong persistence property

In this section, our aim is to prove that any square-free monomial ideal satisfies the symbolic strong persistence property. To achieve this, we start with the definition of symbolic powers of an ideal. It should be noted that the results of this section can be found in [START_REF] Nasernejad | The strong persistence property and symbolic strong persistence property[END_REF].

I (n) = q 1 ∩ • • • ∩ q r ,
where q i is the primary component of I n corresponding to p i .

Remark 4.1.2. In much literature on symbolic powers a different definition is used based on all the primary decomposition, not just the minimal primes, that is,

I (n) = p∈Min(I) (I n R p ∩ R) and I (n) = p∈Ass R (R/I) (I n R p ∩ R).
See for example page 1 of [32] for more information. If all associated primes of I are minimal (as is the case with square-free monomial ideals), then the two definitions of I (n) are the same. Otherwise, they are different. In general, both definitions have advantages. When we take p ranging over Ass R (R/I), we get I (1) = I, while taking p ranging over Min(I) means that I (n) coincides with the intersection of the primary components of I n corresponding to its minimal primes.

Here, we list some properties of symbolic powers in the following remark. (1) I (1) = I;

(2) For all n ≥ 1, The proposition below investigates the symbolic strong persistence property of powers of ideals. Proposition 4.1.5. Let R be a commutative Noetherian ring and I be an ideal of R such that I has a non-zerodivisor element. Then there exists a positive integer s such that I s has the symbolic strong persistence property.

I n ⊆ I (n) ; (3) I a ⊆ I (b) if
Proof. According to Proposition 3.2.6, one can conclude that there exists a positive integer s such that I s is a superficial ideal for I s , that is, I s has the strong persistence property. Now, the claim follows readily from [START_REF] Reyes | On the strong persistence property for monomial ideals[END_REF]Theorem 11].

The following proposition says that if an ideal has the symbolic strong persistence property, then any power of it has the symbolic strong persistence property as well. Proposition 4.1.6. Let I be an ideal in a commutative Noetherian ring R such that I has the symbolic strong persistence property. Then I s has the symbolic strong persistence property for all positive integers s. Proof. Fix s, k ≥ 1. It suffices to prove that ((I s ) (k+1) : R (I s ) (1) ) = (I s ) (k) . Let

Min(I) = {p 1 , . . . , p r }. Set S := R \ r i=1 p i . It follows from [133, Proposition 4.3.23] that I (n) = S -1 I n ∩ R for all n ≥ 1. Since Min(I n ) = Min(I) for all n ≥ 1,
this implies the following equalities

(I s ) (n) = S -1 (I s ) n ∩ R = S -1 I sn ∩ R = I (sn) .
Accordingly, one obtains (I s ) (k+1) = I (sk+s) , (I s ) (k) = I (sk) , and (I s ) (1) = I (s) . Therefore, we get ((I s ) (k+1) : R (I s ) (1) ) = (I (sk+s) : R I (s) ). Because the ideal I has the symbolic strong persistence property, it follows from [START_REF] Reyes | On the strong persistence property for monomial ideals[END_REF]Proposition 12] that (I (sk+s) : R I (s) ) = I (sk) , and so ((I s ) (k+1) : R (I s ) (1) ) = (I s ) (k) , as claimed.

Here, we turn our attention to studying the symbolic strong persistence property of monomial ideals. In fact, one may ask the following question: Question 4.1.7. Does every monomial ideal satisfy the symbolic strong persistence property?

The answer is negative. To see this, we come back to Proposition 3.3.13. Indeed, we proved that for the monomial ideal

I n = (x n , x n-1 y, xy n-1 , y n ) in the polynomial ring R = Q[x, y], n ≥ 4, one has (I a+2 n : I n ) ⊋ I a+1 n for 0 ≤ a ≤ n -4.
On the other hand, since I n = (x, y n ) ∩ (x n-1 , y n-1 ) ∩ (x n , y), we can deduce that I n is a (x, y)-primary monomial ideal. This yields that we have

I (k) n = I k n for all k ≥ 1. Therefore, we get (I (a+2) n : I (1) n ) ⊋ I (a+1) n
for 0 ≤ a ≤ n -4. This means that I n does not satisfy the symbolic strong persistence property.

Based on [START_REF] Reyes | On the strong persistence property for monomial ideals[END_REF]Theorem 11], the strong persistence property implies the symbolic strong persistence property. Moreover, in view of Proposition 3.1.12, one can conclude that the strong persistence property implies the persistence property. Now, one may ask the following questions: (ii) Does normally torsion-freeness imply the symbolic strong persistence property?

Our answers are negative. To accomplish this, consider the monomial ideal y)} for all k ≥ 1, one can conclude that I has the persistence property, and also is normally torsion-free. While, by the argument which has been mentioned before, we obtain I does not satisfy the symbolic strong persistence property.

I n = (x n , x n-1 y, xy n-1 , y n ) in the polynomial ring R = Q[x, y], n ≥ 4. Since Ass R (R/I k n ) = {(x,
In the following, our intent is to show that every square-free monomial ideal satisfies the symbolic strong persistence property. To achieve this, we require Proposition 4.1.9 and Lemma 4.1.10. Proposition 4.1.9. Every power of a primary monomial ideal is primary.

Proof. Assume that Q is a primary monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K, and fix t ≥ 1. Let x i |u 1 • • • u t , where 1 ≤ i ≤ n and u 1 , . . . , u t ∈ G(Q). This yields that x i | u s for some 1 ≤ s ≤ t.
Since Q is a primary monomial ideal, it follows from [133, Proposition 6.1.7] that there exists a positive integer k such that x k i ∈ G(Q), and hence

x kt i ∈ G(Q t ). By setting α := kt, one has x α i ∈ G(Q t ).
Once again, [133, Proposition 6.1.7] implies that Q t is a primary monomial ideal, as required. Lemma 4.1.10. Let Q 1 , . . . , Q r be primary monomial ideals in a polynomial ring

R = K[x 1 , . . . , x n ] over a field K such that √ Q i = Q j for 1 ≤ i = j ≤
Q k+1 i : R r i=1 Q i ) = r i=1 Q k i .
Proof. We give a sketch of the proof. Assume that Q i has the strong persistence property for each i = 1, . . . , r. Fix k ≥ 1, and pick u in

( r i=1 Q k+1 i : R r i=1 Q i ). Put p i := √ Q i for each i = 1, . . . , r. Fix 1 ≤ j ≤ r. One can choose an element such as v i ∈ Q i \ p j for 1 ≤ i = j ≤ r. Let λ be an arbitrary element in Q j . There- fore, uλv 1 • • • v j-1 v j+1 • • • v r ∈ r i=1 Q k+1 i ⊆ Q k+1 j
, and also

v 1 • • • v j-1 v j+1 • • • v r / ∈ p j . Since Q j is primary, Proposition 4.1.9 gives that Q k+1 j is primary. Hence, uλ ∈ Q k+1 j , and so u ∈ (Q k+1 j : R Q j ) = Q k j . Therefore, ( r i=1 Q k+1 i : R r i=1 Q i ) ⊆ r i=1 Q k i . To prove the reverse inclusion, one should note that ( r i=1 Q i )( r i=1 Q k i ) ⊆ r i=1 Q k+1 i , and so r i=1 Q k i ⊆ ( r i=1 Q k+1 i : R r i=1 Q i ).
This finishes our argument.

We are now in a position to express the main result of this subsection. We can now ask the following question: Question 4.1.12. Is there an ideal satisfying the symbolic strong persistence propoerty but does not have the strong persistence property?

The answer is positive. We give such an ideal. Consider the following square-free monomial ideal I in the polynomial R = K[x 1 , . . . , x 12 ] over a field K,

I = (x 1 x 3 x 6 x 8 x 9 x 10 x 11 x 12 , x 2 x 4 x 5 x 7 x 9 x 10 x 11 x 12 , x 1 x 2 x 4 x 5 x 7 x 10 x 11 x 12 , x 2 x 3 x 5 x 6 x 8 x 9 x 11 x 12 , x 1 x 2 x 3 x 6 x 8 x 9 x 11 x 12 , x 2 x 4 x 5 x 6 x 7 x 9 x 11 x 12 , x 1 x 3 x 6 x 7 x 8 x 9 x 10 x 12 , x 2 x 3 x 5 x 7 x 8 x 9 x 10 x 12 , x 2 x 3 x 4 x 5 x 7 x 9 x 10 x 12 , x 1 x 3 x 4 x 5 x 6 x 7 x 10 x 12 , x 1 x 2 x 3 x 4 x 5 x 7 x 10 x 12 , x 1 x 3 x 4 x 6 x 8 x 9 x 10 x 11 , x 1 x 2 x 4 x 5 x 7 x 8 x 10 x 11 , x 1 x 3 x 4 x 5 x 6 x 8 x 10 x 11 , x 1 x 2 x 4 x 6 x 7 x 8 x 9 x 11 , x 1 x 2 x 3 x 4 x 6 x 8 x 9 x 11 ).
Furthermore, using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF] implies that (I 4 : R I) = I 3 . We thus get I does not satisfy the strong persistence property, whereas Theorem 4.1.11 shows that I satisfies the symbolic strong persistence property.

We terminate this argument with the following corollary. In fact, it has been established in Theorem 3.2.42, and we now re-prove it by using Theorem 4.1.11.

Symbolic strong persistence property under monomial operations

The aim of this section is to state some methods for constructing new monomial ideals which have the symbolic strong persistence property based on the monomial ideals which have the symbolic strong persistence property. It should be noted that the results of this section can be found in [START_REF] Khashyarmanesh | Symbolic strong persistence property under monomial operations and strong persistence property of cover ideals[END_REF].

In what follows, we first show that a monomial ideal has the symbolic strong persistence property if and only if its weighted ideal has the symbolic strong persistence property. To see this, one requires to recall the definition of weighted ideals, see Definition 3.2.11.

The following proposition and lemma are necessary for us to prove the subsequent theorem.

(I W : R J W ) = L W .
Proof. This result is a straightforward consequence of the fact that gcd and lcm of two monomials behave well with respect to taking weights. Lemma 4.2.2. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K, and W a weight over R.

Then (I (k) ) W = (I W ) (k) for all k ∈ N.
Proof. It is straightforward, and can be deduced by definitions.

We are ready to state one of the main results of this section in the following theorem. In what follows, our goal is to establish that if a monomial ideal has the symbolic strong persistence property, then its contracted ideal has the symbolic strong persistence property as well, see Definition 3.2.19. To accomplish this, we first need to prove some auxiliary results as follows. Proposition 4.2.4. Let A be a finite set of irreducible monomial ideals whose radicals are mutually incomparable with respect to inclusion. Then Q∈A Q is a minimal primary decomposition.

Proof. Assume that

A := {Q 1 , . . . , Q r }. Let p i = √ Q i for all i.
If there exists j such that i̸ =j Q i ⊆ Q j , then i̸ =j p i ⊆ p j , which implies that p i ⊆ p j for some i = j, a contradiction. Lemma 4.2.5. Let I and J be monomial ideals in R = K[x 1 , . . . , x n ] over a field K, and

1 ≤ i ≤ n. Then (i) If I = Q 1 ∩ • • • ∩ Q r is a minimal primary decomposition of I, then I/x i = x i / ∈G( √ Q j ) Q j and (I/x i ) (1) = x i / ∈G( √ Q j ), √ Q j ∈Min(I) Q j . (ii) I (k) /x i = (I/x i ) (k) for all k ∈ N. Proof. (i) If I = Q 1 ∩• • •∩ Q r is a minimal primary decomposition of I, then Lemma 3.2.20 implies that I/x i = Q 1 /x i ∩ • • • ∩ Q r /x i .
In addition, for each j = 1, . . . , r, we have

Q j /x i =    R if x i ∈ G( Q j ) Q j if x i / ∈ G( Q j ).
Accordingly, we get

I/x i = x i / ∈G( √ Q j ) Q j .
Based on Proposition 4.2.4, one can conclude the minimal primary decomposition of I/x i . Finally, since p ∈ Min(I/x i ) if and only if p ∈ Min(I) and x i / ∈ G(p), we gain the following equality

(I/x i ) (1) = x i / ∈G( √ Q j ), √ Q j ∈Min(I) Q j .
(ii) Assume that

I k = Q 1 ∩ • • • ∩ Q r is
a minimal primary decomposition of I k , where Q j is a minimal prime of I if and only if j ≤ r ′′ , and

x i / ∈ G( Q j ) if and only if j ≤ r ′ ≤ r ′′ . We thus have I (k) = Q 1 ∩ • • • ∩ Q r ′′ . By Lemma 3.2.20, one can conclude that (I/x i ) k = I k /x i = Q 1 /x i ∩ • • • ∩ Q r /x i .
Furthermore, part (i) and the fact that L (k) = (L k ) (1) for every monomial ideal L, yield that

(I/x i ) (k) = Q 1 ∩ • • • ∩ Q r ′ = I (k) /x i .
This terminates the proof.

We are now in a position to express another main result of this section in Theorem 4.2.6. Theorem 4.2.6. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K, and 1 ≤ i ≤ n. If I has the symbolic strong persistence property, then I/x i has the symbolic strong persistence property.

Proof. Suppose that I has the symbolic strong persistence property. Fix k ≥ 1. In the light of (I (k+1) : R I (1) ) = I (k) , it follows from Lemma 3.2.20(iv) that (I (k+1) /x i : R/x i I (1) /x i ) = I (k) /x i . By virtue of Lemma 4.2.5(ii), one can deduce that ((I/x i ) (k+1) : R/x i (I/x i ) (1) ) = (I/x i ) (k) . That is, I/x i has the symbolic strong persistence property. This completes the proof. To see an application of Theorem 4.2.7, one can consider Question 4.2.10. To do this, we begin with the definition of monomial ideals of clutter type in the following definition. Definition 4.2.8. Let I be a non-square-free monomial ideal in a polynomial ring

R = K[x 1 , . . . , x n ] over a field K with G(I) = {u 1 , . . . , u r }. We say that I is of clutter type if √ u i ∤ √ u j (or equivalently, supp(u i ) ⊈ supp(u j )) for each 1 ≤ i = j ≤ r.
Example 4.2.9. The answer is negative. We provide a counterexample. To accomplish this, consider the monomial ideal

Let I = (x 1 x 2 2 x 3 , x 2 x 2 3 x 4 , x 3 x 2 4 x 5 , x 4 x 2 5 x 1 , x 5 x 2 1 x 2 ) be a monomial ideal in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 ,
I = (x 4 1 x 3 , x 3 1 x 2 x 4 , x 1 x 3 2 x 5 , x 4 2 x 6 ) in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ]
over a field K. It is easy to check that I is a non-squarefree monomial ideal of clutter type. Furthermore, using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF] implies that p = (x 1 , x 2 ) ∈ Min(I). On account of Theorem 4.2.7, one can deduce that

I(p) = (x 4 1 , x 3 1 x 2 , x 1 x 3 2 , x 4
2 ) has the symbolic strong persistence property, whereas the monomial ideal I(p) does not satisfy the symbolic strong persistence property since (I(p) (2) : I(p) (1) ) = I(p) (1) . This finishes our argument.

Finally, we want to examine the symbolic strong persistence property under expansion operation, see Definition 2.2.9. Indeed, we show that a monomial ideal has the symbolic strong persistence property if and only if its expansion has the symbolic strong persistence property.

Let us provide the other main result of this section in the subsequent theorem. This means that I * has the symbolic strong persistence property. Conversely, assume that I * has the symbolic strong persistence property. Accordingly, we have

((I * ) (k+1) : R * (I * ) (1) ) = (I * ) (k) . On account

Some relations between SPP and SSPP

In this section, we give some relations between the strong persistence property (SPP for short) and symbolic strong persistence property (SSPP for short). It should be noted that the results of this section can be found in [START_REF] Khashyarmanesh | Symbolic strong persistence property under monomial operations and strong persistence property of cover ideals[END_REF]. We first show Theorem 4. :

Sp 1 I p 1 ) = I k p 1 for all k ≥ 1. Fix k ≥ 1.
Also, for all s, suppose that

I s = Q s,1 ∩ • • • ∩ Q s,t ∩ Q s,t+1 ∩ • • • ∩ Q s,rs ,
is a minimal primary decomposition of I s with Q s,i = p i for each i = 1, . . . , t, and Q s,i is not minimal for each i = t + 1, . . . , r s . We therefore have

I (1) = t i=1 Q 1,i , I (k) = t i=1
Q k,i , and

I (k+1) = t i=1 Q k+1,i .
Since I has the symbolic strong persistence property, one has (I (k+1) : S I (1) ) = I (k) , and hence (I

(k+1) p 1 : Sp 1 I (1) p 1 ) = I (k) p 1 .
It follows also from Proposition 3.3.17 that, for all s, (I s ) p 1 = (Q s,1 ) p 1 and (I (s) ) p 1 = (Q s,1 ) p 1 . We thus gain (I k+1

p 1 : Sp 1 I p 1 ) = I k p 1 , as required.
(⇐) Assume that I p has the strong persistence property for all p ∈ Min(I). Want to show that (I (k+1) : S I (1) ) = I (k) for all k ≥ 1. To accomplish this, fix k ≥ 1. Our strategy is to use [START_REF] Matsumura | Commutative Ring Theory[END_REF]Exercise 6.4]. For this purpose, one has to prove that (I (k+1) q

: Sq I (1) q ) = I (k) q
for all q ∈ Ass S (S/I (k) ). With the notation which has been used in the proof of the forward implication, and by considering the fact that Ass S (S/I (s) ) = Min(I s ) = Min(I) for all s, one can deduce that Ass S (S/I (k) ) = {p 1 , . . . , p t }. Without loss of generality, we need only demonstrate that (I

(k+1) p 1 : Sp 1 I (1) p 1 ) = I (k)
p 1 . Since I p 1 has the strong persistence property, we obtain (I k+1

p 1 : Sp 1 I p 1 ) = I k p 1 . As (I s ) p 1 = (Q s,1 ) p 1 and (I (s) ) p 1 = (Q s,1
) p 1 for all s, one derives that (I

(k+1) p 1 : Sp 1 I (1) p 1 ) = I (k)
p 1 . This terminates the proof.

To show the next corollary, one needs the following proposition. In fact, we showed this result in Corollary 3.3.3, but we re-prove it here by a different way. Proposition 4.3.2. Every irreducible primary monomial ideal has the strong persistence property. In particular, every monomial prime ideal has the strong persistence property.

Proof. Fix k ≥ 1, and assume that Q = (x α 1 i 1 , . . . , x αt it ) is an irreducible primary monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K with α 1 , . . . , α t are positive integers and {x i 1 , . . . , x it } ⊆ {x 1 , . . . , x n }. We have to prove that

(Q k+1 : R Q) = Q k for all positive integers k. Fix k ≥ 1. Because Q k ⊆ (Q k+1 : R Q), it remains to show that (Q k+1 : R Q) ⊆ Q k .
To do this, one needs to verify that

(Q k+1 : R (x α 1 i 1 )) ⊆ Q k .
Here, inspired by Remark 2.3.5, one can deduce the following equalities,

Q k+1 : R (x α 1 i 1 ) = θ 1 +•••+θt=k+1 ((x α 1 i 1 ) θ 1 • • • (x αt it ) θt ) : R (x α 1 i 1 ) = θ 1 +•••+θt=k+1 ((x α 1 i 1 ) θ 1 • • • (x αt it ) θt ) : R (x α 1 i 1 ) = θ 1 =0, θ 2 +•••+θt=k+1 ((x α 2 i 2 ) θ 2 • • • (x αt it ) θt ) + θ 1 ≥1, θ 1 +•••+θt=k+1 ((x α 1 i 1 ) θ 1 -1 • • • (x αt it ) θt ). As ((x α 2 i 2 ) θ 2 • • • (x αt it ) θt ) ⊆ Q k with θ 2 +• • •+θ t = k +1 and ((x α 1 i 1 ) θ 1 -1 • • • (x αt it ) θt ) ⊆ Q k with θ 1 ≥ 1, θ 1 +• • •+θ t = k+1, we conclude that (Q k+1 : R (x α 1 i 1 )) ⊆ Q k , as required.
I = Q 1 ∩ • • • ∩ Q m ∩ Q m+1 ∩ • • • ∩ Q r be a minimal primary decomposition of I such that Min(I) = {p 1 , . . . , p m }, where p i = √ Q i for each i = 1, . . . , m.
If Q i has the strong persistence property for each i = 1, . . . , m, then I has the symbolic strong persistence property. In particular, every square-free monomial ideal has the symbolic strong persistence property.

Proof. Fix k ≥ 1. By virtue of Proposition Proposition 3.3.17, we get

I p i = (Q i ) p i for each i = 1, . . . , m. Since Q i has the strong persistence property for each i = 1, . . . , m, one has (Q k+1 i : S Q i ) = Q k i for each i = 1, . . . , m. Thus, ((Q i ) k+1 p i : Sp i (Q i ) p i ) = (Q i ) k p i
for each i = 1, . . . , m. This implies that I p i has the strong persistence property for each i = 1, . . . , m. Now, the claim follows from Theorem 4.3.1. The last assertion is a direct consequence from Proposition 4.3.2, which says that every monomial prime ideal has the strong persistence property.

We are ready to present the next result in the following theorem.

Theorem 4.3.4. Suppose that I is a square-free monomial ideal in a polynomial ring

R = K[x 1 , . . . , x n ] over a field K. Also, for all s ≥ 2, suppose that Ass R (R/I s ) = Ass R (R/I) ∪ {q} such that q st is
the primary component of the embedded prime q, where t := min{|{x j 1 , . . . , x j ℓ }| :

x j 1 • • • x j ℓ ∈ I}.
Then I has the strong persistence property.

I s = p s 1 ∩ • • • ∩ p s r ∩ q st is
a minimal primary decomposition of I s for all s ≥ 2. Pick an arbitrary monomial u in (I k+1 : R I). Fix 1 ≤ j ≤ r. Since p i ⊈ p j for 1 ≤ i = j ≤ r, this implies that one can choose an element such as v i ∈ p i \ j̸ =i p j . Let λ be an element in p j . Because u ∈ (I k+1 : R I), we have uI

⊆ I k+1 . By virtue of v 1 • • • v j-1 λv j+1 • • • v r ∈ r i=1 p i and I = r i=1 p i , one gains uλv 1 • • • v j-1 v j+1 • • • v r ∈ r i=1 p k+1 i ∩ q (k+1)t . This yields that uλv 1 • • • v j-1 v j+1 • • • v r ∈ p k+1 j . Due to v i / ∈ p j for 1 ≤ i = j ≤ r, we obtain v 1 • • • v j-1 v j+1 • • • v r / ∈ p j .
Thanks to p k+1 j is primary, one has uλ ∈ p k+1 j , and hence u ∈ (p k+1 j : R p j ). It follows from Proposition 4.3.2 that p j has the strong persistence property, and so u ∈ p k j . Accordingly, we have u ∈ r i=1 p k i . To finish the proof, one requires to establish u ∈ q kt . Without loss of generality, let

x 1 • • • x t ∈ r i=1 p i . This implies that ux 1 • • • x t ∈ I k+1 . We thus have ux 1 • • • x t ∈ q (k+1
)t , and therefore there exists some monomial h ∈ G(q (k+1)t ) such that h|ux

1 • • • x t . Assume that h = x α 1 1 • • • x αn n and u = x β 1 1 • • • x βn n with α i ≥ 0 and β i ≥ 0 for each i. Since h ∈ 121
G(q (k+1)t ), this means that h is a minimal generator of q (k+1)t , and by considering the assumption that q is a monomial prime ideal, one can conclude that

α 1 + • • • + α n = (k + 1)t. On account of h|ux 1 • • • x t , this yields that (k + 1)t ≤ β 1 + • • • + β n + t, and hence kt ≤ β 1 + • • • + β n .
Consequently, one can conclude that u ∈ q kt . This implies that u ∈ I k , and the proof is done.

It has already been proved that the cover ideal of any odd cycle graph satsfies the strong persistence property, see Theorem 3.1.11. As an application of Theorem 4.3.4, we re-prove this fact in Theorem 4.3.6. To do this, we need the following auxiliary lemma. Indeed, in Lemma 4.3.5, we give a minimal primary decomposition of the powers of the cover ideal of any odd cycle graph.

J(C 2n+1 ) s = 2n+1 i=1 (x i , x i+1 ) s ∩ m s(n+1) , where m = (x 1 , . . . , x 2n+1 ) is the unique homogeneous maximal ideal in the polyno- mial ring R = K[x 1 , . . . , x 2n+1 ] over a field K. Proof. Since J(C 2n+1 ) = 2n+1 i=1 (x i , x i+1 ), it is sufficient for us to prove that (4.3.1) ( 2n+1 i=1 (x i , x i+1 )) s = 2n+1 i=1 (x i , x i+1 ) s ∩ m s(n+1) .
Let A (respectively, B) denote the ideal on the left-hand (respectively, right-hand) side of (4.3.1). Fix s ≥ 2. We first show that

A ⊆ B. As 2n+1 i=1 (x i , x i+1 ) ⊆ (x i , x i+1 ) for each i = 1, . . . , 2n + 1, this implies that ( 2n+1 i=1 (x i , x i+1 )) s ⊆ 2n+1 i=1 (x i , x i+1
) s . To complete the argument, it suffices to prove that ( 2n+1 i=1 (x i , x i+1 )) s ⊆ m s(n+1) . To see this, consider a minimal generator u in ( 2n+1 i=1 (x i , x i+1 )) s . We thus have u = s i=1 g i , where each g i is a minimal generator of 2n+1 i=1 (x i , x i+1 ). By virtue of

2n+1 i=1 (x i , x i+1
) is exactly the cover ideal of the odd cycle C 2n+1 and because of any minimal geneator of J(C 2n+1 ) corresponds to a minimal vertex cover set of C 2n+1 , and also by considering the fact that any minimal vertex cover set of C 2n+1 has at least n+1 elements, we can deduce that degu = s i=1 degg i ≥ s(n+1). This implies that u ∈ m s(n+1) . Accordingly, one has A ⊆ B.

We now verify that B ⊆ A. For this purpose, select a minimal generator u in B.

Let u := x ℓ 1 1 • • • x ℓ 2n+1
2n+1 with ℓ i ≥ 0 for each i = 1, . . . , 2n + 1. Our strategy is to use [104, Lemma 3.2]. To accomplish this, one has to demonstrate that n+1) , and so

ℓ i + ℓ i+1 ≥ s for eech i = 1, . . . , 2n + 1 and 2n+1 i=1 [(ℓ i + ℓ i+1 ) -s] ≥ s. Fix 1 ≤ i ≤ 2n + 1. It follows from u ∈ B that u ∈ (x i , x i+1 ) s , and so x α i i x s-α i i+1 |u for some 0 ≤ α i ≤ s. This implies that ℓ i + ℓ i+1 ≥ s. To finish the proof, we establish 2n+1 i=1 [(ℓ i + ℓ i+1 ) -s] ≥ s. Since u ∈ B, one has u ∈ m s(
x θ 1 1 • • • x θ 2n+1 2n+1 |u with 2n+1 i=1 θ i = s(n + 1). By virtue of 2n+1 i=1 ℓ i ≥ 2n+1 i=1 θ i , this leads to 2n+1 i=1 [(ℓ i + ℓ i+1 ) -s] = 2n+1 i=1 (ℓ i + ℓ i+1 ) -s(2n + 1) =2 2n+1 i=1 ℓ i -s(2n + 1) ≥2 2n+1 i=1 θ i -s(2n + 1)
=s.

We therefore have u ∈ A, and thus B ⊆ A, as required.

Theorem 4.3.6. The cover ideal of any odd cycle graph satsfies the strong persistence property.

Proof. Let C 2n+1 denote an odd cycle graph with V (C 2n+1 ) = {1, . . . , 2n + 1} and

E(C) = {{i, i + 1} : i = 1, . . . , 2n + 1}, where 2n + 2 = 1. Fix s ≥ 2. It follows from Lemma 4.3.5 that (4.3.2) J(C 2n+1 ) s = ( 2n+1 i=1 (x i , x i+1 )) s = 2n+1 i=1 (x i , x i+1 ) s ∩ m s(n+1) ,
where m = (x 1 , . . . , x 2n+1 ) is the unique homogeneous maximal ideal in the polynomial ring R = K[x 1 , . . . , x 2n+1 ] over a field K. In other words, (4.3.2) is a minimal primary decomposition of J(C 2n+1 ) s . This gives rise to the following equality

Ass R (R/J(C 2n+1 ) s ) = Ass R (R/J(C 2n+1 )) ∪ {m}.
In the light of any minimal geneator of J(C 2n+1 ) corresponds to a minimal vertex cover set of C 2n+1 , and by remembering the fact that any minimal vertex cover set of C 2n+1 has at least n + 1 elements, and also

J(C 2n+1 ) = p∈Ass R (R/J(C 2n+1 )) p, one can derive t = min{|{x j 1 , . . . , x j ℓ }| : x j 1 • • • x j ℓ ∈ J(C 2n+1 )} = n + 1. Now, Theorem 4.
3.4 gives that J(C 2n+1 ) satisfes the strong persistence property.

To establish Lemma 4.3.9, we require to know the following proposition.

Proposition 4.3.7. Suppose that I 1 and I 2 are two monomial ideals in a polynomial ring

R = K[x 1 , . . . , x n ] over a field K such that G(I 1 ) ⊂ R 1 = K[x 1 , . . . , x m ] and G(I 2 ) ⊂ R 2 = K[x m+1 , . . . , x n ] for some positive integer m, 1 ≤ m < n. Then (I 1 I 2 ) (k) = I (k) 1 I (k) 2
for all k ≥ 1.

123

Proof. Fix k ≥ 1. On account of I 1 and I 2 are generated by disjoint sets of variables, [70, Lemma 1.1] implies that I 1 I 2 = I 1 ∩ I 2 . It follows also from the assumption that Min(I 1 ∩ I 2 ) = Min(I 1 ) ∪ Min(I 2 ). Hence, one can conclude the following equalities

(I 1 ∩ I 2 ) (k) = p∈Min(I 1 ∩I 2 ) ((I 1 ∩ I 2 ) k R p ∩ R) = p∈Min(I 1 ) ((I 1 ) k R p ∩ R) ∩ p∈Min(I 2 ) ((I 2 ) k R p ∩ R) = I (k) 1 ∩ I (k) 2 .
We therefore obtain

(I 1 I 2 ) (k) = (I 1 ∩ I 2 ) (k) = I (k) 1 ∩ I (k) 2 = I (k) 1 I (k) 2 .
The theorem below is crucial for us to show Lemma 4.3.9.

) (I + J) (n) = i+j=n I (i) J (j) .
In order to establish Theorem 4.3.10, one needs to prove the following lemma.

Lemma 4.3.9. Let I be a monomial ideal in a polynomial ring

R = K[x 1 , . . . , x n ] over a field K such that I = I 1 R + I 2 R, where G(I 1 ) ⊂ R 1 = K[x 1 , . . . , x m ] and G(I 2 ) ⊂ R 2 = K[x m+1 , . . . , x n ] for some positive integer 1 ≤ m < n. Then I (s) = s i=1 (I (i) 1 + I (s+1-i) 2
) for all s ∈ N.

Proof. Fix s ∈ N, and set L t :=

I (t) 1 + t-1 i=0 I (i) 1 ∩ I (s-i) 2 with 1 ≤ t ≤ s.
In what follows, want to prove that L t = t i=1 (I

(i) 1 + I (s+1-i) 2
). To achieve this, we proceed by induction on t. One can easily see that the assertion is true for the case in which t = 1. Now, suppose, inductively, that t > 1 and that the result has been proved for all r less than t with t ≤ s. It follows also from the inductive hypothesis that L t-1 = t-1 i=1 (I

(i) 1 + I (s+1-i) 2
). In view of Remark 4.1.3, we get

I (t) 1 ⊆ I (t-1) 1
, and so

I (t) 1 ∩ I (t-1) 1 = I (t) 1 . In addition, if 0 ≤ i ≤ t -2 ≤ s, then s + 1 -t ≤ s -i, and thus I (s-i) 2 ⊆ I (s+1-t) 2
. This implies t-2 i=0

I (i) 1 ∩ I (s-i) 2 ⊆ I (s+1-t) 2
, and so

I (s+1-t) 2 ∩ t-2 i=0 I (i) 1 ∩ I (s-i) 2 = t-2 i=0 I (i) 1 ∩ I (s-i) 2
. Now, by considering the fact that

I (t) 1 ∩ t-2 i=0 I (i) 1 ∩ I (s-i) 2 ⊆ t-2 i=0 I (i) 1 ∩ I (s-i) 2
, we get

I (t) 1 ∩ I (t-1) 1 + I (t-1) 1 ∩ I (s-t+1) 2 + I (t) 1 ∩ t-2 i=0 I (i) 1 ∩ I (s-i) 2 + I (s+1-t) 2 ∩ t-2 i=0 I (i) 1 ∩ I (s-i) 2 = I (t) 1 + I (t-1) 1 ∩ I (s-t+1) 2 + t-2 i=0 I (i) 1 ∩ I (s-i) 2
.

124

Hence, one can derive the following equalities

L t = I (t) 1 + t-1 i=0 I (i) 1 ∩ I (s-i) 2 = I (t) 1 + I (t-1) 1 ∩ I (s-t+1) 2 + t-2 i=0 I (i) 1 ∩ I (s-i) 2 = I (t) 1 ∩ I (t-1) 1 + I (t-1) 1 ∩ I (s-t+1) 2 + I (t) 1 ∩ t-2 i=0 I (i) 1 ∩ I (s-i) 2 + I (s+1-t) 2 ∩ t-2 i=0 I (i) 1 ∩ I (s-i) 2 = (I (t) 1 + I (s+1-t) 2
) ∩ (I

(t-1) 1 + t-2 i=0 I (i) 1 ∩ I (s-i) 2 ) = (I (t) 1 + I (s+1-t) 2 ) ∩ L t-1 = (I (t) 1 + I (s+1-t) 2 ) ∩ t-1 i=1 (I (i) 1 + I (s+1-i) 2 ) = t i=1 (I (i) 1 + I (s+1-i) 2
).

This completes the inductive step, and so the claim has been proved by induction. Especially, one can conclude that L s = s i=1 (I

(i) 1 + I (s+1-i) 2
), and hence s i=1 (I

(i) 1 + I (s+1-i) 2 ) = I (s) 1 + s-1 i=0 I (i) 1 ∩ I (s-i) 2 = s i=0 I (i) 1 ∩ I (s-i) 2
.

On account of G(I 1 ) ⊂ R 1 = K[x 1 , . . . , x m ] and G(I 2 ) ⊂ R 2 = K[x m+1 , . . . , x n ], Proposition 4.3.7 yields that I (i) 1 ∩ I (s-i) 2 = I (i) 1 I (s-i) 2
for each i = 0, . . . , s. Therefore, we get s i=1 (I

(i) 1 + I (s+1-i) 2 ) = s i=0 I (i) 1 I (s-i) 2
. Finally, it follows from Theorem 4.3.8 that s i=1 (I

(i) 1 + I (s+1-i) 2 ) = (I 1 + I 2 ) (s) = I (s) , as required.
We are now ready to express and prove Theorem 4.3.10.

Theorem 4.3.10. Let I be a monomial ideal in a polynomial ring

R = K[x 1 , . . . , x n ] over a field K and G(I) = G 1 ∪ • • • ∪ G r such that {x s : x s |m for some m ∈ G i } ∩ {x t : x t |m for some m ∈ G j } = ∅,
for all 1 ≤ i = j ≤ r. Then (G i )R has the symbolic strong persistence property for some 1 ≤ i ≤ r if and only if I has the symbolic strong persistence property.

Proof. It is enough to show our claim only for r = 2. To accomplish this, let

I be a monomial ideal in R = K[x 1 , . . . , x n ] such that I = I 1 R + I 2 R, where G(I 1 ) ⊂ R 1 = K[x 1 , . . . , x m ] and G(I 2 ) ⊂ R 2 = K[x m+1 , . . . , x n ] for some positive integer 1 ≤ m < 125
n. We first show the forward implication. Without loss of generality, assume that I 1 has the symbolic strong persistence property. Also, let G(I

(1)

1 ) = {v 1 , . . . , v t }. Our main aim is to prove that (I (k+1) : R I (1) ) = I (k) for all k. Now, fix k ∈ N. By virtue of the assumption, one has (I

(i) 1 : R I (1) 1 ) = I (i-1) 1
for all i. By applying Lemma 4.3.9, we have the following equalities

(I (k+1) : R I (1) 1 ) = ( k+1 i=1 (I (i) 1 + I (k+2-i) 2 ) : R I (1) 1 ) = k+1 i=1 t j=1 ((I (i) 1 : R v j ) + (I (k+2-i) 2 : R v j )) = k+1 i=1 t j=1 ((I (i) 1 : R v j ) + I (k+2-i) 2 ) = k+1 i=1 (I (k+2-i) 2 + ∩ t j=1 (I (i) 1 : R v j )) = k+1 i=1 (I (k+2-i) 2 + (I (i) 1 : R I (1) 1 )) = k+1 i=2 (I (k+2-i) 2 + I (i-1) 1 ) = k ℓ=1 (I (k+1-ℓ) 2 + I (ℓ) 1 ) = I (k) .
Thanks to Remark 4.1.3, one can derive I (k) ⊆ (I (k+1) : R I (1) ), and by considering this fact that (I (k+1) : R I (1) ) ⊆ (I (k+1) : R I

(1) 1 ), one has (I (k+1) : R I (1) ) = I (k) , as required.

To establish the converse implication, suppose, on the contrary, that I 1 and I 2 do not satisfy the symbolic strong persistence property. This implies that there exist a positive integer k 1 (respectively, k 2 ) and a monomial m 1 (respectively, m 2 ) such that

m 1 ∈ G(I (k 1 +1) 1 : R I (1) 1 ) \ G(I (k 1 ) 1 ) (respectively, m 2 ∈ G(I (k 2 +1) 2 : R I (1) 2 ) \ G(I (k 2 ) 2
)). Take the nonnegative integer a 1 (respectively, a 2 ) such that m 1 ∈ I

(a 1 ) 1 \ I (a 1 +1) 1 (respectively, m 2 ∈ I (a 2 ) 2 \ I (a 2 +1) 2
). In the light of Remark 4.1.3, we deduce that

a 1 ≤ k 1 -1 (respectively, a 2 ≤ k 2 -1). Put m := m 1 m 2 and b := a 1 + a 2 .
Thus, one has m ∈ I (b) . We claim that m / ∈ I (b+1) . Suppose, on the contrary, that m ∈ I (b+1) . Since, by Theorem 4.3.8,

I (b+1) = b+1 ℓ=0 I (ℓ) 1 I (b+1-ℓ) 2
, one obtains m ∈ I , and

m 2 ∈ I (a 2 ) 2 \ I (a 2 +1) 2 126
that a 1 ≥ ℓ and a 2 ≥ b + 1 -ℓ. This gives that a 1 + a 2 ≥ b + 1, which contradicts the fact that b = a 1 + a 2 . Accordingly, m / ∈ I (b+1) . By setting s 1 := k 1 + 1 + a 2 and s 2 := a 1 + k 2 + 1, one can easily see that s i ≥ a 1 + a 2 + 2 for each i = 1, 2. Hence, s := min{s 1 , s 2 } ≥ a 1 + a 2 + 2. In addition, it is routine to check that if u ∈ G(I

(1) 1 ) (respectively, u ∈ G(I (1)
2 )), then mu ∈ I (s 1 ) (respectively, mu ∈ I (s 2 ) ). Therefore, we get m ∈ (I (s) : R I (1) ) \ I (s-1) , which contradicts the assumption that I has the symbolic strong persistence property, as desired.

The following lemma says that, under certain condition, a monomial ideal has the symbolic strong persistence property if and only if its monomial multiple has the symbolic strong persistence property. Proof. To simplify our notation, set L := hI. We first assume that I has the symbolic strong persistence property. This means that (I (k+1) : R I (1) ) = I (k) for all k. In order to complete the argument, one has to show that (L (k+1) : R L (1)

) = L (k) for all k. Fix k ≥ 1. It follows readily from Proposition 4.3.7 that (hI) (k) = h k I (k) .
In view of (L (k+1) : R h) = (h k+1 I (k+1) : R h) = h k I (k+1) , one can conclude that (L (k+1) : R L (1) ) = (h k I (k+1) : R I (1) ). By virtue of gcd(h, u) = 1 for all u ∈ G(I), we get gcd(h, v) = 1 for all v ∈ G(I (1) ). The assumption yields that (L (k+1) : R L (1) ) = h k I (k) = (hI) (k) . Therefore, (L (k+1) : R L (1) ) = L (k) , and the proof is over.

To establish the converse implication, suppose that L has the symbolic strong persistence property. Want to prove that (I (k+1) : R I (1) ) = I (k) for all k. To do this, fix k ≥ 1. Because of L has the symbolic strong persistence property, we get (L (k+1) : R L (1) ) = L (k) , and hence (h k+1 I (k+1) : R hI (1) ) = h k I (k) . This yields that (h k I (k+1) : R I (1) ) = h k I (k) . Moreover, a similar argument gives that (h k I (k+1) : R I (1) ) = h k (I (k+1) : R I (1) ), and so h k (I (k+1) : R I (1) ) = h k I (k) . This implies that (I (k+1) : R I (1) ) = I (k) , that is, I has the symbolic strong persistence property, as claimed.

Chapter 5

On the normally torsion-freeness of monomial ideals

Normally torsion-freeness of path ideals of directed trees

Let H = (V (H), E(H)) be a simple hypergraph on the vertex set V (H) and edge set E(H). The edge ideal of H, denoted by I(H), is the ideal generated by the monomials corresponding to the edges of H. A hypergraph H is called Mengerian if it satisfies a certain min-max equation, which is known as the Mengerian property in hypergraph theory or as the max-flow min-cut property in integer programming. Algebraically, it is equivalent to I(H) being normally torsion-free, see [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 10.3.15] and [START_REF] Villarreal | Monomial algebras[END_REF]Theorem 14.3.6]. Hence, the notion of normally torsion-freeness is important for both Commutative Algebra and Combinatorics.

Let R be a commutative Noetherian ring and I be an ideal of R. In addition, let Ass R (R/I) be the set of all prime ideals associated to I. An ideal I is called normally torsion-free if Ass(R/I k ) ⊆ Ass(R/I) for all k ≥ 1, see [60, Definition 1.4.5]. Normally torsion-free ideals have been a topic of several papers, however, a few classes of these ideals originate from graph theory. Simis, Vasconcelos, and Villarreal showed in [START_REF] Simis | On the ideal theory of graphs[END_REF] that a finite simple graph is bipartite if and only if its edge ideal is normally torsion-free. An analogue of bipartite graphs in higher dimensions is considered to be a hypergraph that avoids "special odd cycles". Such hypergraphs are called balanced , and a well-known result of Fulkerson, Hoffman, and Oppenheim in [START_REF] Fulkerson | On balanced matrices[END_REF] states that balanced hypergraphs are Mengerian. It follows immediately that the edge ideals of balanced hypergraphs are normally torsion-free. However, unlike the case of bipartite graphs, it should be noted that the converse of this statement is not true. More information related to definitions in hypergraph theory can be found in [15].

The main motivation of this section is a conjecture which has been stated in [16]. In fact, in [16, Conjecture 6.1], made up the conjecture below.

Conjecture 5.1.1. Let Γ = (V Γ , E Γ ) be a rooted tree with the corresponding Alexander dual J 2 . Then Ass R (R/J 2 ) = Ass R (R/J s 2 ) for all s ≥ 1.

Our goal in this section is to provide an affirmative answer to above conjecture (see Theorem 5.1.6). It should be noted that the results of this section can be found in [START_REF] Khashyarmanesh | A note on the Alexander dual of path ideals of rooted trees[END_REF].

Also, for a simple finite graph G = (V (G), E(G)), the symbol N G (v) stands for the neighbor set of a vertex v ∈ V (G).

We first recall some definitions from [START_REF] He | Algebraic properties of the path ideal of a tree[END_REF] and [START_REF] West | Introduction to graph theory[END_REF] which will be needed in the following.

Definition 5.1.2. A directed edge of a graph is an assignment of a direction to an edge of a graph. If {w, u} is an edge, we write (w, u) to denote the directed edge, where the direction is from w to u. A graph is a directed graph if each edge has been assigned a direction. A path of length t in a directed graph is a sequence of vertices i 1 , . . . , i t+1 such that e j = (i j , i j+1 ) is a directed edge for j = 1, . . . , t.

Fix a positive integer t and a directed graph G, the path ideal of G of length t is the following monomial ideal

I t (G) := (x i 1 • • • x i t+1 | i 1 , . . . , i t+1 is a path of G of length t),
and also the Alexander dual of I t (G) is defined as follows:

I t (G) ∨ := i 1 ,...,i t+1 is a path of G of length t (x i 1 , . . . , x i t+1 ).
A tree T can be considered as a directed graph by choosing any vertex of T to be the root of the tree, and assigning to each edge the direction "away" from the root. Since T is a tree, the assignment of a direction will always be possible. A rooted tree T is a tree with one vertex chosen as root. If tree T has no such root, then we say that T is unrooted.

To remember the definition and properties of hypergraphs, which will be needed in the rest of this section, one can refer to Section 1.3. In particular, see [START_REF] Francisco | Powers of square-free monomial deals and combinatorics[END_REF][START_REF] Hà | Monomial ideals, edge ideals of hypergraphs, and their graded betti numbers[END_REF] for more information.

To establish Theorem 5.1.6, one has to apply the notions of parameter ideals and corner-elements which were first introduced in [START_REF] Heinzer | Parametric decomposition of monomial ideals (I)[END_REF]. We recall them in the subsequent definitions. We continue with the subsequent proposition which is essential for us to prove Theorem 5.1.6. Proposition 5.1.5. Let T be a rooted tree with root z on the vertex set {z, 1, . . . , 2k} for some positive integer k, and the edge set

E(T ) = {(z, i), (i, i + k) | i = 1, . . . , k}. Then I 2 (T ) ∨ is normally torsion-free. Proof. By virtue of E(T ) = {(z, i), (i, i + k) | i = 1, . . . , k}, one can conclude that I 2 (T ) = (x z x i x i+k | i = 1, . . . , k), and so I 2 (T ) ∨ = k i=1 (x z , x i , x i+k ). This implies that I 2 (T ) ∨ = x z R ′ + k i=1 (x i , x i+k ), where R ′ = K[x z , x 1 , . . . , x 2k
] over a field K. Now, put p i := (x i , x i+k ) for each i = 1, . . . , k, and L := k i=1 p i . Fix an arbitrary positive integer s. If p ∈ Ass R ′ (R ′ /(I 2 (T ) ∨ ) s ), it follows from [50, Lemma 3.4] that p = (x z ) + q for some q ∈ Ass R (R/L s ), where R = K[x 1 , . . . , x 2k ]. Thus, it is sufficient to show that L is normally torsion-free. As p 1 , . . . , p k are generated by disjoint non-empty subsets of {x 1 , . . . , x 2k }, according to [START_REF] Ratliff | On prime divisors of I n , n large[END_REF]Remark 2.8], for all positive integers r 1 , . . . , r k , one has the following equalities

Ass R (R/p r 1 1 • • • p r k k ) = k i=1 Ass R (R/p i ) = {p 1 , . . . , p k }. Furthermore, [113, Theorem 2.7] implies that p r 1 1 • • • p r k k = p r 1 1 ∩ • • • ∩ p r k k for all positive integers r 1 , . . . , r k . Since L = p 1 ∩ • • • ∩ p k , we obtain Ass R (R/L s ) = {p 1 , . . . , p k } = Ass R (R/L).
This means that L is normally torsion-free, and so I 2 (T ) ∨ is also normally torsionfree, as claimed.

We are now ready to express the main result of this section in the following theorem.

Theorem 5.1.6. Let T be a rooted tree. Then I 2 (T ) ∨ is normally torsion-free.

Proof. Let R = K[x v | v ∈ V (T )
] be the polynomial ring over a field K. To simplify our notation, we put J 2 (T ) := I 2 (T ) ∨ . We argue by induction on n := |V (T )|. If n ≤ 2, then there is nothing to prove. So let n = 3, say V (T ) := {z, 1, 2} with z as the root. If deg T z = 2, then there is nothing to verify. If deg T z = 1, then we get J 2 (T ) = (x z , x 1 , x 2 ). Since Ass(J 2 (T )) = Ass(J 2 (T ) s ) for all s, one can conclude that J 2 (T ) is normally torsion-free. Therefore, the claim holds in the case in which n = 3. Now, suppose, inductively, that n > 3 and that the result has been proved for any rooted tree with the size of its vertex set is less than n. Assume that V (T ) = {z, 1, . . . , n -1}, z is the root of T , and L T denotes the set of leaves of T . If L T = V (T ) \ {z}, then there is nothing to show. Hence, one may assume that

L T = V (T ) \ {z}. Furthermore, if v ′ ∈ N T (z) ∩ L T ,
then the variable x v ′ does not appear in any of the variables of generators of I 2 (T ), and so we can assume that N T (z) ∩ L T = ∅. Therefore, one may consider the following two cases:

Case 1. The tree T has no subtree which is a claw or a path with length three. Then T is a rooted tree with root z on the vertex set {z, 1, . . . , 2k} for some positive integer k, and the edge set E(T ) = {(z, i), (i, i + k) | i = 1, . . . , k}. On account of Proposition 5.1.5, one can conclude that I 2 (T ) ∨ is normally torsion-free, and the proof is over.

Case 2. The tree T has a subtree T 1 which is a claw with V (T 1 ) = {f, v, w, g}, w ∈ L T , and

E(T 1 ) = {(f, v), (v, w), (v, g)},
E(H) = {{x i 1 , x i 2 , x i 3 } : i 1 , i 2 , i 3 is a path of T of length two}.
Also, the edge ideal I(H) of the hypergraph H is equal to the path ideal I 2 (T ). Therefore, Proposition 2.7 in [START_REF] Francisco | Powers of square-free monomial deals and combinatorics[END_REF] implies that

J 2 (T ) = I 2 (T ) ∨ = I(H) ∨ = J(H),
where J(H) denotes the cover ideal of the hypergraph H. Once again, we can make the finite simple hypergraph L corresponding to G, with V (L) = V (G) and

E(L) = {{x j 1 , x j 2 , x j 3 } : j 1 , j 2 , j 3 is a path of G of length two}.
Furthermore, the edge ideal I(L) of the hypergraph L is equal to the path ideal I 2 (G). Thus, Proposition 2.7 in [START_REF] Francisco | Powers of square-free monomial deals and combinatorics[END_REF] yields that

J 2 (G) = I 2 (G) ∨ = I(L) ∨ = J(L),
where J(L) denotes the cover ideal of the hypergraph L. In the following, our aim is to demonstrate that

Ass(J(H) s ) = Ass(J(L) s ) ∪ {(x w , x v , x f )},
for all s. Now, fix s ∈ N. We first show that

Ass(J(L) s ) ∪ {(x w , x v , x f )} ⊆ Ass(J(H) s ).
To do this, let p = (x i 1 , . . . , x ir ) be an arbitrary element of Ass(J(L) s ). It follows from [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11] that p ∈ Ass(K[p]/J(L p ) s ), where

K[p] = K[x i 1 , . . . , x ir ]
and L p is the induced hypergraph of L on the vertex set {i 1 , . . . , i r } ⊆ V (L). As L p = H p , we obtain p ∈ Ass(K[p]/J(H p ) s ). This implies that p ∈ Ass(J(H) s ). By virtue of (x w , x v , x f ) ∈ Min(J(H)), one derives (x w , x v , x f ) ∈ Ass(J(H) s ), and so

Ass(J(L) s ) ∪ {(x w , x v , x f )} ⊆ Ass(J(H) s ).
For completing the argument, it is sufficient to verify the reverse inclusion. To achieve this, assume that p = (x i 1 , . . . , x ir ) is an arbitrary element of Ass(J(H) s ) with {i 1 , . . . , i r } ⊆ V (H). If {i 1 , . . . , i r } ⊆ V (G), similar to discussion above, Lemma 2.11 in [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF] yields that p ∈ Ass(J(L) s ), and the proof is complete. We thus let w ∈ {i 1 , . . . , i r }. We next show that x v ∈ p and

x f ∈ p. Suppose, on the contrary, that x v / ∈ p or x f / ∈ p. Thanks to p ∈ Ass(J(H) s ), we get m p ∈ Ass(J(H) s (p)), where m p is the graded maximal ideal of R(p). Because 1), and so (x w , x v , x f ) / ∈ Ass(J(H)(p)). On account of the only minimal prime ideal which contains x w is (x w , x v , x f ), one can easily see that the variable x w does not appear in any of the variables of generators of (J(H)(p)) s , while x w ∈ m p . This leads to a contradiction, and hence one can conclude that (x w , x v , x f ) ⊆ p. Let r ≤ n -1. Hence, Lemma 2.11 in [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF] implies that p ∈ Ass(K[p]/J(H p ) s ), where K[p] = K[x i 1 , . . . , x ir ]. On the other hand, in view of the proof of Lemma 2.4 in [17], when t = 2, one derives H p is connected. This means that H p is a rooted subtree with the size of its vertex set is less than n. We see immediately from the inductive hypothesis that J(H p ) is normally torsion-free. Therefore, p = (x w , x v , x f ), and the proof is over. To conclude the proof, it is sufficient for us to demonstrate the case r = n is impossible. On the contrary, assume that p is the unique homogeneous maximal ideal of R. It follows from p ∈ Ass(J(H) s ) that there exists a monomial h ∈ R with h / ∈ J(H) s such that p = (J(H) s : h), and so ph ⊆ J(H) s . This implies that x w h ∈ J(H) s , and hence there exist monomials u 1 , . . . , u s ∈ G(J(H)) such that

J(H) s (p) = (J(H)(p)) s , one derives m p ∈ Ass((J(H)(p)) s ). It follows from x v / ∈ p or x f / ∈ p that, in R(p), we have (x w , x v , x f ) = (
x w h = u 1 • • • u s Q for some monomial Q ∈ R. If x w |Q, then h ∈ J(H) s ,
which is a contradiction. Therefore, x w |u k for some 1 ≤ k ≤ s, and thus h ∈ J(H) s-1 . Since J(H) s ⊆ J(L) s , one has p ⊆ (J(L) s : h). If p = (J(L) s : h), then x w h ∈ J(L) s , and so there exists an element A ∈ G(J(L) s ) such that A|x w h. As gcd(A, x w ) = 1, we obtain A|h, and thus h ∈ J(L) s . This leads to a contradiction. Therefore, (J(L) s : h) = R, and hence h ∈ J(L) s . By virtue of h / ∈ J(H) s and p = (J(H) s : h), we get h is a J(H) s -corner-element. It follows from [START_REF] Moore | Monomial Ideals and Their Decompositions[END_REF]Corollary 6.3.3] that J(H) s ⊆ P R (h). This yields that (J(H) s : h) ⊆ (P R (h) : h), and hence p ⊆ (P R (h) : h). Based on [93, Lemma 6.1.3], one has h / ∈ P R (h), and so p = (P R (h) : h).

Let h = x α 1 1 • • • x αn n 132 with nonnegative integers α 1 , . . . , α n . Then P R (h) = (x α 1 +1 1 , . . . , x αn+1 n
). We next prove that h|( n i=1 x i ) s-1 . Suppose, on the contrary, that h ∤ ( n i=1 x i ) s-1 . This implies that there exists some x i in h such that α i ≥ s. Therefore, the exponent of x i in hx i is at least s + 1. Due to hx i ∈ J(H) s , one can conclude that there exist monomials u 1 , . . . , u s ∈ G(J(H)) such that

hx i = u 1 • • • u s Q for some monomial Q in R.
Thanks to, for each j = 1, . . . , s, u j is a square-free monomial, one has the exponent of x i in each u j is at most one. Accordingly, we get x i |Q, and hence h ∈ J(H) s , which is a contradiction. Therefore, h|( n i=1 x i ) s-1 . In the following, our aim is to verify that h = ( n i=1 x i ) s-1 . Since h ∈ J(H) s-1 , [93, Exercise 6.1.12] derives J(H) s-1 ⊈ P R (h), and thus there exists an element

M ∈ G(J(H) s-1 ) such that M / ∈ P R (h). It follows from M ∈ G(J(H) s-1 ) that there exist monomials A 1 , . . . , A s-1 ∈ G(J(H)) such that M = A 1 • • • A s-1 . This implies that x α i +1 i ∤ A 1 • • • A s-1 for each i = 1, . . . , n.
In the light of, for each j = 1, . . . , s -1, A j is a square-free monomial, one obtains the exponent of x i in A j , for each i = 1, . . . , n, is at most one. This yields that α i + 1 > s -1 for each i = 1, . . . , n, and so α i > s -2 for each i = 1, . . . , n. It follows from h|

( n i=1 x i ) s-1 that α i ≤ s -1 for each i = 1, . . . , n. Consequently, α i = s -1 for each i = 1, . . . , n, and thus h = ( n i=1 x i ) s-1 . Because h ∈ J(L) s , [ 93
, Exercise 6.1.12] gives that J(L) s ⊈ (x s 1 , . . . , x s n ), and hence there exist monomials

B 1 , . . . , B s ∈ G(J(L)) such that x s i ∤ B 1 • • • B s for each i = 1, . . . , n.
On the other hand, by Proposition 1.2.1 in [START_REF] Herzog | Monomial Ideals[END_REF], it is routine to check that if u ∈ G(J(L)) with x v |u or x f |u, then u ∈ J(H), but if x v ∤ u and x f ∤ u, then ux w ∈ J(H). Let x v |B i for each i = 1, . . . , ℓ, x f |B i for each i = ℓ + 1, . . . , t, and x v ∤ B i and x f ∤ B i for each i = t + 1, . . . , s. Therefore, we obtain B 1 , . . . , B t ∈ J(H), and B t+1 x w , . . . , B s x w ∈ J(H). This implies that

B 1 • • • B s x s-t w ∈ J(H) s . Since h /
∈ J(H) s , Exercise 6.1.12 in [START_REF] Moore | Monomial Ideals and Their Decompositions[END_REF] yields that J(H) s ⊆ P R (h), and hence J(H) s ⊆ (x s 1 , . . . , x s n ). Accordingly, there exists a positive integer j with 1 ≤ j ≤ n such that

x s j |B 1 • • • B s x s-t w . If j = w, then x s j |B 1 • • • B s , which contradicts that x s i ∤ B 1 • • • B s for each i = 1, . . . , n.
Hence, j = w, and so t = 0. Let B i = α∈W i x α for each i = 1, . . . , s. In view of the definition, we deduce that W i is a minimal vertex cover of L for each i = 1, . . . , s.

As {x g , x f , x v } ∈ E(L), one derives {x g , x f , x v } ∩ {x α | α ∈ W i } = ∅ for each i = 1, . . . , s. Due to, for each i = 1, . . . , s, x v ∤ B i and x f ∤ B i , we get x g |B i for each i = 1, . . . , s. This gives that x s g |B 1 • • • B s .
This leads to a contradiction. So, the case r = n is impossible. Hence, we gain

Ass(J(H) s ) = Ass(J(L) s ) ∪ {(x w , x v , x f )}. Since |V (G)| = |V (T )| -1, the inductive hypothesis implies that J 2 (G) = I 2 (G) ∨
is normally torsion-free, and hence J(L) is so. Accordingly, one can deduce that Ass(J(L) s ) = Ass(J(L)) for all s. Due to Ass(J(H)) = Ass(J(L)) ∪ {(x w , x v , x f )}, we get Ass(J(H) s ) = Ass(J(H)) for all s. Therefore, J(H) is normally torsion-free, and thus J 2 (T ) is so. This completes the inductive step, and so the theorem has been proved by induction.

We end this section with an example that illustrates Theorem 5.1.6. To accomplish this, we consider an important class of rooted trees which are called broom graphs.

Example 5.1.7. ([19,Definition 4.3]) A broom graph Γ of type t consists of a handle, which is a directed path x 0 , . . . , x s , such that every edge in Γ (not on the handle) has the form (x i , y j ) for some i ≥ s + 1 -t and y j ∈ V (Γ) \ {x 0 , . . . , x s }.

Suppose that Γ is a broom graph with x 0 as the root and vertex set

V (Γ) = {x 0 , x 1 , x 2 , x 3 , x 4 ,
E(Γ) = {(x 0 , x 1 ), (x 1 , x 2
), (x 2 , x 3 ), (x 3 , x 4 ), (x 3 , y 1 ), (x 3 , y 2 ), (x 3 , y 3 ), (x 4 , y 4 ), (x 4 , y 5 ), (x 4 , y 6 ), (x 4 , y 7 )}.

Based on Theorem 5.1.6, one can easily see that the square-free monomial ideal

I 2 (Γ) ∨ = (x 0 , x 1 , x 2 ) ∩ (x 1 , x 2 , x 3 ) ∩ (x 2 , x 3 , x 4 ) ∩ (x 2 , x 3 , y 1 ) ∩ (x 2 , x 3 , y 2 ) ∩ (x 2 , x 3 , y 3 ) ∩ (x 3 , x 4 , y 4 ) ∩ (x 3 , x 4 , y 5 ) ∩ (x 3 , x 4 , y 6 ) ∩ (x 3 , x 4 , y 7 ),
is normally torsion-free.

Normally torsion-freeness and the Alexander dual

In this section, our aim is to show that the Alexander dual of the monomial ideal generated by the paths of maximal lengths in an unrooted starlike tree (respectively, a rooted starlike tree) has the persistence property (respectively, is normally torsionfree). It should be noted that the results of this section can be found in [START_REF] Nasernejad | On the Alexander dual of the path ideals of rooted and unrooted trees[END_REF]. Throughout this section, for a simple finite graph G = (V (G), E(G)), the symbol L G stands for the set of leaves of G (i.e., the set of vertices of degree one in G). We also denote the distance between two vertices u and v in V (G) by d(u, v).

We now explain one of the main results in this section in the following theorem.

Theorem 5.2.1. Let T be a rooted starlike tree on the vertex set {z, 1, . . . , n} with root z. Let I be the monomial ideal corresponding to T which is generated by the paths of maximal lengths such that every path is naturally oriented away from z, and corresponding Alexander dual J. Then the ideal J is normally torsion-free.

Proof. Suppose that deg T z = k. Hence, T \{z} has exactly k connected components, say L 1 , . . . , L k , where each component is a path graph with |V (L i )| = h i for any i = 1, . . . , k. Set h 0 := 0 and for all i = 1, . . . , k, let

V (L i ) := {h 1 + • • • + h i-1 + j : j = 1, . . . , h i }.
This implies that E(T ) is given by

{(z, i-1 t=1 h t + 1), ( i-1 t=1 h t + j, i-1 t=1 h t + j + 1) : i = 1, . . . , k, j = 1, . . . , h i -1}.
In the light of Lemma 2.2.5, one can conclude that x z ∈ G(J). Assume that G(J) = {x z , u 1 , . . . , u r }. Fix an arbitrary positive integer s.

If p ∈ Ass R ′ (R ′ /J s ), where R ′ = K[x z , x 1 , . . . ,
x n ] over a field K, it follows from [50, Lemma 3.4] that p = (x z )+q for some q ∈ Ass R (R/I s ), where R = K[x 1 , . . . , x n ] and I = (u 1 , . . . , u r ). Thus, it is sufficient to show that I is normally torsion-free. Now, put

p i := (x h 1 +•••+h i-1 +j : j = 1, . . . , h i ),
for all i = 1, . . . , k. As p 1 , . . . , p k are generated by disjoint non-empty subsets of {x 1 , . . . , x n }, according to [START_REF] Ratliff | On prime divisors of I n , n large[END_REF]Remark 2.8], for all positive integers r 1 , . . . , r k , one has the following equalities

Ass R (R/p r 1 1 • • • p r k k ) = k i=1 Ass R (R/p i ) = {p 1 , . . . , p k }.
Furthermore, [START_REF] Ratliff | On prime divisors of I n , n large[END_REF]Theorem 2.7]

implies that p r 1 1 • • • p r k k = p r 1 1 ∩ • • • ∩ p r k k for all positive integers r 1 , . . . , r k . Since I = p 1 ∩ • • • ∩ p k , we obtain Ass R (R/I s ) = {p 1 , . . . , p k } = Ass R (R/I).
Therefore, I is normally torsion-free, and so J is also normally torsion-free, as claimed.

Corollary 5.2.2. Suppose that T is a rooted symmetric starlike tree on the vertex set V (T ) = {z, 1, . . . , n} with root z such that every path is naturally oriented away from z, and the following edge set

E(T ) = {(z, i), (kj + i, kj + k + i) | i = 1, . . . , k and j = 0, . . . , m -1},
such that n = k(m + 1) for some k ∈ N and m ∈ N 0 . Suppose also that

J m+1 := k i=1 (x z , x i , x k+i , . . . , x mk+i ).
Then J m+1 is normally torsion-free.

In order to verify another main result in this section, we need the following lemma. To recall the notion of the expansion operator on monomial ideals refer to Definition 2.2.9.

Lemma 5.2.3. Let I be a monomial ideal of R. Then I is normally torsion-free if and only if

I * is.
Proof. For necessity, consider p ∈ Ass R (R/I n ) for an arbitrary n ∈ N. Due to Proposition 2.2.10, it follows that p * ∈ Ass R * (R * /(I n ) * ). According to Lemma 2.2.11(iii), we have

(I n ) * = (I * ) n . Hence, p * ∈ Ass R * (R * /(I * ) n ).
By hypothesis, we deduce that p * ∈ Ass R * (R * /I * ). In view of Proposition 2.2.10, one can conclude that p ∈ Ass R (R/I). Therefore, I is a normally torsion-free ideal of R. The sufficiency part follows from in a similar way, and the proof is complete.

In the following example we clarify the importance of Lemma 5.2.3.

Example 5.2.4. Consider the following monomial ideal

J := (x 1,1 x 3,1 x 4,1 , x 1,1 x 3,1 x 4,2 , x 1,1 x 3,2 x 4,1 , x 1,1 x 3,2 x 4,2 , x 1,2 x 3,1 x 4,1 , x 1,2 x 3,1 x 4,2 , x 1,2 x 3,2 x 4,1 , x 1,2 x 3,2 x 4,2 , x 1,1 x 5,1 , x 1,2 x 5,1 , x 1,1 x 5,2 , x 1,2 x 5,2 , x 1,1 x 5,3 , x 1,2 x 5,3 , x 2,1 x 3,1 x 4,1 , x 2,1 x 3,1 x 4,2 , x 2,1 x 3,2 x 4,1 , x 2,1 x 3,2 x 4,2 ),
in the following polynomial ring, where K is a field,

R * = K[x 1,1 , x 1,2 , x 2,1 , x 3,1 , x 3,2 , x 4,1 , x 4,2 , x 5,1 , x 5,2 , x 5,3 ]. Now, set p 1 := (x 1,1 , x 1,2
), p 2 := (x 2,1 ), p 3 := (x 3,1 , x 3,2 ), p 4 := (x 4,1 , x 4,2 ), and p 5 := (x 5,1 , x 5,2 , x 5,3 ). Thus, it is routine to check that

J = p 1 p 3 p 4 + p 1 p 5 + p 2 p 3 p 4 .
Consider the square-free monomial ideal 136

I := (x 1 x 3 x 4 , x 1 x 5 , x 2 x 3 x 4 ) in the polyno- mial ring R = K[x 1 , x 2 , x 3 , x 4 ,
It is clear that G is a bipartite graph with the following edge ideal

I(G) := (x 1 x 2 , x 1 x 3 , x 1 x 4 , x 3 x 5 , x 4 x 5 ),
and the following cover ideal

J(G) := (x 1 , x 2 ) ∩ (x 1 , x 3 ) ∩ (x 1 , x 4 ) ∩ (x 3 , x 5 ) ∩ (x 4 , x 5 ).
One can easily compute that J(G) = I. On the other hand, according to [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6], it follows that J(G) is normally torsion-free. Hence, I is normally torsion-free and Lemma 5.2.3 implies that the ideal J is also normally torsion-free.

The subsequent definition is essential for us to provide Theorem 5.2.7.

Definition 5.2.5. An (k 1 , k 2 , . . . , k r )-banana tree is a graph obtained by connecting one leaf of each of an k i -star graph, for all i = 1, . . . , r, with a single root vertex that is distinct from all the stars.

It should be observed that, for all i = 1, . . . , r, the number k i in the definition of an (k 1 , k 2 , . . . , k r )-banana tree refers to the total number of vertices in the associated star graph. Furthermore, it is necessary to note that when k 1 = • • • = k r = k, we get an (r, k)-banana tree, as defined by Chen et al. [26].

We next give two examples which illustrate our definitions.

Examples 5.2.6. (i) Suppose that T is the tree which is shown in Figure 5. One can easily see that T is a (4, 5, 7)-banana tree. (ii) Assume that T is the tree which is shown in Figure 6. It is routine to check that T is a (3, 5)-banana tree. We are now in a position to prove another main result in this section.

Theorem 5.2.7. Suppose that T is a rooted (k 1 , k 2 , . . . , k r )-banana tree on the vertex set

V (T ) = {i ∈ N : i = 1, 2, . . . , k 1 + k 2 + • • • + k r + 1}
with vertex 1 chosen as root, s 0 := 0, s i := i t=1 k t , and the edge set E(T ) is given by {(1, s i +2), (s i +2, s i +3), (s i +3, s i +j) : i = 0, 1, . . . , r-1 and j = 4, 5, . . . , k i+1 +1}.

Suppose also that

I 2 := (x 1 x s i +2 x s i +3 , x s i +2 x s i +3
x s i +j : i = 0, 1, . . . , r -1 and j = 4, 5, . . . , k i+1 + 1}, and J 2 is the Alexander dual of I 2 . Then the ideal J 2 is normally torsion-free.

Proof. It follows from the hypothesis that

J 2 = r-1 i=0 k i+1 +1 j=4 (x 1 , x s i +2 , x s i +3 ) ∩ (x s i +2 , x s i +3 , x s i +j ) . Suppose that R = K[x i : 1 ≤ i ≤ r t=1 k t + 1] and set q i := (x s i +2 , x s i +3 ) for all i = 0, 1, . . . , r -1. Thus, J 2 = r-1 i=0 (q i + x 1 k i+1 +1 j=4
x s i +j R). Now, put p 1 := x 1 R, p s i +2 := q i for all i = 0, 1, . . . , r -1, and p s i +j := x s i +j R for all i = 0, 1, . . . , r -1 and j = 4, 5, . . . , k i+1 + 1. One can deduce that

J 2 = r-1 i=0 (p s i +2 + p 1 k i+1 +1 j=4 p s i +j ).
Let F be the following monomial ideal with

k 1 + k 2 + • • • + k r -r + 1 variables F := r-1 i=0 (x s i +2 R + x 1 k i+1 +1 j=4 x s i +j R).
Accordingly, one can easily see that J 2 is the expansion of F . Our next aim is to show that F is normally torsion-free. To do this, consider the graph G on the following vertex set V (G) {s i +2 : i = 0, 1, . . . , r-1}∪{1, s i +j : i = 0, 1, . . . , r-1 and j = 4, 5, . . . , k i+1 +1}, and the following edge set E(G)

{{x s i +2 , x 1 }, {x s i +2 , x s i +j } : i = 0, 1, . . . , r -1 and j = 4, 5, . . . , k i+1 + 1}. Since r-1 i=0 (x s i +2 R + x 1 k i+1 +1 j=4 x s i +j R) = r-1 i=0 (x s i +2 , x 1 ) ∩ r-1 i=0 k i+1 +1 j=4 (x s i +2 , x s i +j ), one has G = (V (G), E(G)
) is a bipartite graph such that F is the cover ideal of G. On the other hand, according to [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6], it follows that F is normally torsion-free, and Lemma 5.2.3 implies that the ideal J 2 is also normally torsion-free, as claimed.

We conclude this section with the corollary below. Corollary 5.2.8. Suppose that T is a rooted (n, k)-banana tree on the vertex set V (T ) = {i ∈ N : i = 1, 2, . . . , nk + 1} with vertex 1 chosen as root and the edge set E(T ) is given by {(1, ik +2), (ik +2, ik +3), (ik +3, ik +j) : i = 0, 1, . . . , n-1 and j = 4, 5, . . . , k +1}.

Suppose also that

I 2 := (x 1 x ik+2 x ik+3 , x ik+2 x ik+3
x ik+j : i = 0, 1, . . . , n -1 and j = 4, 5, . . . , k + 1}, and J 2 is the Alexander dual of I 2 . Then the ideal J 2 is normally torsion-free.

Normally torsion-freeness under some monomial operators

A few examples of normally torsion-free monomial ideals appear from graph theory. Already, in [START_REF] Simis | On the ideal theory of graphs[END_REF], it has been proved that a finite simple graph G is bipartite if and only if its edge ideal is normally torsion-free. Moreover, by [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF], it is well-known that the cover ideals of bipartite graphs are normally torsion-free. In addition, in [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF], it has been verified that every transversal polymatroidal ideal is normally torsion-free. However, normally torsion-free square-free monomial ideals have been studied in [START_REF] Hà | Embedded associated primes of powers of square-free monomial ideals[END_REF][START_REF] Simis | On the ideal theory of graphs[END_REF][START_REF] Sullivant | Combinatorial symbolic powers[END_REF], but little is known for the normally torsion-free monomial ideals which are not square-free, for example see [START_REF] Olteanu | Normally torsion-free lexsegment ideals[END_REF]. In this section, we continue this argument and by using some monomial operators such as expansion, weighted, monomial multiple, monomial localization, contraction, and deletion, we introduce several methods for constructing new normally torsion-free monomial ideals based on the monomial ideals which have normally torsion-freeness. It should be noted that the results of this section can be found in [START_REF] Sayedsadeghi | Normally torsion-freeness of monomial ideals under monomial operators[END_REF].

We start our argument with the proposition below, which says that for an arbitrary ideal in a commutative Noetherian ring, there exists some power of it, which is normally torsion-free. Proposition 5.3.1. Let I be an ideal in a commutative Noetherian ring S. Then there exists a positive integer t such that I t is a normally torsion-free ideal.

Proof. Assume that t is the index of stability of I, i.e., Ass R (R/I m ) = Ass R (R/I t ) for all m ≥ t. Fix k ∈ N, and consider p ∈ Ass R (R/(I t ) k ). The assumption yields that Ass R (R/I tk ) = Ass R (R/I t ), and so p ∈ Ass R (R/I t ). This means that I t is normally torsion-free, as required.

The following lemma guarantees that any power of a normally torsion-free squarefree monomial ideal, is normally torsion-free as well.

Lemma 5.3.2. Let I be a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ]. If I is normally torsion-free, then, for all positive integers s, I s is normally torsion-free.

Proof. Assume that I is a normally torsion-free square-free monomial ideal of R. Fix positive integers k and s, and pick p ∈ Ass R (R/(I s ) k ). It follows from the assumption that Ass R (R/I sk ) ⊆ Ass R (R/I), and so p ∈ Ass R (R/I). Since I is a square-free monomial ideal, one has Ass R (R/I) = Ass R (R/I s ). This yields that p ∈ Ass R (R/I s ), and hence Ass R (R/(I s ) k ) ⊆ Ass R (R/I s ), that is, I s is normally torsion-free, as desired.

In the following, we focus on the normally torsion-freeness of the sum of two monomial ideals. For this purpose, we first recall the following well-known result.

R = R 1 ⊗ k R 2 .
We are now ready to express Theorem 5.3.4, which is related to the normally torsion-freeness of the sum of two normally torsion-free monomial ideals. Theorem 5.3.4. Let I be a monomial ideal in R = K[x 1 , . . . , x n ] such that I =

I 1 R + I 2 R, where G(I 1 ) ⊂ R 1 = K[x 1 , . . . , x m ] and G(I 2 ) ⊂ R 2 = K[x m+1 , . . . , x n ]
for some positive integer 1 ≤ m < n. If I 1 and I 2 are normally torsion-free, then I is so.

Proof. Assume that I 1 and I 2 are normally torsion-free monomial ideals. Fix s ∈ N, and let p ∈ Ass R (R/I s ). This implies that p = (I s : R c) for some monomial c in R.

Certainly, one can write

c = c 1 c 2 with c 1 ∈ R 1 and c 2 ∈ R 2 . By virtue of Lemma 3.2.7, one has I s = s i=1 (I i 1 +I s+1-i 2
). Accordingly, we obtain the following equalities p = (I s : R c)

= ( s i=1 (I i 1 + I s+1-i 2 ) : R c) = s i=1 (I i 1 + I s+1-i 2 : R c) = s i=1 ((I i 1 : R c 1 c 2 ) + (I s+1-i 2 : R c 1 c 2 )) = s i=1 ((I i 1 : R 1 c 1 ) + (I s+1-i 2 : R 2 c 2 )).
Consequently, one can conclude that there exists a positive integer r with 1 ≤ r ≤ s such that p = (I r 1 : R 1 c 1 ) + (I s+1-r 2 : R 2 c 2 ). On the other hand, one can write p = p 1 R + p 2 R such that p 1 ⊂ R 1 and p 2 ⊂ R 2 are monomial prime ideals. This gives that (I r 1 : R 1 c 1 ) = p 1 and (I s+1-r

2 : R 2 c 2 ) = p 2 . Therefore, p 1 ∈ Ass R 1 (R 1 /I r 1 ) and p 2 ∈ Ass R 2 (R 2 /I s+1-r 2
). Based on I 1 and I 2 are normally torsion-free, we have In what follows, we introduce several methods for constructing new classes of monomial ideals which have normally torsion-freeness. To accomplish this purpose, we begin with the first method. Thus, we need to recall the definition of the expansion operator on monomial ideals, which has been stated in Definition 2.2.9.

p 1 ∈ Ass R 1 (R 1 /I 1 ) and p 2 ∈ Ass R 2 (R 2 /I 2 ).
The subsequent theorem reconsiders a powerful tool for producing new normally torsion-free monomial ideals. Indeed, it says that a monomial ideal is normally torsion-free if and only if its expansion is normally torsion-free.

Theorem 5.3.5. (see Lemma 5.2.3) Let I be a monomial ideal of R. Then I is normally torsion-free if and only if I * is.

Here, we state the second method for constructing new normally torsion-free monomial ideals. To achieve this, one requires to recall Definition 3.2.11. Some basic properties of the weighted operator are given in Lemma 3.2.12, which are essential for us to prove Theorem 5.

(i) If I is a monomial ideal of R such that Q ⊊ I, then there exist posi- tive integers α r+1 , . . . , α n such that x ℓ 1 -1 i 1 • • • x ℓr-1 ir x α r+1 i r+1 • • • x αn in ∈ I, where {i 1 , . . . , i r , i r+1 , . . . , i n } = {1, . . . , n}. (ii) If I and J are two monomial ideals of R such that Q = I ∩ J, then Q = I or Q = J. Proof. (i) It follows from Q ⊊ I that there exists a monomial u ∈ R such that u ∈ I \ Q. Assume that u = x α 1 i 1 • • • x αr ir x α r+1 i r+1 • • • x αn in for some nonnegative integers α 1 , . . . , α n . Because u / ∈ Q, one can conclude that x ℓ j
i j ∤ u for each j = 1, . . . , r. This implies that ℓ i > α i for each i = 1, . . . , r. Hence, we get ℓ i -1 ≥ α i for each i = 1, . . . , r, and so

x ℓ 1 -1 i 1 • • • x ℓr-1 ir x α r+1 i r+1 • • • x αn in = (x ℓ 1 -1-α 1 i 1 • • • x ℓr-1-αr ir ).u ∈ I. (ii)
We prove the contrapositive of this claim, i.e., if Q ⊊ I and Q ⊊ J, then Q ⊊ I ∩ J. Because Q ⊊ I, part (i) implies that there exist positive integers λ r+1 , . . . , λ n such that x ℓ 1 -1

i 1 • • • x ℓr-1 ir x λ r+1 i r+1 • • • x λn in ∈ I.
Once again, by virtue of Q ⊊ J, part (i) gives that there exist positive integers θ r+1 , . . . , θ n such that

x ℓ 1 -1 i 1 • • • x ℓr-1 ir x θ r+1 i r+1 • • • x θn in ∈ J. Now, set v := x ℓ 1 -1 i 1 • • • x ℓr-1 ir x β r+1 i r+1 • • • x βn in ,
where

β i := max{λ i , θ i } for each i = r + 1, . . . , n. It is routine to check that v ∈ I ∩ J \ Q. Therefore, Q ⊊ I ∩ J, as required.
Corollary 5.3.7. Let Q be an irreducible monomial ideal in a polymonial ring R = K[x 1 , . . . , x n ], and I 1 , . . . , I s be some monomial ideals of R.

If Q = I 1 ∩ • • • ∩ I s , then Q = I j for some j with 1 ≤ j ≤ s.
Proof. Follow by induction on s, and apply Proposition 5.3.6. Proposition 5.3.8. Let Q be an irreducible monomial ideal and Q 1 , . . . , Q s be some primary monomial ideals in a polymonial ring

R = K[x 1 , . . . , x n ]. If Q 1 ∩ • • • ∩ Q s ⊆ Q, then Q j ⊆ Q for some j with 1 ≤ j ≤ s. Proof. Due to Q 1 ∩ • • • ∩ Q s ⊆ Q, one derives Q 1 ∩ • • • ∩ Q s + Q = Q.
Here, our aim is to show that

Q 1 ∩ • • • ∩ Q s + Q = (Q 1 + Q) ∩ • • • ∩ (Q s + Q). Since Q 1 ∩ • • • ∩ Q s ⊆ Q i for each i = 1, . . . , s, one has Q 1 ∩ • • • ∩ Q s + Q ⊆ Q i + Q for each i = 1, . . . , s. This implies that Q 1 ∩ • • • ∩ Q s + Q ⊆ (Q 1 + Q) ∩ • • • ∩ (Q s + Q). Conversely, we verify that (Q 1 + Q) ∩ • • • ∩ (Q s + Q) ⊆ Q 1 ∩ • • • ∩ Q s + Q. To do this, consider u ∈ G((Q 1 + Q) ∩ • • • ∩ (Q s + Q)). This yields that u ∈ Q i + Q for each i = 1, . . . , s. Hence, one can conclude that u ∈ Q i or u ∈ Q for each i = 1, . . . , s. If u ∈ Q, then u ∈ Q 1 ∩ • • • ∩ Q s + Q,
and the argument is complete. Thus, let u / ∈ Q. This implies that u ∈ Q i for each i = 1, . . . , s, and so

u ∈ Q 1 ∩ • • • ∩ Q s . Consequently, one has u ∈ Q 1 ∩ • • • ∩ Q s + Q. Therefore, we get Q 1 ∩ • • • ∩ Q s + Q = (Q 1 + Q) ∩ • • • ∩ (Q s + Q). Accordingly, one obtains Q = (Q 1 +Q)∩• • •∩(Q s +Q).
It follows now from Corollary 5.3.7 that there exists a positive integer j with 1 ≤ j ≤ s such that

Q j + Q = Q. This gives that Q j ⊆ Q, as claimed.
The following lemma probes the relation between the set of associated primes of a monomial ideal and the set of associated primes of its weighted. ). This concludes that (Q i ) W is a (x i 1 , . . . , x ir )-primary ideal of R. Accordingly, one can deduce that

I W = (Q 1 ) W ∩ • • • ∩ (Q s ) W is a presentation of I W .
To complete the proof, one has to demonstrate that this presentation is irredundant. On the contrary, assume that this presentation is not irredundant, that is, there exists a positive integer j with 1 ≤ j ≤ s such that

(Q 1 ) W ∩ • • • ∩ (Q j-1 ) W ∩ (Q j+1 ) W ∩ • • • ∩ (Q s ) W ⊆ (Q j ) W .
Now, Proposition 5.3.8 yields that there exists a positive integer 1 ≤ t ≤ s with t = j such that (Q t ) W ⊆ (Q j ) W . Thus, we get Q t ⊆ Q j , which shows that Q j is redundant in the presentation of I, a contradiction. Hence, one derives that

I W = (Q 1 ) W ∩ • • • ∩ (Q s ) W is the irredundant presentation of I W . This leads to Ass R (R/I W ) = Ass R (R/I)
, and the proof is over.

We are now ready to prove Theorem 5.3.10 by using Lemmas 3.2.12 and 5.3.9. In fact, it tells us that a monomial ideal is normally torsion-free if and only if its weighted is normally torsion-free. Proof. For the forward implication, assume that I is a normally torsion-free monomial ideal of R. Fix k ∈ N, and consider p ∈ Ass R (R/I k W ). On account of Lemma 3.2.12(ii), one has I k W = (I k ) W . This gives that p ∈ Ass R (R/(I k ) W ). Due to Lemma 5.3.9, we obtain p ∈ Ass R (R/I k ). Since I is normally torsion-free, one can conclude that p ∈ Ass R (R/I). Once again, Lemma 5.3.9 implies that p ∈ Ass R (R/I W ). This means that I W is a normally torsion-free monomial ideal of R.

For the converse, assume that I W is a normally torsion-free monomial ideal of R. Fix k ∈ N, and pick p ∈ Ass R (R/I k ). In the light of Lemma 5.3.9, one derives p ∈ Ass R (R/(I k ) W ). It follows from Lemma 3.2.12(ii) that

I k W = (I k ) W . This yields that p ∈ Ass R (R/I k W )
. Because I W is normally torsion-free, one can deduce that p ∈ Ass R (R/I W ). Once again, Lemma 5.3.9 concludes that p ∈ Ass R (R/I), and thus I is a normally torsion-free monomial ideal of R.

The subsequent lemma says that a monomial ideal is normally torsion-free if and only if its monomial multiple is normally torsion-free. Proof. For necessity, assume that I is normally torsion-free, and fix k ∈ N. Thanks to Theorem 2.1.27, we get Ass R (R/hI) = Ass R (R/I) ∪ {(x j 1 ), . . . , (x js )}, and also

Ass R (R/(hI) k ) = Ass R (R/I k ) ∪ {(x j 1 ), . . . , (x js )}. Since I is normally torsion-free, one has Ass R (R/I k ) ⊆ Ass R (R/I). Hence, we get Ass R (R/(hI) k ) ⊆ Ass R (R/hI),
that is, hI is normally torsion-free, as claimed.

To prove the sufficiency, let hI be normally torsion-free, and fix k ∈ N. This yields that Ass R (R/(hI) k ) ⊆ Ass R (R/hI). It follows from Theorem 2.1.27 that

Ass R (R/I k ) ∪ {(x j 1 ), . . . , (x js )} ⊆ Ass R (R/I) ∪ {(x j 1 ), . . . , (x js )}. Pick p ∈ Ass R (R/I k ). Hence, we obtain p ∈ Ass R (R/I) ∪ {(x j 1 ), . . . , (x js )}. If p ∈
Ass R (R/I), then the proof is complete. Thus, let p = (x jt ) for some positive integer t with 1 ≤ t ≤ s. This means that p ∈ Min(I k ), and by virtue of Min(I k ) = Min(I), one derives p ∈ Min(I). Consequently, one has p ∈ Ass R (R/I). Therefore, I is normally torsion-free, as required.

In order to prove Theorem 5.3.12, we need to recall the notion of the monomial localization of a monomial ideal with respect to a monomial prime ideal as has been introduced in Definition 3.1.14.

Theorem 5.3.12. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ], and p ∈ V * (I). If I is normally torsion-free, then I(p) is so.

Proof. Assume that I is a normally torsion-free monomial ideal of R. Fix k ∈ N, and consider q ∈ Ass R(p) (R(p)/I(p) k ). By Lemma 3.2.15(ii), one deduces that I(p) k = I k (p), and so q ∈ Ass R(p) (R(p)/I k (p)). It follows now from Lemma 3.2.15(viii) that q ∈ Ass R (R/I k ) and q ⊆ p. Due to I is normally torsion-free, we gain q ∈ Ass R (R/I). In the light of q ∈ Ass R (R/I) and q ⊆ p, Lemma 3.2.15(viii) yields that q ∈ Ass R(p) (R(p)/I(p)). Therefore, one can conclude that I(p) is normally torsion-free, as desired.

It should be noted that Theorem 5.3.12 helps us to construct new normally torsion-free monomial ideals inspired by some well-known normally torsion-free monomial ideals, as will be mentioned in Corollary 5.3.13.

Corollary 5.3.13. Let I be a monomial ideal in R 1 = K[x 1 , . . . , x r ] with G(I) = {u 1 , . . . , u m }, and J 1 , . . . , J m be some monomial ideals in R 2 = K[x r+1 , . . . , x n ]. If L := u 1 J 1 R + • • • + u m J m R is normally torsion-free, where R = K[x 1 , . . . , x n ], then I is so. Proof. Put p := ( m i=1 supp(u i ))R.
It is obvious that p is a monomial prime ideal of R, and L ⊆ p. By virtue of Theorem 5.3.12, one derives L(p) is normally torsion-free. Because L(p) = I, we get I is normally torsion-free, as claimed.

It is necessary to observe that the following corollary allows us to refute the normally torsion-freeness of some monomial ideals based on some non-normally torsionfree monomial ideals.

Corollary 5.3.14. Let I be a monomial ideal in R 1 = K[x 1 , . . . , x r ] with G(I) = {u 1 , . . . , u m }, and J 1 , . . . , J m be some monomial ideals in

R 2 = K[x r+1 , . . . , x n ]. If I is not normally torsion-free, then L := u 1 J 1 R + • • • + u m J m R is not normally torsion-free, where R = K[x 1 , . . . , x n ].
To state the subsequent result, one has to recall the definition of the contraction operator, which has been stated in Definition 3.2.19.

Note that the contraction I/x j is exactly the monomial localization of I with respect to p = m \ {x j }, where m = (x 1 , . . . , x n ) is the graded maximal ideal of If I is normally torsion-free, then I/x j is so.

R = K[x 1 , . . . , x n ],
To express Theorem 5.3.16, we require to recall the definition of the deletion operator, which has been stated in Definition 3.2.19.

Neighborhood hypergraphs and their transversal hypergraphs

The edge ideals and cover ideals of bipartite graphs are known to be normally torsion-free, see [START_REF] Simis | On the ideal theory of graphs[END_REF] and [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]. The bipartite graphs are characterized as the graphs that avoid any cycle of odd lengths. With the help of Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], one can verify that N I(C 8 ) and DI(C 8 ) are not normally torsion-free. This shows that one cannot directly extend the results of [START_REF] Simis | On the ideal theory of graphs[END_REF] and [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF] in the case of closed neighborhood ideals and dominating ideals of bipartite graphs. However, if we consider the neighborhood hypergraph N (G) of a graph G such that there are no hypergraphic cycles in N (G), then one obtains a similar statement as in the case of edge ideals and cover ideals of bipartite graphs. To establish this, we recall some definitions and notions from hypergraph theory. For more details about these concepts, we refer the reader to [15]. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Algebraic implications of neighborhood hypergraphs and their transversal hypergraphs[END_REF].

Let H be a simple hypergraph with n vertices and m edges, and let M be the incidence matrix of H. A hypergraph H is called Mengerian if for all c ∈ Z n + , we have the following equality

min{a • c : a ∈ Z n + , M • a ≥ 1} = max{b • 1 : b ∈ Z m + , M T • b ≤ c},
where Z + denotes the nonnegative integers. Algebraically, it is equivalent to I(H) being normally torsion-free; it can be found in [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 10.3.15] and [START_REF] Villarreal | Monomial algebras[END_REF]Theorem 14.3.6]. We next define some of the well-known classes of Mengerian hypergraphs.

Definition 5.4.1. A cycle of length

k ≥ 2 in a hypergraph H is a sequence C : x 1 , E 1 , x 2 , E 2 , . . . , x k , E k , x 1 such that 1. E 1 , . . . , E k are distinct edges of H, 2. x 1 , . . . , x k are distinct vertices of H, 3. x 1 , x k ∈ E k and x i , x i+1 ∈ E i for all i = 1, . . . , k -1.
The vertices of C are x 1 , . . . , x k and the edges of C are E 1 , . . . , E k . Following [START_REF] Herzog | Monomial Ideals[END_REF]Section 10.3.4], a cycle C of length k ≥ 3 is said to be special if all the edges of C contain exactly two vertices of C. Note that if we reduce to the case of graphs, then the special cycles of length k ≥ 3 are just the chordless cycles.

A hypergraph H is called balanced if every cycle of H of odd length k ≥ 3 has an edge containing at least three vertices of the cycle. Moreover, H is called totally balanced if every cycle of H of length k ≥ 3 has an edge containing at least three vertices of the cycle, see [15,Chapter 5]. In other words, H is balanced if it does not have special odd cycles and H is totally balanced if it does not have any special cycles. It follows from the definition that every totally balanced hypergraph is also balanced.

Remark 5.4.2. A well-known result of Fulkerson, Hoffman, and Oppenheim in [START_REF] Fulkerson | On balanced matrices[END_REF] states that balanced hypergraphs are Mengerian. In particular, totally balanced hypergraphs are also Mengerian. Furthermore, [15,Theorem 19 It is natural to ask that if there exists a complete characterization of graphs for which their associated neighborhood hypergraphs are balanced or totally balanced. Such graphs are characterized in [START_REF] Farber | Characterization of strongly chordal graphs[END_REF] and [23]. To describe these graphs we need to recall some definitions from [START_REF] Farber | Characterization of strongly chordal graphs[END_REF] and [23]. Definition 5.4.5. Let n ≥ 3. An incomplete trampoline is a chordal graph G on 2n vertices whose vertex set can be partitioned into two sets, W = {y 1 , . . . , y n } and U = {u 1 , . . . , u n } such that no vertices of W are adjacent to each other, and for each i and j, w i is adjacent to u i if and only if i = j or i ≡ j + 1 (mod n). In addition, an incomplete trampoline is called trampoline if the induced subgraph on U is a complete graph. If n is odd, then (incomplete) trampoline is called an odd (incomplete) trampoline.

Trampolines on 2n vertices are sometimes also mentioned as n-sun graphs. It is shown in [23,Theorem 2] that for a given graph G, we have N (G) is balanced if and only if G is chordal with no induced odd incomplete trampoline. Once again, we can deduce from Remark 5.4.2 and Proposition 5.4.3 the next theorem. Theorem 5.4.6. Let G be a chordal graph with no induced odd incomplete trampoline. Then the neighborhood ideal and the dominating ideal of G is normally torsion-free. Remark 5.4.7. It is crucial to mention that there are Mengerian neighborhood hypergraphs that are not balanced. To see this, suppose that C 6 = (V (C 6 ), E(C 6 )) is the even cycle graph of order 6 on the vertex set V (C 6 ) = {x 1 , . . . , x 6 } and the edge set E(C 6 ) = {{x i , x i+1 } : i = 1, . . . , 5} ∪ {{x 6 , x 1 }}. We can view the C 9 in a similar way. The cycle graphs C 6 and C 9 are not chordal, and hence [23,Theorem 2] tells us N (C 6 ) and N (C 9 ) are not balanced. Indeed, the edges of N (C 6 ) are

N [x 1 ] = {x 1 , x 2 , x 6 }, N [x 2 ] = {x 1 , x 2 , x 3 }, N [x 3 ] = {x 2 , x 3 , x 4 }, N [x 4 ] = {x 3 , x 4 , x 5 }, N [x 5 ] = {x 4 , x 5 , x 6 }, N [x 6 ] = {x 5 , x 6 , x 1 },
and the sequence C :

x 1 , N [x 2 ], x 3 , N [x 4 ], x 5 , N [x 6 ],
x 1 gives a special odd cycle in the hypergraph N (C 6 ). One can construct a special odd cycle in N (C 9 ) in a similar fashion. In Subsection 5.4.2, we will show that not only N I(C 6 ) and N I(C 9 ) are normally torsion-free, but also their Alexander duals DI(C 6 ) and DI(C 9 ) are normally torsion-free. This highlights that the converse of Theorem 5.4.6 is not true. Now, we turn our attention to the characterization of graphs for which the associated neighborhood hypergraphs are totally balanced. To accomplish this, one has to recall the following definitions.

A strong elimination ordering of G is an ordering v 1 , . . . , v n of V (G) with the property that for each i, j, k, and l,

if i < j, k < l with v k , v l ∈ N G [v i ] and v k ∈ N G [v j ], then v l ∈ N G [v j ].
Graphs that admit a strong elimination ordering are called strongly chordal graphs. The strongly chordal graphs were introduced by Farber in [START_REF] Farber | Characterization of strongly chordal graphs[END_REF]. They form a subclass of chordal graphs, because a strong elimination ordering is also a perfect elimination ordering, by letting i = k. Among many other characterizations of strongly chordal graphs, it is shown in [START_REF] Farber | Characterization of strongly chordal graphs[END_REF] that a graph G is strongly chordal if and only if it is chordal and does not contain any induced trampoline. In [START_REF] Farber | Characterization of strongly chordal graphs[END_REF]Theorem 3.3], a polynomial procedure to recognize strongly chordal graphs and to construct strong elimination orderings is given. Some of the special subclasses of strongly chordal graphs are trees and interval graphs. It is proved in [START_REF] Farber | Characterization of strongly chordal graphs[END_REF]Theorem 5.2] and re-proved in [23,Theorem 2] that a graph G is strongly chordal if and only if N (G) is totally balanced. We provide a neat and straightforward proof of the forward implication in the next theorem. Proof. Let G be a strongly chordal graph. To simplify the notation in the following text, for any p, q ∈ V (G) we write p < q if p appears before q in the strong elimination ordering of G. Let

C : x 1 , E 1 , x 2 , E 2 , . . . , x k , E k , x 1 , be a cycle in N (G) with k ≥ 3. Let v 1 , . . . , v k ∈ G be such that E i = N [v i ] for all i = 1, . . . , k.
We can rearrange the edges in C such that v 1 < v i for all i = 2, . . . , k, and

x 2 < x 1 . Since x 1 , x 2 ∈ E 1 = N [v 1 ] and x 2 ∈ E 2 = N [v 2 ], it follows that x 1 ∈ E 2 .
This shows that C is not a special cycle, and so N (G) is totally balanced, as claimed.

Totally balanced hypergraphs have further nice algebraic implications. It follows from [START_REF] Herzog | Standard graded vertex cover algebras, cycles and leaves[END_REF]Theorem 2.3] that a simple hypergraph H is a simplicial forest if and only if H is totally balanced. Simplicial forests were introduced by Faridi in [START_REF] Faridi | The facet ideal of a simplicial complex[END_REF] in the language of simplicial complexes. The facet ideals of simplicial forests possess many nice properties as noted in [START_REF] Faridi | The facet ideal of a simplicial complex[END_REF] and [START_REF] Faridi | Simplicial trees are sequentially Cohen-Macaulay[END_REF]. For example, Faridi proved in [40, Corollaries 5.5 and 5.6] that if ∆ is a simplicial forest, then its facet ideal I(∆) is sequentially Cohen-Macaulay and I(∆) ∨ is componentwise linear. More details and related definitions can be found in [START_REF] Faridi | The facet ideal of a simplicial complex[END_REF] and [START_REF] Faridi | Simplicial trees are sequentially Cohen-Macaulay[END_REF].

We summarize the discussion above together with Remark 5.4.2 and Proposition 5.4.3, in the next theorem. (iv) DI(G) is componentwise linear.

The t-path hypergraphs of path graphs

Let P n be a path on n vertices. It follows from [START_REF] He | Algebraic properties of the path ideal of a tree[END_REF]Theorem 2.7] that H t (P n ) is a simplicial forest and using [START_REF] Herzog | Standard graded vertex cover algebras, cycles and leaves[END_REF]Theorem 3.2] it can be seen that H t (P n ) is totally balanced. Consequently, I t (P n ) is normally torsion-free. Furthermore, we conclude from Remark 5.4.2 that I t (P n ) ∨ is normally torsion-free. We provide another proof of these facts by using algebraic tools. To do this, we first introduce a technique to construct new normally torsion-free square-free monomial ideals based on the existing ones. This technique will be implemented in the subsequent result.

= I + vR. If J \ x t = m \ x t for some 1 ≤ t ≤ n, then we can write J = x t L + m \ x t . If L = R, then J = m, and the assertion holds. Assume that L = R. Let h ∈ G(L). If x α | h for some α ∈ {1, . . . , n} \ {t}, this gives that h ∈ m \ x t ,
and hence L ⊆ m \ x t . We thus get J = m \ x t , and the proof is over. Thus, we assume that J \ x t = m \ x t for all t = 1, . . . , n. In what follows, we first show that v ∈ p \ p 2 for any p ∈ Min(J). Pick an arbitrary element p ∈ Min(J). Due to v ∈ J and J ⊆ p, we have v ∈ p. On the contrary, assume that v ∈ p 2 . Because v is a square-free monomial, this yields that |supp(p) ∩ supp(v)| ≥ 2. If gcd(v, u m ) = 1, then gcd(v, u i ) = 1 for all i = 1, . . . , m-1. In this case, the claim can be deduced directly from Theorem 5.3.4. Let supp(gcd(v, u m )) = {x i 1 , . . . , x i k } with {i 1 , . . . , i k } ⊆ {1, . . . , n}. Here, one may consider the following cases: If c ≤ d (respectively, c > d), then the assumption yields that I ⊆ p \ {x λ } (respectively, I ⊆ p \ {x ir }), and by virtue of v ∈ p \ {x λ } (respectively, v ∈ p \ {x ir }), this gives rise to J ⊆ p \ {x λ } (respectively, J ⊆ p \ {x ir }), which contradicts the minimality of p. Consequently, we conclude that x λ / ∈ {x i 1 , . . . , x i k }, and thus x λ / ∈ supp(I). In the light of v ∈ p \ {x λ }, this leads to J ⊆ p \ {x λ }, a contradiction to the minimality of p.

Case 1. x ir ∈ p for some x ir ∈ {x i 1 , . . . , x i k }. Since |supp(p) ∩ supp(v)| ≥ 2, we can take x λ ∈ supp(p) ∩ supp(v) with x λ = x ir . We claim that x λ / ∈ {x i 1 , . . . , x i k }. Suppose, on the contrary, that x λ ∈ {x i 1 , . . . , x i k }. Let c := min{j : x ir | u j ,

Case 2. x ir /

∈ p for any x ir ∈ {x i 1 , . . . , x i k }. Select two distinct variables x α , x β ∈ supp(p) ∩ supp(v). Accordingly, we have x α , x β / ∈ supp(I). We thus get I ⊆ p \ {x α }. On account of v ∈ p \ {x α }, one can conclude that J ⊆ p \ {x α }. This contradicts the minimality of p.

Therefore, we deduce that v / ∈ p 2 . In addition, note that J \ x t = I \ x t for any x t ∈ supp(v). It follows also from Theorem 5.3.16 that I \ x t is normally torsion-free for any x t ∈ supp(v), and so J \ x t is normally torsion-free for any x t ∈ supp(v). Fix s ≥ 1 and x t ∈ supp(v). Suppose, on the contrary, that m \ x t ∈ Ass(R/(J \ x t ) s ). In view of Ass(R/(J \ x t ) s ) = Min(J \ x t ), we derive that m \ x t ∈ Min(J \ x t ), and thus

J \ x t = m \ x t ,
which is a contradiction. We therefore have m \ x t / ∈ Ass(R/(J \ x t ) s ) for all s and x t ∈ supp(v). It follows now from Theorem 10.2.6 that J is normally torsion-free, and the proof is complete.

(ii)-(vi) can be shown similar to the proof of Corollary 6.2.3.

As an application of Theorem 5.4.10, we give the following result.

Proposition 5.4.11. The t-path ideals of path graphs are normally torsion-free, for all t ≥ 0.

Proof. Let P n be a path graph with the vertex set V (P n ) = {x 1 , . . . , x n } and the edge set

E(P n ) = {{x i , x i+1 } : i = 1, . . . , n -1}. Then the t-path ideal of P n is given by I t (P n ) = (x i x i+1 • • • x i+t-1 : i = 1, . . . , n -t + 1). The proof is by induction on n. If n = t, then I t (P n ) = (x 1 x 2 • • • x t )
, and there is nothing to show.

Let n > t and that the claim has been proved for any value less than n. Now, put

I := (x i x i+1 • • • x i+t-1 : i = 1, . . . , n -t), v := x n-t+1 x n-t+2 • • • x n-1 x n , and
u i := x i x i+1 • • • x i+t-1 for all i = 1, . . . , n -t.
It is easily seen that gcd(v, u i ) = 1 for all i = 1, . . . , n -2t + 1, and gcd(v,

u i ) = x n-t+1 x n-t+2 • • • x t+i-1 for all i = n -2t + 2, . . . , n -t.
In particular, this implies that gcd(v, u i ) divides gcd(v, u i+1 ) for all i = 1, . . . , n -t -1. It follows also from the inductive hypothesis that I is normally torsion-free. Thanks to I t (P n ) = I + vR, where R = K[x 1 , . . . , x n ], we deduce immediately the assertion from Theorem 5.4.

10(i).

In what follows, our aim is to study the normally torsion-freeness of the Alexander duals of 3-path ideals of path graphs. For this, we will need the following proposition.

Proposition 5.4.12. Suppose that I and J are two normally torsion-free squarefree monomial ideals in a polynomial ring R = K[x 1 , . . . , x n ] over a field K such that supp(I) ∩ supp(J) = ∅. Then I ∩ J = IJ is normally torsion-free.

Proof. Due to supp(I) ∩ supp(J) = ∅, we deduce from [60, Proposition 1.2.1] that IJ = I ∩ J. In particular, we have Min(I ∩ J) = Min(I) ∪ Min(J). On the other hand, we know from [60, Theorem 1.4.6] that a square-free monomial ideal L is normally torsion-free if and only if L k = L (k) for all k ≥ 1, where L (k) denotes the k-th symbolic power of L. Hence, it is enough for us to show that (I ∩J) k = (I ∩J) (k) for all k ≥ 1. To achieve this, fix k ≥ 1. Since I and J are normally torsion-free, one can conclude that I k = I (k) and J k = J (k) . Thus, we get the following equalities

(I ∩ J) (k) = p∈Min(I∩J) ((I ∩ J) k R p ∩ R) = p∈Min(I) (I k R p ∩ R) ∩ p∈Min(J) (J k R p ∩ R) = I (k) ∩ J (k) = I k ∩ J k = (I ∩ J) k .
This shows that I ∩ J = IJ is normally torsion-free, as claimed.

We are now ready to prove the following proposition. Lemma 5.4.13. The Alexander duals of 3-path ideals of path graphs are normally torsion-free square-free monomial ideals.

Proof. Let P n be a path graph with the vertex set V (P n ) = {x 1 , . . . , x n } and the edge set E(P n ) = {{x i , x i+1 } : i = 1, . . . , n -1}. Then the Alexander dual of the 3-path ideal of the path graph P n , denoted by I 3 (P n ) ∨ , is given by I 3 (P n ) ∨ = n-2 i=1 (x i , x i+1 , x i+2 ). We show the claim by induction on n. Since every monomial prime ideal is normally torsion-free, this implies that the assertion is true for the case in which n = 3. Let n > 3 and that the claim has been shown for any value less than n. Our strategy is to use Theorem 10.2.6. For this purpose, set p i := (x i , x i+1 , x i+2 ) for all i = 1, . . . , n -2, and

v := [ n-1 3 ] k=0 x 1+3k .
It is easy to check that |supp(v) ∩ supp(p i )| = 1 for all i = 1, . . . , n -2. This yields that v ∈ p i \ p 2 i for all i = 1, . . . , n -2. To complete the argument, one has to verify that m \ x i / ∈ Ass(R/(I 3 (P n ) ∨ \ x i ) s ) for all s and x i ∈ supp(v), where m = (x 1 , . . . , x n ). Fix s ≥ 1 and x i ∈ supp(v). One can check that I 3 (P n ) ∨ \ x i , after deleting the redundant components, can be viewed as J 1 ∩J 2 ∩Γ, where J 1 = ℓ 1 r=1 q r (respectively, J 2 = ℓ 2 t=1 q ′ t ) is the Alexander dual of a 3-path ideal of a path graph such as P 1 (respectively, P 2 ) with V (P 1 ) < n (respectively, V (P 2 ) < n), V (P 1 ) ∩ V (P 2 ) = ∅ and Γ = λ c=1 (x αc , x βc ) such that for any 1 ≤ r ≤ ℓ 1 , 1 ≤ t ≤ ℓ 2 , and

1 ≤ c = d ≤ λ, we have |supp(q r ) ∩ supp(x αc , x βc )| ≤ 1, |supp(q ′ t ) ∩ supp(x αc , x βc )| ≤ 153
1, and |supp(x αc , x βc ) ∩ supp(x α d , x β d )| = 1. Furthermore, the inductive hypothesis gives that J 1 and J 2 are normally torsion-free, and Proposition 5.4.12 implies that J 1 ∩ J 2 is normally torsion-free. It follows now by repeated use of Lemma 6.4.9 that J 1 ∩ J 2 ∩ Γ is normally torsion-free, and so I 3 (P n ) ∨ \ x i is normally torsion-free. This means that m \ x i / ∈ Ass(R/(I 3 (P n ) ∨ \ x i ) s ), as required. We conclude from Theorem 10.2.6 that I 3 (P n ) ∨ is normally torsion-free. This completes the inductive step, and therefore the claim has been shown by induction.

Associated primes of dominating ideals of cycles

In this subsection, we will mainly discuss the normally torsion-freeness of DI(C n ) and the stable sets of associated primes of DI(C n ). In [8, Theorem 3.9], Alilooee and Banerjee showed that the 3-path ideal of a graph G is normally torsion-free if and only if G is a path graph P k for some k ≥ 3 or G is a cycle C 3k when k = 1, 2, 3. Observe that the 3-path ideal of a cycle C n with n ≥ 3 coincides with N I(C n ). With this observation, we can rephrase [8, Theorem 3.9] in the following way.

Proposition 5.4.14. Let C n be a cycle graph with n ≥ 3. Then N I(C n ) is normally torsion-free if and only if n ∈ {3, 6, 9}.

As the first main result of this subsection, in Theorem 5.4.16, we will prove that the statement above also holds for DI(C n ). To show this, we first consider the case of DI(C 6 ) and DI(C 9 ) and verify that they are normally torsion-free. Proof. (i) Suppose that E(C 6 ) = {{x i , x i+1 } : i = 1, . . . , 5} ∪ {{x 6 , x 1 }}. Then the dominating ideal of C 6 is given by

DI(C 6 ) = (x 1 , x 2 , x 3 ) ∩ (x 2 , x 3 , x 4 ) ∩ (x 3 , x 4 , x 5 ) ∩ (x 4 , x 5 , x 6 ) ∩ (x 5 , x 6 , x 1 ) ∩ (x 6 , x 1 , x 2 ) = (x 3 x 6 , x 2 x 5 , x 1 x 4 , x 2 x 4 x 6 , x 1 x 3 x 5 ).
We want to demonstrate that DI(C 6 ) is normally torsion-free by using Theorem 10.2.6. For this purpose, set v := x 3 x 6 and L := DI(C 6 ). One can easily see that v ∈ p \ p 2 for any p ∈ Min(L). Here, we must show that m \ x i / ∈ Ass(R/(L \ x i ) s ) for all s and x i ∈ {x 3 , x 6 }. Fix s ≥ 1. First, we establish m \ x 3 / ∈ Ass(R/(L \ x 3 ) s ), where m = (x 1 , . . . , x 6 ). According to the definition of deletion, we obtain

L \ x 3 =(x 1 , x 2 ) ∩ (x 2 , x 4 ) ∩ (x 4 , x 5 ) ∩ (x 5 , x 6 , x 1 ) =(x 2 x 5 , x 1 x 4 , x 2 x 4 x 6 ).
It follows from Theorem 5.3.4 and Lemma 5.3.11 that x 4 (x 1 , x 2 x 6 ) is normally torsion-free. Because gcd(x 2 x 5 , x 1 x 4 ) = 1 and gcd(x 2 x 5 , x 2 x 4 x 6 ) = x 2 , one can derive from Theorem 5.4.10 that L \ x 3 is normally torsion-free, and so m \ x 3 = (x 1 , x 2 , x 4 , x 5 , x 6 ) / ∈ Ass(R/(L \ x 3 ) s ). Now, we show that m \ x 6 / ∈ Ass(R/(L \ x 3 ) s ). A similar computation gives that

L \ x 6 =(x 1 , x 2 ) ∩ (x 2 , x 3 , x 4 ) ∩ (x 4 , x 5 ) ∩ (x 5 , x 1 ) =(x 2 x 5 , x 1 x 4 , x 1 x 3 x 5 ).
On account on Theorem 5.3.4 and Lemma 5.3.11, we get x 1 (x 4 , x 3 x 5 ) is normally torsion-free. As gcd(x 2 x 5 , x 1 x 4 ) = 1 and gcd(x 2 x 5 , x 1 x 3 x 5 ) = x 5 , Theorem 5.4.10 implies that L \ x 6 is normally torsion-free, and hence m \ x 6 = (x 1 , x 2 , x 3 , x 4 , x 5 ) / ∈ Ass(R/(L \ x 6 ) s ). It follows now from Theorem 10.2.6 that DI(C 6 ) is normally torsion-free, as claimed.

(ii) Let E(C 9 ) = {{x i , x i+1 } : i = 1, . . . , 8} ∪ {{x 9 , x 1 }}. Then the dominating ideal of C 9 is given by

DI(C 9 ) = (x 1 , x 2 , x 3 ) ∩ (x 2 , x 3 , x 4 ) ∩ (x 3 , x 4 , x 5 ) ∩ (x 4 , x 5 , x 6 ) ∩ (x 5 , x 6 , x 7 ) ∩ (x 6 , x 7 , x 8 ) ∩ (x 7 , x 8 , x 9 ) ∩ (x 8 , x 9 , x 1 ) ∩ (x 9 , x 1 , x 2 ).
Our strategy is to use Theorem 10.2.6. To do this, put v := x 3 x 6 x 9 and L := DI(C 9 ). It is routine to investigate that v ∈ p \ p 2 for any p ∈ Min(L). To complete the argument, one has to verify that m \ x i / ∈ Ass(R/(L \ x i ) s ) for all s and x i ∈ {x 3 , x 6 , x 9 }. Fix s ≥ 1. We first prove that m \ x 3 / ∈ Ass(R/(L \ x 3 ) s ), where m = (x 1 , . . . , x 9 ). It follows from the definition of deletion that

L \ x 3 = (x 1 , x 2 ) ∩ (x 2 , x 4 ) ∩ (x 4 , x 5 ) ∩ (x 5 , x 6 , x 7 ) ∩ (x 6 , x 7 , x 8 ) ∩ (x 7 , x 8 , x 9 ) ∩ (x 8 , x 9 , x 1 ). Set α := x 1 x 4 x 7 and A := L \ x 3 . It is easy to see that α ∈ p \ p 2 for any p ∈ Min(A). We show that Q \ x i / ∈ Ass(R/(A \ x i ) s ) for all s and x i ∈ {x 1 , x 4 , x 7 }, where Q = m \ x 3 . Fix s ≥ 1. We first prove that Q \ x 1 / ∈ Ass(R/(A \ x 1 ) s ). Direct computation gives that A \ x 1 = (x 6 , x 7 , x 8 ) ∩ (x 5 , x 6 , x 7 ) ∩ (x 8 , x 9 ) ∩ (x 4 , x 5 ) ∩ (x 2 ).
It follows from Lemma 5.4.13 that (x 6 , x 7 , x 8 ) ∩ (x 5 , x 6 , x 7 ) is normally torsionfree, and by virtue of Lemma 6.4.9, we get (x 6 , x 7 , x 8 )∩(x 5 , x 6 , x 7 )∩(x 8 , x 9 )∩(x 4 , x 5 ) is normally torsion-free. According to Proposition 5.4.12, one has A \ x 1 is normally torsion-free, and so Q \ x 1 / ∈ Ass(R/(A \ x 1 ) s ). By a similar argument, one can show that A \ x 4 = (x 2 ) ∩ (x 5 ) ∩ (x 6 , x 7 , x 8 ) ∩ (x 7 , x 8 , x 9 ) ∩ (x 8 , x 9 , x 1 ), and A \ x 7 = (x 1 , x 2 ) ∩ (x 2 , x 4 ) ∩ (x 4 , x 5 ) ∩ (x 5 , x 6 ) ∩ (x 6 , x 8 ) ∩ (x 8 , x 9 ), are normally torsion-free. Hence, we get Q \ x 4 / ∈ Ass(R/(A \ x 4 ) s ) and Q \ x 7 / ∈ Ass(R/(A \ x 7 ) s ). We deduce from Theorem 10.2.6 that A = L \ x 3 is normally torsion-free, and so m \ x 3 / ∈ Ass(R/(L \ x 3 ) s ). In what follows, we give the sketch of the proof of normally torsion-freeness of B := L \ x 6 and C := L \ x 9 . We can rapidly conclude the following

L \ x 6 = (x 4 , x 5 ) ∩ (x 5 , x 7 ) ∩ (x 7 , x 8 ) ∩ (x 1 , x 2 , x 3 ) ∩ (x 2 , x 3 , x 4 ) ∩ (x 1 , x 8 , x 9 ) ∩ (x 1 , x 2 , x 9 ), and
L \ x 9 = (x 7 , x 8 ) ∩ (x 8 , x 1 ) ∩ (x 1 , x 2 ) ∩ (x 2 , x 3 , x 4 ) ∩ (x 3 , x 4 , x 5 ) ∩ (x 4 , x 5 , x 6 ) ∩ (x 5 , x 6 , x 7 ).
It is routine to check that α = x 1 x 4 x 7 ∈ p \ p 2 for any p ∈ Min(B) (respectively, for any p ∈ Min(C)). Furthermore, we have the following

B \ x 1 = (x 4 , x 5 ) ∩ (x 5 , x 7 ) ∩ (x 7 , x 8 ) ∩ (x 8 , x 9 ) ∩ (x 9 , x 2 ) ∩ (x 2 , x 3 ), B \ x 4 = (x 5 ) ∩ (x 7 , x 8 ) ∩ (x 2 , x 3 ) ∩ (x 1 , x 8 , x 9 ) ∩ (x 1 , x 2 , x 9 ), B \ x 7 = (x 5 ) ∩ (x 8 ) ∩ (x 1 , x 2 , x 3 ) ∩ (x 2 , x 3 , x 4 ) ∩ (x 1 , x 2 , x 9 ), C \ x 1 = (x 8 ) ∩ (x 2 ) ∩ (x 3 , x 4 , x 5 ) ∩ (x 4 , x 5 , x 6 ) ∩ (x 5 , x 6 , x 7 ), C \ x 4 = (x 7 , x 8 ) ∩ (x 8 , x 1 ) ∩ (x 1 , x 2 ) ∩ (x 2 , x 3 ) ∩ (x 3 , x 5 ) ∩ (x 5 , x 6 ),
and

C \ x 7 = (x 8 ) ∩ (x 1 , x 2 ) ∩ (x 5 , x 6 ) ∩ (x 2 , x 3 , x 4 ) ∩ (x 3 , x 4 , x 5 ).
By using Lemma 5.4.13, Theorem 5.3.4, and Lemma 6.4.9, we can conclude that B \ x 1 , B \ x 4 , and B \ x 7 (respectively, C \ x 1 , C \ x 4 , and C \ x 7 ) are normally torsion-free. This gives rise to m \ x 6 / ∈ Ass(R/(I \ x 6 ) s ) (respectively, m \ x 9 / ∈ Ass(R/(I \ x 9 ) s ). Finally, Theorem 10.2.6 gives that DI(C 9 ) is normally torsionfree.

We are now ready to establish the first main result of this subsection in the following theorem. Proof. We first assume that n ∈ {3, 6, 9}.

Theorem 5.4.16. Let

C n = (V (C n ), E(C n )) be a cycle graph of order n ≥ 3. Then DI(C n ) t = DI(C n ) (t
Let V (C n ) = {x 1 , . . . , x n } and E(C n ) = {{x i , x i+1 } : i = 1, . . . , n-1}∪{{x n , x 1 }}. It is easy to see that DI(C 3 ) = (x 1 , x 2 , x 3 )
is normally torsion-free. According to Lemma 5.4.15, it follows that DI(C 6 ) and DI(C 9 ) are also normally torsion-free. To complete the proof, we need to show that DI(C n ) t = DI(C n ) (t) for some t ≥ 1 when n / ∈ {3, 6, 9}. To do this, we divide the proof into three cases. We will repeatedly use the following facts in our proof.

(i) The domination number of C 3k is k, and the domination number of C 3k+1 and C 3k+2 is k + 1, see [START_REF] Hernández-Gómez | Transitivity on Minimum Dominating Sets of Paths and Cycles[END_REF] for more information.

(ii) The associated primes of DI(C n ) correspond to the closed neighborhoods of the vertices of C n . Now, we proceed with the proof.

Case 1.

Let n = 3k for some k ≥ 4. It follows from [START_REF] Hernández-Gómez | Transitivity on Minimum Dominating Sets of Paths and Cycles[END_REF]Lemma 4] that there are exactly three minimal dominating sets of C n of cardinality k, given by A = {x 1 , x 4 , x 7 , . . . , x 3k-2 }, B = {x 2 , x 5 , x 8 , . . . , x 3k-1 }, and

C = {x 3 , x 6 , x 9 , . . . , x 3k }.
Our aim is to construct an element f with deg f = 2k +1 such that none of A or B or C is in supp(f ). To do this, set h := x 4 x 8 k j=4 x 3j and g := 3k i=1 x i . We claim that f = g/h / ∈ DI(C 3k ) 2 but f ∈ DI(C 3k ) (2) . It is routine to check that deg h = k -1, and hence deg f = 3k -k + 1 = 2k + 1. In addition, from the constructions of A, B, and C, we know that the only elements of degree k in G(DI(C 3k )) are

{ x i ∈A x i , x i ∈B x i , x i ∈C x i }. As x 4 , x 8 , x 12 /
∈ supp(f ), we deduce that x i ∈A x i ∤ f, x i ∈B x i ∤ f, and x i ∈C x i ∤ f . Therefore, none of the elements of degree k in DI(C 3k ) divides f , and consequently

f / ∈ DI(C 3k ) 2 .
In the light of [60, Proposition 1.4.4], we obtain

DI(C 3k ) (2) = p∈Ass(DI(C 3k )) p 2 .
To see f ∈ DI(C 3k ) (2) , note that the closed neighborhood set of each vertex of C 3k is covered twice by supp(f ). Indeed, for each x i ∈ C 3k , we have either 2) , and so the assertion is true for C 4 . Let n = 3k + 1 for some k ≥ 2. Let f ′ :=

x i x i+1 ∈ supp(f ), or x i-1 x i ∈ supp(f ), or x i-1 x i+1 ∈ supp(f ). Case 2. It follows from x 1 x 2 x 3 ∈ DI(C 4 ) (2) \DI(C 4 ) 2 that DI(C 4 ) 2 = DI(C 4 ) (
x 3k+1 k-1 i=0 x 3i+1 x 3i+2 . Then deg f ′ = 2k + 1.
By virtue of the minimal degree of any generator in DI(C 3k+1 ) 2 is 2k + 2, this shows that f ′ / ∈ DI(C 3k+1 ) 2 . Moreover, the closed neighborhood set of each vertex of C 3k+1 is covered twice by supp(f ′ ), which yields that f ′ ∈ DI(C 3k+1 ) (2) .

Case 3. Let n = 3k + 2 for some k ≥ 1 and put f ′′ := 3k+2 i=1 x i . Then deg f ′′ = 3k + 2. In view of the minimal degree of any generator in DI(C 3k+2 ) 3 is 3k + 3, we get f ′′ / ∈ DI(C 3k+2 ) 3 . Furthermore, the closed neighborhood set of each vertex of C 3k+2 is covered three times by supp(f ′′ ), which gives that f ′′ ∈ DI(C 3k+2 ) (3) .

Accordingly, the discussion above shows that if n / ∈ {3, 6, 9}, then DI(C n ) t = DI(C n ) (t) for some t ≥ 1. This finishes the proof.

In general, it is well-known that the cover ideals of even cycle graphs are always normally torsion-free (cf. [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6]). It has already been shown in Theorem 6.2.1 that the cover ideals of odd cycle graphs are nearly normally torsion-free (to recall the definition of nearly normally torsion-freeness refer to Section 1.1). Furthermore, the associated primes of powers of cover ideals of odd cycle graphs have already been studied in [START_REF] Nasernejad | Associated primes of powers of cover ideals under graph operations[END_REF], refer to Proposition 3.1. 15.

In what follows, our aim is to establish that a similar result as above holds for the associated primes of powers of dominating ideals of cycle graphs of any order. Before stating the next theorem, one has to recall the definition of the monomial localization of a monomial ideal with respect to a monomial prime ideal in Definition 3.1.14.

We are now in a position to state our next result related to the associated primes of powers of dominating ideals of cycle graphs. Theorem 5.4.17. Let C n be a cycle graph of order n with V (C n ) = {x 1 , . . . , x n } and

E(C n ) = {{x i , x i+1 } : i = 1, . . . , n -1} ∪ {{x n , x 1 }}. Let DI(C n ) be the dominating ideal of C n in the polynomial ring R = K[x 1 , . . . , x n ].
Then the following statements hold.

(i) The monomial ideal DI(C n ) satisfies the persistence property.

(ii) For all s ≥ 1, we have

Ass(DI(C n ) s ) ⊆ {(x i , x i+1 , x i+2 ) : i = 1, . . . , n} ∪ {m}, where m = (x 1 , . . . , x n ) and x n+1 (respectively, x n+2 ) represents x 1 (respec- tively, x 2 ). (iii) Ass ∞ (DI(C n )) ⊆ {(x i , x i+1 , x i+2 ) : i = 1, . . . , n} ∪ {m}.
Proof. For convenience of notation, put L := DI(C n ). Since all claims are true for DI(C 3 ) = (x 1 , x 2 , x 3 ), we assume that n ≥ 4. We first show that Ass(L s ) ⊆ Min(L) ∪ {m} for all s ≥ 1. To see this, fix s ≥ 1, and pick an arbitrary element p ∈ Ass(L s ). If p = m, then there is nothing to show. We thus assume that p = m. Without loss of generality, one may assume that p = (x i : i = 1, . . . , r), where 3 ≤ r < n. Then we get L(p) = ℓ i=1 I ∨ 3 (P i ), where P 1 , . . . , P ℓ are some disjoint paths in C n . By virtue of Lemma 5.4.13, one can immediately conclude that I ∨ 3 (P i ) is normally torsion-free for all i = 1, . . . , ℓ. In the light of Proposition 5.4.12, we get L(p) is normally torsion-free, and hence p ∈ Min(L(p)). On the other hand, based on Lemma 3.2.15(viii), we know that p ∈ Ass(L s ) if and only if p ∈ Ass(R(p)/L(p) s ). Consequently, one obtains p ∈ Ass(R(p)/L(p) s ). We therefore get p = (x α , x α+1 , x α+2 ) for some 1 ≤ α ≤ r -2, and thus p ∈ Min(L). Accordingly, we derive that Ass(L s ) ⊆ Min(L) ∪ {m} for all s ≥ 1. This proves (ii).

We now show that L has the persistence property. Note that, in view of Theorem 3.2.34 and Proposition 3.1.12, every normal monomial ideal has the persistence property. Hence, it is sufficient for us to demonstrate that L is normal. Our strategy is to use Theorem 6.5.1. To accomplish this, set

A := n-3 i=1 (x i , x i+1 , x i+2 ) and B := (x n-1 , x n-2 ) ∩ (x n-1 , x 1 ) ∩ (x 1 , x 2 ).

Then we have

L = A ∩ (x n R + B) = x n A + A ∩ B.
Due to A is the Alexander dual of the 3-path ideal of a path graph P = (V (P ), E(P )) with the vertex set V (P ) = {x 1 , . . . , x n-1 } and the following edge set

E(P ) = {{x i , x i+1 } : i = 1, . . . , n -2},
Lemma 5.4.13 gives that A is normally torsion-free, and by [60, Theorem 1.4.6], we deduce that A is normal. Also, note that A + A ∩ B = A is normal. In what follows, our aim is to establish A ∩ B is normal. It is easy to check that

A ∩ B = n-4 i=2 (x i , x i+1 , x i+2 ) ∩ (x n-1 , x n-2 ) ∩ (x n-1 , x 1 ) ∩ (x 1 , x 2 ) =x 1 ( n-4 i=2 (x i , x i+1 , x i+2 ) ∩ (x n-1 , x n-2 )) +x 2 R ∩ n-4 i=3 (x i , x i+1 , x i+2 ) ∩ (x n-1 , x n-2 ) ∩ (x n-1 , x 1 ).
To simplify the notation, put C := n-4 i=2 (x i , x i+1 , x i+2 ) ∩ (x n-1 , x n-2 ) and

D := x 2 R ∩ n-4 i=3 (x i , x i+1 , x i+2 ) ∩ (x n-1 , x n-2 ) ∩ (x n-1 , x 1 ).
This gives rise to A∩B = x 1 C+D. As the monomial ideal n-4 i=2 (x i , x i+1 , x i+2 ) can be viewed as the Alexander dual of the 3-path ideal of a path graph, by Lemma 5.4.13, is normally torsion-free, and on account of Lemma 6.4.9, we can conclude that C is normally torsion-free, and based on [60, Theorem 1.4.6], is normal. Moreover, since D ⊆ C, one has C + D = C is normal as well. Now, a similar argument shows that the monomial ideal n-4 i=3 (x i , x i+1 , x i+2 ) is normally torsion-free, and after two times using Lemma 6.4.9, we can derive that n-4 i=3 (x i , x i+1 , x i+2 ) ∩ (x n-1 , x n-2 )∩ (x n-1 , x 1 ) is normally torsion-free. It follows now from Lemma 7.4.5 that D is normal. Thanks to Theorem 6.5.1, we get A ∩ B = x 1 C + D is normal. Due to A and A ∩ B are normal, Theorem 6.5.1 yields that L = x n A + A ∩ B is normal, and so has the persistence property, as we claimed in (i).

Finally, combining (i) and (ii) leads us to

Ass ∞ (DI(C n )) ⊆ {(x i , x i+1 , x i+2 ) : i = 1, . . . , n} ∪ {m}.
This demonstrates (iii), and the proof is complete.

Chapter 6 On the nearly normally torsion-freeness of monomial ideals

In this chapter, we deal with the concept of nearly normally torsion-free monomial ideals. Recall that a monomial ideal I in a polynomial ring R = K[x 1 , . . . , x n ] over a field K is called nearly normally torsion-free if there exist a positive integer k and a monomial prime ideal p such that Ass R (R/I m ) = Min(I) for all 1 ≤ m ≤ k, and Ass R (R/I m ) ⊆ Min(I) ∪ {p} for all m ≥ k + 1, see [9, Definition 2.1]. This notion generalizes normally torsion-freeness such that every normally torsion-free monomial ideal is nearly normally torsion-free. More recently, in [9, Theorem 2.3], the author characterized all connected graphs whose edge ideals are nearly normally torsion-free. The main purpose of this chapter is to continue this argument. In particular, we present some new classes of nearly normally torsion-free monomial ideals.

Some classes of nearly normally torsion-free ideals

In this section, our aim is to give additional results related to nearly normally torsion-free monomial ideals. To achieve this, one has to recall the definition of the monomial localization of a monomial ideal with respect to a monomial prime ideal, see Definition 3.1.14. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Classes of normally and nearly normally torsion-free monomial ideals[END_REF]. Lemma 6.1.1. Let I be a nearly normally torsion-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. Then there exists a monomial prime ideal p ∈ V * (I) such that I(p \ {x i }) is normally torsion-free for all x i ∈ p.

Proof. Since I is nearly normally torsion-free, there exists a positive integer k and a monomial prime ideal p such that Ass R (R/I m ) = Min(I) for all 1 ≤ m ≤ k, and Ass R (R/I m ) ⊆ Min(I) ∪ {p} for all m ≥ k + 1. We claim that I(p \ {x i }) is normally torsion-free for all x i ∈ p. To do this, fix x i ∈ p, and set q := p \ {x i }. We need to show that Ass R(q) (R(q)/(I(q)) ℓ ) ⊆ Ass R(q) (R(q)/I(q)) for all ℓ. Fix ℓ ≥ 1. It follows from Lemma 3.2.15 that Ass R(q) (R(q)/(I(q)) ℓ ) = Ass R(q) (R(q)/I ℓ (q)) = {Q : Q ∈ Ass R (R/I ℓ ) and Q ⊆ q}.

Pick an arbitrary element Q ∈ Ass R(q) (R(q)/(I(q)) ℓ ). Thus, Q ∈ Ass R (R/I ℓ ) and Q ⊆ q. Since q = p \ {x i }, this yields that Q = p; thus, one must have Q ∈ Min(I). Hence, Q ∈ Ass R (R/I), and so Q ∈ Ass R(q) (R(q)/I(q)). Therefore, I(p \ {x i }) is normally torsion-free. Lemma 6.1.2. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K such that Ass R (R/I) = Min(I). Let I(m \ {x i }) be normally torsionfree for all i = 1, . . . , n, where m = (x 1 , . . . , x n ). Then I is nearly normally torsionfree.

Proof. In the light of Ass R (R/I) = Min(I), it is enough for us to show that Ass R (R/I k ) ⊆ Min(I) ∪ {m} for all k ≥ 2. To achieve this, fix k ≥ 2, and take an arbitrary element Q ∈ Ass R (R/I k ). If Q = m, then the proof is over. Hence, let Q = m. On account of Q is a monomial prime ideal, this implies that Q ⊆ m \ {x j } for some x j ∈ m. Put q := m \ {x j }. Because I(q) is normally torsion-free, we thus have Ass R(q) (R(q)/(I(q)) k ) ⊆ Ass R(q) (R(q)/I(q)). Also, one can deduce from Lemma 3.2.15 that Q ∈ Ass R(q) (R(q)/(I(q)) k ), and so Q ∈ Ass R(q) (R(q)/I(q)). This yields that Q ∈ Ass R (R/I), and hence Q ∈ Min(I). This completes the proof.

As an immediate consequence of Lemma 6.1.2, we obtain the following corollary: Corollary 6.1.3. Let I be a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. Let I(m \ {x i }) be normally torsion-free for all i = 1, . . . , n, where m = (x 1 , . . . , x n ). Then I is nearly normally torsion-free.

As an application of Lemma 6.1.2, we give a class of nearly normally torsion-free monomial ideals in the subsequent theorem. To see this, one needs to recall from [START_REF] Herzog | Monomial ideals with primary components given by powers of monomial prime ideals[END_REF] that a monomial ideal is said to be a monomial ideal of intersection type when it can be presented as an intersection of powers of monomial prime ideals. Proposition 6.1.4. Let I = p∈Ass R (R/I) p dp ⊂ R = K[x 1 , . . . , x n ] be a monomial ideal of intersection type such that for any 1 ≤ i ≤ n, there exists unique p ∈ Ass R (R/I) with I(m \ {x i }) = p dp , where m = (x 1 , . . . , x n ). Then I is nearly normally torsion-free. Example 6.1.6. Let R = K[x 1 , x 2 , x 3 ] be the polynomial ring over a field K and I := (x 4 2 , x 1 x 3 2 , x 3 1 x 2 , x 4 1 x 3 ) be a monomial ideal of R. On account of

I = (x 1 , x 4 2 ) ∩ (x 3 1 , x 3 2 ) ∩ (x 4 1 , x 2 ) ∩ (x 2 , x 3 ), one can conclude that Ass R (R/I) = Min(I) = {(x 1 , x 2 ), (x 2 , x 3 )}. We claim that Ass R (R/I m ) = Min(I) ∪ {(x 1 , x 2 , x 3 )} for all m ≥ 2. To prove this claim, fix m ≥ 2.
In what follows, we verify that (x 1 , x 2 , x 3 ) ∈ Ass R (R/I m ). To establish this, we show that (I m : R v) = (x 1 , x 2 , x 3 ), where v := x 3m-1 1

x m+1 2 .
To see this, one may consider the following statements:

(i) Since vx 1 = x 3m 1 x m+1 2 = (x 3 1 x 2 ) m x 2 and x 3 1 x 2 ∈ I, we get vx 1 ∈ I m , and so x 1 ∈ (I m : R v); (ii) Due to vx 2 = x 3m-1 1 x m+2 2 = (x 3 1 x 2 ) m-1 (x 2 1 x 3 2
) and x 2 1 x 3 2 ∈ I, one has vx 2 ∈ I m , and hence x 2 ∈ (I m : R v);

(iii) As vx 3 = x 3m-1 1 x m+1 2 x 3 = (x 3 1 x 2 ) m-2 (x 5 1 x 3 2
x 3 ) and x 5 1 x 3 2 x 3 ∈ I 2 , we obtain vx 3 ∈ I m , and thus x 3 ∈ (I m : R v).

Consequently, (x 1 , x 2 , x 3 ) ⊆ (I m : R v). For the converse, let v ∈ I m . This leads to there exist monomials h 1 , . . . , h m ∈ G(I) such that x 3 ∤ h i for each i = 1, . . . , m, and

h 1 • • • h m |x 3m-1 1 x m+1 2
. Hence, we have the following equality (6.1.3)

h 1 • • • h m = (x 4 2 ) α 1 (x 1 x 3 2 ) α 2 (x 3 1 x 2 ) α 3 ,
for some nonnegative integers α 1 , α 2 , α 3 with α 1 + α 2 + α 3 = m. Particularly, one can derive from (6.1.3) that (6.1.4)

4α 1 + 3α 2 + α 3 ≤ m + 1, and
(6.1.5) α 2 + 3α 3 ≤ 3m -1.
Since 3 i=1 α i = m, we obtain from (6.1.4) that m + 3α 1 + 2α 2 ≤ m + 1, and so 3α 1 + 2α 2 ≤ 1. We thus have α 1 = α 2 = 0, and so α 3 = m. Moreover, it follows from (6.1.5) that 3m ≤ 3m -1, which is a contradiction. Therefore, v / ∈ I m , and so (I m : R v) = (x 1 , x 2 , x 3 ). This gives rise to (x 1 , x 2 , x 3 ) ∈ Ass R (R/I m ) for all m ≥ 2. Hence, Ass R (R/I m ) = Min(I) ∪ {(x 1 , x 2 , x 3 )} for all m ≥ 2. This means that I has the persistence property and is a nearly normally torsion-free ideal. On the other hand, using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF] shows that (I 2 : R I) = I. We therefore deduce that I does not satisfy the strong persistence property.

Nearly normally torsion-freeness and cover ideals

In the following, our goal is to characterize all finite simple connected graphs such that their cover ideals are nearly normally torsion-free, see Theorem 6.2.1. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Classes of normally and nearly normally torsion-free monomial ideals[END_REF].

To understand the proof of Theorem 6.2.1, we first review some notations from [START_REF] Jha | A note on outerplanarity of product graphs[END_REF] as follows:

Let G = (V (G), E(G)) be a finite simple connected graph. For any x, y ∈ V (G), an (x, y)-path is simply a path between the vertices x and y in G. Also, for a vertex subset W of a graph G, W will denote the subgraph of G induced by W .

Let G be an almost bipartite graph, that is to say, G has only one induced odd cycle subgraph, say C 2k+1 . For each i ∈ V (C 2k+1 ), let

A i = {x ∈ V (G)| x = i
and for all j ∈ V (C 2k+1 ), i appears on every (x, j)-path}.

Based on [75, Page 540], it should be noted that the set A i may be empty for some i, and it follows from [75, Lemma 2.3] that if A i = ∅, then the induced subgraph A i is bipartite in its own right. Also, for every edge e = {i, j} ∈ E(C 2k+1 ), let

B e = {x ∈ V (G) \ V (C 2k+1 )| for every m ∈ V (C 2k+1 ) \ {i, j}
, there is an (x, m)-path in which i appears but j does not, and an (x, m)-path in which j appears but i does not}.

Once again, according to [75, Page 541], one must note that the set B e may be empty for some e, and it can be deduced from [75, Lemma 2.3] that if B e = ∅, then the induced subgraph B e is bipartite in its own right. Furthermore, the A i 's and B e 's are all mutually disjoint. In other words, each vertex of G is in exactly one of the following sets: (i) V (C 2k+1 ), (ii) A i for some i, and (iii) B e for some e. Moreover, for each i, j ∈ V (C 2k+1 ) with i = j, and e, e ′ ∈ E(C 2k+1 ) with e = e ′ , one can easily derive from the definitions that there is no path between any two vertices of A i and B e , or A i and A j , or B e and B e ′ .

As an example, consider the following graph G from [START_REF] Jha | A note on outerplanarity of product graphs[END_REF].

= K[x α : α ∈ V (G)] and R ′ = K[x α : α ∈ V (H)].
Proof. For convenience of notation, set I := J(G) and J := J(H). We first prove that Ass R (R/I s ) ∪ {(x v , x w )} ⊆ Ass R ′ (R ′ /J s ) for all s. Fix s ≥ 1, and assume that p = (x i 1 , . . . , x ir ) is an arbitrary element of Ass R (R/I s ). According to [45, Lemma 2.11], we get p ∈ Ass(K[p]/J(G p ) s ), where K[p] = K[x i 1 , . . . , x ir ] and G p is the induced subhypergraph of G on the vertex set {i 1 , . . . , i r } ⊆ V (G). Since G p = H p , we have p ∈ Ass(K[p]/J(H p ) s ). This yields that p ∈ Ass R ′ (R ′ /J s ). On account of

(x v , x w ) ∈ Ass R ′ (R ′ /J s ), one derives Ass R (R/I s ) ∪ {(x v , x w )} ⊆ Ass R ′ (R ′ /J s ).
To complete the proof, it is enough for us to show the reverse inclusion. Assume that [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11] implies that p ∈ Ass R (R/I s ), and the proof is done. Thus, let w ∈ {i 1 , . . . , i r }. It follows from [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Corollary 4.5] that the associated primes of J(H) s will correspond to critical chromatic subhypergraphs of size s + 1 in the s-th expansion of H. This means that one can take the induced subhypergraph on the vertex set {i 1 , . . . , i r }, and then form the s-th expansion on this induced subhypergraph, and within this new hypergraph find a critical (s + 1)-chromatic hypergraph. Notice that since this expansion cannot have any critical chromatic subgraphs, this implies that H p must be connected. Hence, v ∈ {i 1 , . . . , i r }. Without loss of generality, one may assume that i 1 = v and i 2 = w. Thanks to w is only connected to v in the hypergraph H, and because this induced subhypergraph is critical, if we remove the vertex w, one can color the resulting hypergraph with at least s colors. This leads to that w has to be adjacent to at least s vertices. But the only thing w is adjacent to is the shadows of w and the shadows of v, and so one has a clique among these vertices. Accordingly, w and its neighbors will form a clique of size s + 1. Since a clique is a critical graph, it follows that we do not need any element of {i 3 , . . . , i r } or their shadows when making the critical (s + 1)-chromatic hypergraph. Consequently, we obtain p = (x v , x w ), as required.

p = (x i 1 , . . . , x ir ) is an arbitrary element of Ass R ′ (R ′ /J s ) with {i 1 , . . . , i r } ⊆ V (H). If {i 1 , . . . , i r } ⊆ V (G), then
Before stating the next result, it should be noted that one can always view a square-free monomial ideal as the cover ideal of a simple hypergraph. In fact, assume that I is a square-free monomial ideal, and I ∨ denotes its Alexander dual. Also, let H denote the hypergraph corresponding to I ∨ . Then, we have I = J(H), where J(H) denotes the cover ideal of the hypergraph H. Consult [START_REF] Francisco | Powers of square-free monomial deals and combinatorics[END_REF] for further details and information.

For instance, consider the following square-free monomial ideal in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 ] over a field K,

I = (x 1 x 2 x 3 , x 2 x 3 x 4 , x 3 x 4 x 5 , x 4 x 5 x 1 , x 5 x 1 x 2 ).
(iii) By virtue of [60, Theorem 1.4.6], every normally torsion-free square-free monomial ideal is normal. Now, the assertion can be deduced from (i).

(iv) Due to Theorem 3.2.34, every normal monomial ideal has the strong persistence property, and hence the claim follows readily from (iii).

(v) It is shown in [START_REF] Herzog | Persistence and stability properties of powers of ideals[END_REF] or Proposition 3.1.12 that the strong persistence property implies the persistence property. Therefore, one can derive the assertion from (iv).

(vi) Thanks to [START_REF] Reyes | On the strong persistence property for monomial ideals[END_REF]Theorem 11], the strong persistence property implies the symbolic strong persistence property, and thus the claim is an immediate consequence of (iv).

The case of t-spread principal Borel ideals

In this section, we focus on normally torsion-free and nearly normally torsion-free t-spread principal Borel ideals. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Classes of normally and nearly normally torsion-free monomial ideals[END_REF].

We first recall the definition of t-spread monomial ideals from [START_REF] Ene | t-spread strongly stable monomial ideals[END_REF].

Definition 6.3.1. Let t ≥ 0 and R = K[x 1 , . . . , x n ] be a polynomial ring over a field K. A monomial x i 1 x i 2 • • • x i d ∈ R with i 1 ≤ i 2 ≤ • • • ≤ i d is called t-spread if i j -i j-1 ≥ t for all j = 2, . . . , d. A monomial ideal in R is called a t-
= x i 1 x i 2 • • • x i d ∈ R, we have x j 1 x j 2 • • • x j d ∈ G(B t (u)) if and only if j 1 ≤ i 1 , . . . , j d ≤ i d and j k -j k-1 ≥ t for k ∈ {2, .
. . , d}. More information can be found in [START_REF] Ene | t-spread strongly stable monomial ideals[END_REF].

To see an application of Corollary 6.1.

x i x n , x i x n-1 , . . . , x i x i+t , x i-1 x n , x i-1 x n-1 , . . . , x i-1 x i+t-1 , x i-2 x n , x i-2 x n-1 , . . . , x i-2 x i+t-2 , . . . , x 1 x n , x 1 x n-1 , . . . , x 1 x t+1 .
Our strategy is to use Corollary 6.1.3. To do this, one has to show that I(m \ {x z }) is normally torsion-free for all z = 1, . . . , n, where m = (x 1 , . . . , x n ). First, let

1 ≤ z ≤ i -1. Direct computation gives that I(m \ {x z }) = (x α x β : α = 1, . . . , z -1, β = t + 1, . . . , z + t -1, β -α ≥ t) + (x n , x n-1 , . . . , x z+t ).
One can deduce that

(x α x β : α = 1, . . . , z -1, β = t + 1, . . . , z + t -1, β -α ≥ t)
is the edge ideal of a bipartite graph with the following vertex set

{1, . . . , z -1} ∪ {t + 1, . . . , z + t -1},
and so is normally torsion-free. Also, we know that every monomial prime ideal is normally torsion-free. Moreover, on account of the two ideals (x n , x n-1 , . . . , x z+t ) and (x α x β : α = 1, . . . , z -1, β = t + 1, . . . , z + t -1, β -α ≥ t) do not have any common variables, we conclude from Theorem 5.3.4 that I(m \ {x z }) is normally torsion-free as well.

Here, we consider the case z = i. Then we have

I(m \ {x i }) = (x α x β : α = i -t + 1, . . . , i -1, β = i + 1, . . . , i + t -1, β -α ≥ t) + (x n , x n-1 , . . . , x i+t ) + (x 1 , x 2 , . .

. , x i-t ).

A similar technique shows that I(m \ {x i }) is again normally torsion-free. Now, let i + 1 ≤ z ≤ n. It is routine to check that

I(m \ {x z }) = (x α x β : α = z -t + 1, . . . , i, β = z + 1, . . . , n, β -α ≥ t) + (x 1 , x 2 , . . . , x z-t ).
It is easy to see that ) is given below:

(x α x β : α = z -t + 1, . . . , i, β = z + 1, . . . , n, β -α ≥ t)
x 1 x 4 x 1 x 5 x 2 x 5 x 1 x 6 x 2 x 6 x 3 x 6 x 1 x 7 x 2 x 7 x 3 x 7 x 4 x 7
The monomial localizations of I at m \ {x k }, for each k = 1, . . . 7, are listed below:

I(m \ {x 1 }) = (x 4 , x 5 , x 6 , x 7 ); I(m \ {x 2 }) = (x 5 , x 6 , x 7 ) + (x 1 x 4 ); I(m \ {x 3 }) = (x 6 , x 7 ) + (x 1 x 4 , x 1 x 5 ,
I(m \ {x 7 }) = (x 1 , x 2 , x 3 , x 4 ).
Therefore, for each k = 1, . . . , 7, the monomial localization I(m \ {x k }) is either a monomial prime ideal, or sum of a monomial prime ideal and an edge ideal of a bipartite graph, which have no common variables. In each of the above cases,

I(m \ {x k }) is a normally torsion-free ideal. Hence, it follows from Corollary 6.1.3 that I is nearly normally torsion-free.

We further investigate the class of ideals of type B t (u) in the context of normally torsion-freeness and nearly normally torsion-freeness. Given a, b ∈ N, we set

[a, b] := {c ∈ N : a ≤ c ≤ b}. Let u := x i 1 x i 2 • • • x i d-1 x i d and (6.3.1) A k := [(k -1)t + 1, i k ], for each k = 1, . . . d.
We can describe the minimal generators of I = B t (u) in the following way: if

x j 1 • • • x j d ∈ G(I)
, then for each k = 1, . . . , d, we have j k ∈ A k . For example, a complete list of minimal generators of B 2 (x 3 x 5 x 7 ) is given below:

x 1 x 3 x 5 x 1 x 3 x 6 x 1 x 3 x 7 x 1 x 4 x 6 x 2 x 4 x 6 x 1 x 4 x 7 x 2 x 4 x 7 x 1 x 5 x 7 x 2 x 5 x 7 x 3 x 5 x 7
Here,

A 1 = [1, 3], A 2 = [3, 5],
and

A 3 = [5, 7]. Let u = x i 1 x i 2 • • • x i d-1 x i d with i d-1 ≤ (d -1)t, then it forces us to have i k ≤ kt, for each k = 1, . . . , d -2 because u is a t-spread monomial. In this case, A i ∩ A j = ∅.
For example, a complete list of minimal generators of B 3 (x 3 x 6 x 9 ) is given below:

x 1 x 4 x 7 x 1 x 4 x 8 x 1 x 4 x 9 x 1 x 5 x 8 x 2 x 5 x 8 x 1 x 5 x 9 x 2 x 5 x 9 x 1 x 6 x 9 x 2 x 6 x 9 x 3 x 6 x 9
Here, A

If u = x i 1 x i 2 • • • x i d-1 x i d with i d-1 ≤ (d -1)t, then B t (u)

is an edge ideal of a d-uniform d-partite hypergraph whose vertex partition is given by

A 1 , . . . , A d . Recall that H is a d-partite hypergraph if its vertex set V (H) is a disjoint union of sets V 1 , . . . , V d such that if E is an edge of H, then |E ∩ V i | ≤ 1.

Moreover, a hypergraph is d-uniform if each edge of H has size d. In particular, if

H is a d-uniform d-partite hypergraph with vertex partition V 1 , . . . , V d , then |E| = d and |E ∩ V i | = 1 for each E ∈ E(H).
Let A t denote the family of all ideals of the form B t (u) such that B t (u) is an edge ideal of a d-partite d-uniform hypergraph for some d. From the above discussion it follows that B t (u) ∈ A t if and only if there exists some d such that

u = x i 1 x i 2 • • • x i d-1 x i d with i d-1 ≤ (d -1)t. Let H be a hypergraph. A sequence v 1 , E 1 , v 2 , E 2 , . . . , v s , E s , v s+1 = v 1 of distinct edges and vertices of H is called a cycle in H if v i , v i+1 ∈ E i for all i = 1, . . . , s.
Such a cycle is called special if no edge contain more than two vertices of the cycle. After translating the language of simplicial comlexes to hypergraphs, it can be seen from [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 10.3.16] that if a hypergraph does not have any special odd cycles, then its edge ideal is normally torsion-free. The following example shows that hypergraphs with edge ideals in A t , may contain special odd cycles. Example 6.3.7. Let H be the hypergraph whose edge ideal is I = B 3 (x 3 x 6 x 9 x 12 ). Then the following sequence of the vertices and the edges of H gives a special odd cycle.

1, {1, 4, 9, 12}, 9, {2, 5, 9, 12}, 5, {1, 5, 8, 11}, 1

Here, remember the definition of the deletion operator, which has been stated in Definition 3.2.19.

Remark 6.3.8. Let u = x i 1 x i 2 • • • x i d-1 x i d with i d-1 ≤ (d -1)t, and I = B t (u) ⊂ R. Note that we have I ∈ A t . 1. If i k < kt, for some 1 ≤ k ≤ d -1, then the variables x i k +1 , . . . , x kt do not
appear in supp(I). For example, a complete list of minimal generators of

B 3 (x 2 x 5 x 9
) is given below:

x 1 x 4 x 7 x 1 x 4 x 8 x 1 x 4 x 9 x 1 x 5 x 8 x 2 x 5 x 8 x 1 x 5 x 9 x 2 x 5 x 9
Here, A 1 = [1, 2], A 2 = [4, 5], and A 3 = [7, 9], and x 3 , x 6 / ∈ supp(B 3 (x 2 x 5 x 9 )).

2. If i d < n, then x i d +1 , . . . , x n / ∈ supp(I). Hence, we can always assume that i d = n.

If i

k > (k -1)t + 1, that is, |A k | > 1 for some k = 1, . . . , d, then I \ x i k = B t (v)
where v is chosen with the following property:

v = x j 1 • • • x j d ∈ G(I \ x i k ) and for any other w = x l 1 • • • x l d ∈ G(I \ x i k ), we have l k ≤ j k . For example, B 3 (x 2 x 5 x 9 ) \ x 5 = B 3 (x 1 x 4 x 9 )
. Therefore, we conclude that I \ x i k ∈ A t for all x i k with k = 1, . . . , d.

The definition of A k immediately implies that

|A k | = 1 if and only if i k = (k -1)t + 1. Moreover, if i k = (k -1)t + 1 for some 1 ≤ k ≤ d, then i j = (j -1)t + 1 for all j ≤ k because u is a t-spread monomial. In this case, x 1 x t+1 • • • x (k-1)t+1
divides every minimal generator of B t (u), and hence

B t (u) = x 1 x t+1 • • • x (k-1)
t+1 J, where J can be identified as a t-spread principal Borel ideal generated in degree d -k in its ambient polynomial ring. Hence, In what follows, our aim is to show that for any fixed t, all ideals in A t are normally torsion-free. It is known from [10,Corollary 2.6] that B t (u) satisfies the persistence property and the Rees algebra R(B t (u)) is a normal Cohen-Macaulay domain. It is a well-known fact that for any nonzero graded ideal I ⊂ R = K[x 1 , . . . , x n ], if R(I) is Cohen-Macaulay, then lim k→∞ depth(R/I k ) = n -ℓ(I), for example see [START_REF] Herzog | Monomial Ideals[END_REF]Proposition 10.3.2], where ℓ(I) denotes the analytic spread of I, that is, the Krull dimension of the fiber ring R(I)/mR(I). This leads us to the following corollary which will be used in the subsequent results. Note that due to Remark 6.3.8(1), one needs to pay attention to the ambient ring of B t (u). From now on, by the ambient ring, we mean the polynomial ring R containing B t (u) such that all variables in R appear in supp(B t (u)). Corollary 6.3.9.

J = B t (v) ∈ A t with v = u/x 1 x t+1 • • • x (k-1)t+1 . For example, if u = x 1 x 4 x 9 x 12 , then B 3 (u) = x 1 x 4 J with J = B t (v) ⊂ K[x 7 , . . . ,
Let I = B t (u) ⊂ R = K[x i : x i ∈ supp(I)]. If ℓ(I) < n, then lim k→∞ depth(R/I k ) = 0. In particular, if ℓ(I) < n, then m / ∈ Ass(R/I k )
, where m is the unique graded maximal ideal of R.

Next, we compute ℓ(I) for all I ∈ A

(i) For some 1 ≤ k ≤ d, |A k | ≥ 2 if and only if A k ⊆ V (Γ). Moreover, if |A k | ≥ 2, then the induced subgraph of Γ on A k is a complete graph. (ii) Let i k < kt + 1 for some 1 ≤ k ≤ d -1.
Then Γ does not contain any edge with one endpoint lies in A r and the other endpoint lies in A s for any

1 ≤ r < s ≤ k. Proof. (i) We first show that if A k ⊆ V (Γ), then |A k | ≥ 2. Assume that |A k | = 1
for some k. Then from Remark 6.3.8(4), we see that i k = (k -1)t + 1. In this case, each generator of B t (u) is a multiple of x i k , and according to the definition of Γ, we deduce that

A k = {i k } ⊆ V (Γ).
Conversely, we verify that if

|A k | ≥ 2, for some k, then A k ⊆ V (Γ) and the induced subgraph on A k is a complete graph. Take any f, h ∈ A k with f < h. If k = 1, then f < h ≤ i 1 and the monomials v = x h (u/x i 1 ) and v ′ = x f (u/x i 1 ) belong to G(I) because I is t-spread strongly stable. Hence, {f, h} ∈ Γ, as required. A similar argument shows that if |A d | ≥ 2, then A d ⊆ V (Γ) and the induced subgraph on A d is also a complete graph. Now, assume that 1 < k < d. Due to |A k | ≥ 2, we deduce that i k > (k -1)t + 1 and (k -1)t + 1 ≤ f < h ≤ i k . By using the fact that I is t-spread strongly stable, we see that v = x 1 • • • x (k-2)t+1 x f x kt+1 • • • x i d ∈ G(I) and v ′ = x 1 • • • x (k-2)t+1 x h x kt+1 • • • x i d ∈ G(I). Moreover, we have v = (v ′ /x h )x f ,
and so {f, h} ∈ E(Γ), as required.

(ii) If |A r | = 1 or |A s | = 1, then by using (i), we see that the statement holds trivially. Assume that |A r | ≥ 2 and |A s | ≥ 2. Since i k < kt + 1, we have i j < jt + 1 for all j = 1, . . . , k because u is a t-spread monomial. In this case, A r ∩ A s = ∅ for all 1 ≤ r < s ≤ k. Moreover, |A r | ≤ t for all r = 1, . . . , k. Take f ∈ A r and h ∈ A s for some 1 < r < s ≤ k. Then for any u ∈ G(I) with x h |u, we have

(u/x h )x f / ∈ G(I).
This can be easily verified because (u/x h )x f is not a t-spread monomial as it contains two variables with indices in A r and |A r | ≤ t. Hence, we do not have any edge in Γ of the form {f, h} where f ∈ A r and h ∈ A s . Proposition 6.3.12.

Let I = B t (x i 1 x i 2 • • • x i d-1 x i d ) ⊂ R = K[x i : x i ∈ supp(I)] with i d-1 ≤ (d -1)t. Then ℓ(I) < n = |supp(I)|. In particular, m / ∈ Ass(R/I k ) for all k ≥ 1,
where m is the unique graded maximal ideal of R.

Proof. If I is a principal ideal, then the assertion holds trivially. Therefore, we may assume that I is not a principal ideal. To show that ℓ(I) < n, from the equality in (6.3.2), it is enough to prove that the linear relation graph Γ of I has more than one connected components. Let u =

x i 1 x i 2 • • • x i d-1 x i d and A k = [(k -1)t + 1, i k ] for k = 1, . . . ,
ℓ(I) = |A d | -1 + 1 = |A d | < n.
Otherwise, Γ has at least two connected components and again we obtain ℓ(I) < n, as required. Then the assertion m / ∈ Ass(R/I k ), for all k ≥ 1, follows from Corollary 6.3.9.

We now state the second main result of this section in the next theorem. Theorem 6.3.13.

Let I = B t (x i 1 x i 2 • • • x i d-1 x i d ) ⊂ R = K[x i : x i ∈ supp(I)] with i d-1 ≤ (d -1)t. Then I is normally torsion-free.
Proof. We may assume that i k = (k -1)t + 1 for all 1 ≤ k ≤ d. Otherwise, from Remark 6.3.8(4), it follows that I = wB t (v), where w is the product of all variables for which i k = (k -1)t + 1, v = u/w, and B t (v) is a t-spread principal Borel ideal in its ambient ring. Then from Lemma 5.3.11, it follows that I is normally torsionfree if and only if B t (v) is normally torsion-free. Therefore, one may reduce the discussion to B t (v) whose generators are not a multiple of a fixed monomial.

Let

u = x i 1 x i 2 • • • x i d-1 x i d with i 1 > 1.
To show that I = B t (u) is normally torsion-free, we will use Theorem 10.2.6. It can be seen from [9, Theorem 4.2] that u ∈ p \ p 2 for all p ∈ Min(I). We proceed by induction on n := |supp(I)|. Let n = 2. Then d ≤ 2. Hence, i 1 ≤ t, and so we can view I as the edge ideal of a bipartite graph, and so I is normally torsion-free. Let n > 2. Recall that the family of ideals A t is defined by: B t (u) ∈ A t if and only if there exists some d such that

u = x i 1 x i 2 • • • x i d-1 x i d with i d-1 ≤ (d -1)t.
From Remark 6.3.8(3), it follows that I \ x i k ∈ A t for all k = 1, . . . , d. It follows now from the inductive hypothesis that I \ x i k is normally torsion-free for all k = 1, . . . , d. This completes the inductive step, and the proof is over.

In the subsequent theorem, we prove the converse of Theorem 6.3.13 to obtain the complete characterization of normally torsion-free t-spread principal Borel ideals. Theorem 6.3.14.

Let I = B t (x i 1 x i 2 • • • x i d-1 x i d ) ⊂ R = K[x i : x i ∈ supp(I)]. Then

I is normally torsion-free if and only if i d-1 ≤ (d-1)t. In other words, I is normally torsion-free if and only if I can be viewed as an edge ideal of a d-uniform d-partite hypergraph.

Proof. On account of Theorem 6.3.13, it is enough to show that if i d-1 > (d -1)t, then I is not normally torsion-free. Let k be the smallest integer for which i k > kt. Then by remembering this fact that i k+1 -i k ≥ t, we get i k+1 > (k + 1)t. Continuing this process, we obtain i j > jt for all j = k, . . . , d, and as we explained in Remark 6.3.8 (5), one has |A j | ≥ 2 for all j = k, . . . , d. The sets A j 's are defined in (6.3.1). It follows from Lemma 6.3.11 that

A = A k ∪ A k+1 ∪ • • • ∪ A d ⊂ V (Γ),
where Γ is linear relation graph of I and for each j = k, . . . , d, the induced subgraph of Γ on A j is a complete graph. Moreover, i j > jt for each j = k, . . . , d gives that i j ∈ A j ∩A j+1 = ∅ for each j = k, . . . , d-1. Therefore, we conclude that the induced subgraph on A is connected.

Set p := (x (k-1)t+1 , . . . , x i d ). Here, the crucial observation is that if we take the monomial localization of I at p (that is, we map all variables x i to 1, where

i ∈ A 1 ∪ • • • ∪ A k-1
), then we are reducing the degree of each generator of

B t (u) by k -1. It is because A ∩ B = ∅, where B = A 1 ∪ • • • ∪ A k-1 .
Hence, I(p) can be viewed as a t-spread principal Borel ideal by a shift of indices of variables. More precisely, each j ∈ A is shifted to j -(k -1)t. Therefore,

I(p) = B t (x j 1 • • • x j k ), where t < j 1 < • • • < j k .
The linear relation graph of I(p) is isomorphic to the induced subgraph of Γ on the vertex set A. Then we can conclude that This shows that p ∈ Ass(R/I k ) for some k > 1. Moreover, p / ∈ Ass(R/I) because of [9, Theorem 4.2]. Hence, we conclude that I is not normally torsion-free.

ℓ(I(p)) = |supp(B t (x j 1 • • • x j k ))| = dim(R(p)),
In the next example, we illustrate the construction of p as in the proof of Theorem 6.3.14. Example 6.3.15. Let u = x 2 x 7 x 10 x 13 be a 3-spread monomial and I = B 3 (u). Here, i 1 = 2, i 2 = 7, i 3 = 10, and i 4 = 13. Moreover, i 1 = 2 < t = 3, but i 2 = 7 > 2.3. Set p := (x 4 , x 5 , . . . , x 13 ) as in the proof of Theorem 6.3.14. Then the minimal generators of I(p) are listed below. In the following table, u → v indicates that u ∈ G(I(p)) and v is the monomial obtained by shifting each j ∈ [4,13] to j -3.

x 4 x 7 x 10 → x 1 x 4 x 7 x 4 x 7 x 11 → x 1 x 4 x 8 x 4 x 7 x 12 → x 1 x 4 x 9 x 4 x 7 x 13 → x 1 x 4 x 10 x 4 x 8 x 11 → x 1 x 5 x 8 x 5 x 8 x 11 → x 2 x 5 x 8 x 4 x 8 x 12 → x 1 x 5 x 9 x 5 x 8 x 12 → x 2 x 5 x 9
x 4 x 8 x 13 → x 1 x 5 x 10 x 5 x 8 x 13 → x 2 x 5 x 10

x 4 x 9 x 12 → x 1 x 6 x 9 x 5 x 9 x 12 → x 2 x 6 x 9 x 6 x 9 x 12 → x 3 x 6 x 9

x 4 x 9 x 13 → x 1 x 6 x 10 x 5 x 9 x 13 → x 2 x 6 x 10 x 6 x 9 x 13 → x 3 x 6 x 10

x 4 x 10 x 13 → x 1 x 7 x 10 x 5 x 10 x 13 → x 2 x 7 x 10 x 6 x 10 x 13 → x 3 x 7 x 10 x 7 x 10 x 13 → x 4 x 7 x 10 Therefore, I(p) can be viewed as a 3-spread principal Borel ideal B 3 (x 4 x 7 x 10 ). A direct computation with Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF] shows that (x 1 , . . . , x 10 ) is an associated prime of the third power of B 3 (x 4 x 7 x 10 ). Consequently, p is also an associated prime of the third power of I.

When u is a monomial of degree 3, then we have the following outcome. (i) if b < 2t + 1, then I is normally torsion-free.

(ii) if a = 1 and b ≥ 2t + 1, then I is nearly normally torsion-free.

( Let p 1 = (x t+1 , . . . , x i d ). As shown in the last part of the proof of Theorem 6.3.14, p 1 ∈ Ass(R/I k ), for some k > 1, but p 1 / ∈ Ass(R/I). Let p 2 = (x 1 , x t+2 , x t+3 , . . . , x n ); then from [9, Theorem 4.2], we have p 2 / ∈ Ass(R/I). We claim that I(p 2 ) can be viewed as the t-spread principal Borel ideal B t (x r x n ), where r > t. Indeed, by substituting x t+1 = 1, all t-spread monomials of the form x 1 x t+1 x c ′ , with c ′ ∈ A 3 are reduced to x 1 x c ′ . Therefore, the monomials

(R/I k ), for some k > 1, but p 1 , p 2 / ∈ Ass(R/I). Recall from (6.3.1) that if v ∈ G(I), then v = x a ′ x b ′ x c ′ with a ′ ∈ A 1 = [1, a], b ′ ∈ A 2 = [t +
x 1 x b ′ x c ∈ G(I) with b ′ ∈ [t + 2, b] and c ′ ∈ [2t + 2, n] do not appear in G(I(p 2 )). Since a > 1, we have 2 ∈ A 1 . Moreover, if v = x a ′ x b ′ x c ′ ∈ G(I) with a ′ ≥ 2, then b ′ ≥ t + 2 and c ′ ≥ 2t + 2 because v is a t-spread monomial . By substituting x i = 1, for all i ∈ [2, a], all monomials of the form x i x b ′ x c ′ with b ′ ∈ [t + 2, b] and c ′ ∈ [2t + 2, n] are reduced to x b ′ x c ′ .
This shows that I(p 2 ) is generated in degree 2 by t-spread monomials of the following form:

x 1 x c ′ with c ′ ∈ A 3 = [2t + 1, n]; x b ′ x c ′ with b ′ ∈ [t + 2, b] and c ′ ∈ [2t + 2, n] and c ′ -b ′ ≥ t.
Note that supp(I(p 2 )) = {x 1 , x t+2 , x t+3 , . . . , x n }. Moreover, if we shift the indices of variables in {x t+2 , x t+3 , . . . , x n } by k → k -t, then I(p 2 ) can be viewed as a tspread principal Borel ideal B t (x r x s ), where s = n -t and r = b -t > t because b ≥ 2t + 1. Moreover, the linear relation graph of B t (x r x s ) is a connected with |supp(I(p 2 ))| vertices. Hence, lim k→∞ depth(R(p 2 )/I(p 2 ) k ) = 0, and consequently p 2 ∈ Ass(R/I k ) for some k > 1. This finishes our argument.

Closed neighborhood and dominating ideals of classes of trees

Let G be a simple graph. The closed neighborhood ideal N I(G) of G is generated by square-free monomials that correspond to the closed neighborhoods of the vertices of G, whereas, the dominating ideal DI(G) of G is generated by the monomials that correspond to the dominating sets of G (see Section 1.2 for the formal definitions). As shown in [START_REF] Sharifan | Closed neighborhood ideal of a graph[END_REF], N I(G) and DI(G) are the Alexander dual of each other, a similar relation that exists between edge ideals and cover ideals of G.

Domination in graphs was mathematically formulated by Berge and Ore in 1960's and has been widely studied by many researchers due to its enormous and growing applications in various fields including computer sciences, operations research, linear algebra and optimization. Let G be a simple graph with the vertex set V (G). A set S ⊆ V (G) is known as dominating set of G if every vertex in V (G) \ S is adjacent to at least one vertex in S. We refer to [START_REF] Haynes | Fundamentals of Domination in Graphs[END_REF] for further concepts related to the domination in graphs.

In this section, our main goal is to further extend the study on closed neighborhood ideals and dominating ideals. In particular, we focus on normally torsionfreeness and certain stability property of these ideals. Moreover, we establish that the closed neighborhood ideals and the dominating ideals of star graphs are normally torsion-free. To do this, we will first prove several results of general nature. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Dominating ideals and closed neighborhood ideals of graphs[END_REF].

The next theorem will be used frequently to formulate proofs of some main results of this section. It provides a way to create new normally torsion-free ideals based on the existing ones. Theorem 6.4.1. Let I be a normally torsion-free square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] and h be a square-free monomial in R. Let there exist two variables x r and x s with 1 ≤ r = s ≤ n such that gcd(h, u) = 1 or gcd(h, u) = x r or gcd(h, u) = x r x s for all u ∈ G(I). Then the following statements hold:

(i) I + hR is normally torsion-free.

(ii) I + hR is nearly normally torsion-free.

(iii) I + hR is normal.

(iv) I + hR has the strong persistence property.

(v) I + hR has the persistence property.

(vi) I + hR has the symbolic strong persistence property.

Proof. (i) For convenience of notation, put

L := I + hR. If L \ x k = m \ x k for some 1 ≤ k ≤ n, then one can write L = x k J + m \ x k . If J = R, then L = m,
and there is nothing to prove. Let J = R, and take an arbitrary element v ∈ G(J). If x ℓ | v for some ℓ ∈ {1, . . . , n} \ {k}, then v ∈ m \ x k , and so J ⊆ m \ x k . This implies that L = m\x k , and hence the assertion holds. We thus assume that L\x k = m\x k for all k = 1, . . . , n. We claim that h ∈ p\p 2 for any p ∈ Min(L). Take an arbitrary element p ∈ Min(L). Since h ∈ L and L ⊆ p, one has h ∈ p. Suppose, on the contrary, that h ∈ p 2 . Due to h is square-free, this gives that |supp(h) ∩ supp(p)| ≥ 2. We observe the following: (i) If x s ∈ supp(u) for some u ∈ G(I), then x r ∈ supp(u) as well. It is due to the assumption on gcd(h, u) with u ∈ G(I). (ii) At most one of x r and x s can be in supp(p). Indeed, if both x r , x s ∈ supp(p), then x r , x s ∈ supp(h) ∩ supp(p). From (i), we see that u ∈ p \ {x s } for all u ∈ G(I). Also, h ∈ p \ {x s }. Hence, L ⊂ p \ {x s }, a contradiction to the minimality of p.

In order to establish our claim, we have the following cases to discuss:

Case 1. x r ∈ p. Take any x i ∈ supp(h) ∩ supp(p) such that x r = x i . Then

x s = x i due to (ii). From the assumption on gcd(h, u) with u ∈ G(I) it follows that x i / ∈ supp(I). Therefore, I ⊂ p \ {x i }. Since h ∈ p \ {x i }, we conclude that L ⊂ p \ {x i }, a contradiction to the minimality of p.

Case 2. x s ∈ p. By mimicking the same argument as in Case 1, we again obtain a contradiction to the minimality of p.

Case 3. x r /

∈ p and x s / ∈ p. Take any x i , x j ∈ supp(h) ∩ supp(p). Then x i , x j / ∈ supp(I), due to the assumption on gcd(h, u) with u ∈ G(I). It yields that I ⊂ p \ {x i }. Since h ∈ p \ {x i }, we conclude that L ⊂ p \ {x i }, again a contradiction to the minimality of p.

This shows that our claim holds true. To complete the proof, note that for all

x k ∈ supp(h), one has L \ x k = I \ x k .
Based on Theorem 5.3.16, we gain I \ x k is normally torsion-free as well. This leads to L \ x k is normally torsion-free. Fix

s ≥ 1. Suppose, on the contrary, that m \ x k ∈ Ass(R/(L \ x k ) s ) for some k. Because Ass(R/(L \ x k ) s ) = Min(L \ x k ), we get m \ x k ∈ Min(L \ x k )
, and so

L \ x k = m \ x k , which is a contradiction. Therefore, m \ x i / ∈ Ass(R/(I \ x i ) s
) for all s and x i ∈ supp(h). Consequently, the assertion can be readily concluded from Theorem 10.2.6.

(ii) -(vi) can be shown similar to the proof of Corollary 6.2.3.

As an immediate consequence of Theorem 6.4.1, we give the following corollary.

Corollary 6.4.2. The path ideals corresponding to path graphs of length two are normally torsion-free.

Proof. Let P = (V (P ), E(P )) denote a path graph with the vertex set V (P ) = {x 1 , . . . , x n } and the edge set

E(P ) = {{x i , x i+1 } : i = 1, . . . , n -1} ∪ {{x n , x 1 }}.
Hence, the path ideal corresponding to the path graph P of length two is given by

L := (x i x i+1 x i+2 : i = 1, . . . , n -2).
We proceed by induction on n. If n = 3, then L = (x 1 x 2 x 3 ), and there is nothing to prove. Let n > 3 and that the claim has been proved for n -1. Set h := x n-2 x n-1 x n and I := (x i x i+1 x i+2 : i = 1, . . . , n -3). One can easily check that, for each

u ∈ G(I), we have gcd(h, u) = 1 or gcd(h, u) = x n-2 or gcd(h, u) = x n-2 x n-1 .
It follows now from the inductive hypothesis that I is normally torsion-free. Since

L = I + hR, where R = K[x 1 , . . . , x n ],
we can derive the assertion from Theorem 6.4.1.

To see another application of Theorem 6.4.1, we give the following lemma.

Lemma 6.4.3. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple graphs such that V (H) = V (G) ∪ {w} with w / ∈ V (G), and E(H) = E(G) ∪ {{v, w}} for some vertex v ∈ V (G). If N I(G) is normally torsion-free and j∈N

G [v] x j / ∈ G(N I(G)), then N I(H) is normally torsion-free. Proof. Let N I(G) be a normally torsion-free monomial ideal. It is routine to check that N I(H) = N I(G) + (x v x w )R, where R = K[x α : α ∈ V (H)]. Also, one can easily see that either gcd(x v x w , u) = 1 or gcd(x v x w , u) = x v for all u ∈ G(N I(G)).
Therefore, the claim follows immediately from Theorem 6.4.1.

We are ready to state the first main result as an immediate corollary of Theorem 6.4.1 and Lemma 6. In what follows, we investigate the closed neighborhood ideals related to the cone of a graph. (v) Proposition 3.1.12 together with (iv) yield that DI(H) has the persistence property.

(vi) This assertion follows promptly from Theorem 4.1.11.

Our next goal is to show that the dominating ideals of star graphs are normally torsion-free. To do this, we first prove some results of general nature. The following result is a generalized form of Theorem 6.2.2. , t, andE

Theorem 6.4.8. Assume that G = (V (G), E(G)) and H = (V (H), E(H)) are two finite simple hypergraphs such that

V (H) = V (G) ∪ {w 1 , . . . , w t } with w i / ∈ V (G) for each i = 1, . . .
(H) = E(G) ∪ {{v, w 1 , . . . , w t }} for some vertex v ∈ V (G). Then Ass R ′ (R ′ /J(H) s ) = Ass R (R/J(G) s ) ∪ {(x v , x w 1 , . . . , x wt )}, for all s, where R = K[x α : α ∈ V (G)] and R ′ = K[x α : α ∈ V (H)].
Proof. For convenience of notation, set I := J(G) and J := J(H). We first prove that Ass R (R/I s ) ∪ {(x v , x w 1 , . . . , x wt )} ⊆ Ass R ′ (R ′ /J s ) for all s. Fix s ≥ 1, and assume that p = (x i 1 , . . . , x ir ) is an arbitrary element of Ass R (R/I s ). According to [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11], we get p ∈ Ass(K[p]/J(G p ) s ), where

K[p] = K[x i 1 , . . . , x ir ] and G p is the induced subhypergraph of G on the vertex set {i 1 , . . . , i r } ⊆ V (G). Since G p = H p , we have p ∈ Ass(K[p]/J(H p ) s ). This yields that p ∈ Ass R ′ (R ′ /J s ). On account of (x v , x w 1 , . . . , x wt ) ∈ Ass R ′ (R ′ /J s ), one derives Ass R (R/I s ) ∪ {(x v , x w 1 , . . . , x wt )} ⊆ Ass R ′ (R ′ /J s ).
To complete the proof, it is enough for us to show the reverse inclusion. Assume that [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11] implies that p ∈ Ass R (R/I s ), and the proof is done. Thus, let {w 1 , . . . , w t }∩{i 1 , . . . , i r } = ∅. It follows from [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Corollary 4.5] that the associated primes of J(H) s will correspond to critical chromatic subhypergraphs of size s + 1 in the s-th expansion of H. This means that one can take the induced subhypergraph on the vertex set {i 1 , . . . , i r }, and then form the s-th expansion on this induced subhypergraph, and within this new hypergraph find a critical (s + 1)-chromatic hypergraph. Notice that since this expansion cannot have any critical chromatic subgraphs, this implies that H p must be connected. Without loss of generality, one may assume that i 1 = v and i 2 = w 1 , i 3 = w 2 , . . . , i t+1 = w t . Thanks to w 1 , . . . , w t are connected to v in the hypergraph H, and because this induced subhypergraph is critical, if we remove any vertex w k for some 1 ≤ k ≤ t, one can color the resulting hypergraph with at least s colors. This leads to that w k has to be adjacent to at least s vertices. But the only things w k is adjacent to are the shadows of w i for each i = 1, . . . , t, and the shadows of v, and so one has a clique among these vertices. Accordingly, w k and its neighbors will form a clique of size s + 1. Since a clique is a critical graph, it follows that we do not need any element of {i t+2 , . . . , i r } or their shadows when making the critical (s + 1)-chromatic hypergraph. Hence, we obtain p = (x v , x w 1 , . . . , x wt ). This finishes the proof. Lemma 6.4.9. Let I be a normally torsion-free square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] with G(I) ⊂ R. Then the ideal

p = (x i 1 , . . . , x ir ) is an arbitrary element of Ass R ′ (R ′ /J s ) with {i 1 , . . . , i r } ⊆ V (H). If {i 1 , . . . , i r } ⊆ V (G), then
L := IS ∩ (x n , x n+1 , x n+2 , . . . , x m ) ⊂ S = R[x n+1 , x n+2 , . . . , x m ],
is normally torsion-free.

Proof. It is well-known that one can view the square-free monomial ideal I as the cover ideal of a simple hypergraph H such that the hypergraph H corresponds to I ∨ , where I ∨ denotes the Alexander dual of I. Then we have I = J(H), where J(H) denotes the cover ideal of the hypergraph H. Fix k ≥ 1. On account of Theorem 6.4.8, we get the following equality

Ass S (S/L k ) = Ass R (R/J(H) k ) ∪ {(x n , x n+1 , x n+2 , . . . , x m )}.
Because I is normally torsion-free, one derives that Ass R (R/J(H) k ) = Min(J(H)), and hence Ass S (S/L k ) = Min(J(H)) ∪ {(x n , x n+1 , x n+2 , . . . , x m )}. This gives rise to Ass S (S/L k ) = Min(L). Therefore, L is normally torsion-free, as claimed.

Lemma 6.4.10. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple graphs such that

V (H) = V (G) ∪ {w} with w / ∈ V (G), and
E(H) = E(G) ∪ {{v, w}} for some vertex v ∈ V (G) and j∈N G [v] x j / ∈ G(N I(G)). If DI(G)
DI(H) = DI(G)∩(x v , x w )R, where R = K[x α : α ∈ V (H)]
. Now, we can conclude the assertion from Lemma 6.4.9.

We are in a position to give the second main result of this section in the following corollary, which is related to dominating ideals of star graphs. Corollary 6.4.11. The dominating ideals of star graphs are normally torsion-free.

Proof. We use the induction on the number of vertices together with Lemma 6.4.10.

Closed neighborhood ideals and dominating ideals of cycles

As we know, the edge ideals and the cover ideals of bipartite graphs are known to be normally torsion-free, see [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF][START_REF] Simis | On the ideal theory of graphs[END_REF]. In particular, the edge ideals and the cover ideals of even cycles are normally torsion-free. However, this behavior changes when we consider the odd cycles. The cover ideals of odd cycles happen to be nearly normally torsion-free, see [START_REF] Nasernejad | Associated primes of powers of cover ideals under graph operations[END_REF], but edge ideals of odd cycles do not admit such tamed behavior for the set of their associated primes. Given these facts, it is natural to expect some irregularities for the closed neighborhood ideals and dominating ideals of even and odd cycles. It can be verified by using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF] that in general, the closed neighborhood ideals of cycles, regardless of the parity of their lengths, are neither normally torsion-free nor nearly normally torsion-free. However, in this section, we will show that the closed neighborhood ideals of cycles admit the strong persistence property. On the other side, as another main result of this section, we will show that the dominating ideals of cycles are nearly normally torsion-free. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Dominating ideals and closed neighborhood ideals of graphs[END_REF].

To establish above-mentioned results, we begin by proving the following theorem which gives an inductive way to study the normality of a square-free monomial ideal. Theorem 6.5.1. Let I and H be two normal square-free monomial ideals in a polynomial ring R

= K[x 1 , . . . , x n ] such that I + H is normal. Let x c ∈ {x 1 , . . . , x n } be a variable with gcd(v, x c ) = 1 for all v ∈ G(I) ∪ G(H). Then L := I + x c H is normal.
Proof. Let G(I) = {u 1 , . . . , u s } and G(H) = {h 1 , . . . , h r }. Since gcd(v, x c ) = 1 for all v ∈ G(I)∪G(H), without loss of generality, one may assume that

x c = x 1 ∈ K[x 1 ] and
G(I) ∪ G(H) = {u 1 , . . . , u s , h 1 , . . . , h r } ⊆ K[x 2 , . . . , x n ].
We must show that L t = L t for all integers t ≥ 1. For this purpose, it is enough to prove that L t ⊆ L t . Let α be a monomial in L t and write α = x b 1 δ with x 1 ∤ δ and δ ∈ R. On account of [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.2], α k ∈ L tk for some integer k ≥ 1. Write (6.5.1)

α k = x bk 1 δ k = s i=1 u p i i x q+ε 1 r j=1 h q j j β,
with s i=1 p i = p, r j=1 q j = q, p + q = tk, ε ≥ 0, and β is some monomial in R such that x 1 ∤ β. Because x 1 ∤ β, x 1 ∤ δ, and gcd(v, x 1 ) = 1 for all v ∈ G(I) ∪ G(H), one can conclude that bk = q + ε. Accordingly, by virtue of (6.5.1), we obtain

δ k = s i=1 u p i i r j=1 h q j j β ∈ (I + H) tk .
This leads to δ ∈ (I + H) t . Thanks to I + H is normal, we can conclude that (I + H) t = (I + H) t , and so δ ∈ (I + H) t . Therefore, one can write (6.5.2)

δ = s i=1 u l i i r j=1 h z j j γ,
with s i=1 l i = l, r j=1 z j = z, l + z = t, and γ is some monomial in R. Note that x 1 ∤ γ as x 1 ∤ δ. Due to x bk 1 δ k ∈ L tk , it follows immediately from (6.5.2) that s i=1

u l i k i x bk 1 r j=1 h z j k j γ k ∈ L tk = (I + x 1 H) tk .
Consequently, we conclude that bk ≥ zk, that is, b ≥ z. This gives rise to

x b 1 δ = s i=1 u l i i x b 1 r j=1 h z j j γ ∈ (I + x 1 H) t .
We thus get α ∈ L t , and the proof is over.

We here state the third main result in the next theorem, which is related to the closed neighborhood ideals of cycles. Theorem 6.5.2. Let C n be a cycle graph of order n. Then the following statements hold:

(i) N I(C n ) is normal. (ii) N I(C n ) has the strong persistence property. (iii) N I(C n ) has the persistence property. Proof. (i) Let C n = (V (C n ), E(C n )) be a cycle graph of order n with V (C n ) = {x 1 , . . . , x n } and E(C n ) = {{x i , x i+1 } : i = 1, . . . , n -1} ∪ {{x n , x 1 }}. Then the closed neighborhood ideal of C n is given by N I(C n ) = (x i x i+1 x i+2 : i = 1, . . . , n) ⊂ R = K[x 1 , . . . , x n ],
where x n+1 (respectively, x n+2 ) represents x 1 (respectively, x 2 ). If n = 3, then N I(C 3 ) = (x 1 x 2 x 3 ), and so there is nothing to prove. Thus, let n ≥ 4. Put H := (x 2 x 3 , x n-1 x n , x 2 x n ) and I := (x i x i+1 x i+2 : i = 2, . . . , n -2). One can easily see that N I(C n ) = I +x 1 H. Our strategy is to use Theorem 6.5.1 to complete the proof.

To do this, we first show that I, H, and

I + H are normal. Assume that G is a path graph with V (G) = {x 2 , x 3 , x n-1 , x n } and E(G) = {{x 2 , x 3 }, {x n-1 , x n }, {x 2 , x n }}.
It is routine to check that I(G) = H, where I(G) denotes the edge ideal of G. Since, by [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6], the edge ideal of any path graph is normally torsion-free, and by remembering this fact that every normally torsion-free square-free monomial ideal is normal, we deduce that H is a normal square-free monomial ideal. Now, assume that P is a path graph with V (P ) = {x 2 , x 3 , . . . , x n-1 , x n } and E(P ) = {{x i , x i+1 } : i = 2, . . . , n -1}. It is not hard to check that I = I 3 (P ), where I 3 (P ) denotes the path ideal of length 2 of P . It follows readily from Corollary 6.4.2 that I = I 3 (P ) is normally torsion-free, and so is normal. To complete the proof, we show that I + H is normal. To accomplish this, we note that

I + H = (x 2 x 3 , x n-1 x n , x 2 x n , x i x i+1 x i+2 : i = 3, . . . , n -3). Set A := (x 3 , x n ) and B := (x n-1 x n , x i x i+1 x i+2 : i = 3, . . . , n -3). Notice that I + H = B + x 2 A. It is clear that A is a normal ideal.
Furthermore, it follows from Corollary 6.4.2 and Theorem 6.4.1 that B is normally torsion-free, and so is normal.

In addition, we have

B + A = (x 3 , x n , x i x i+1 x i+2 : i = 4, . . . , n -3).
One can easily conclude from Corollary 6.4.2 and Theorem 6.4.1 that B + A is normally torsion-free, and hence is normal. By virtue of Theorem 6.5.1, we deduce that B+x 2 A is normal, and so I +H is normal as well. Finally, note that gcd(v, x 1 ) = 1 for all v ∈ G(I) ∪ G(H). This finishes the proof.

The claims (ii) and (iii) can be shown similar to the proof of Corollary 6.2.3.

The neighborhood ideals of cycles are particularly nice because they are generated by monomial of the same degree. This fact together with Theorem 6.5.2 enables us to study the depth of powers of N I(C n ). For this purpose, we need to recall the definition of linear relation graph, see Definition 6.3.10, and also the following result from [START_REF] Herzog | Persistence and stability properties of powers of ideals[END_REF].

depth(R/I t ) ≤ n -t -1 for t = 1, . . . , r -s.
In order to employ the above theorem, we first analyze the linear relation graph of

N I(C n ). Let V (C n ) = [n] and E(C n ) = {{1, 2}, {2, 3}, . . . , {n -1, n}, {n, 1}}.
We set the following notations.

1. u i = j∈N [i] x j . In simple words, u i is the monomial that corresponds to the closed neighborhood of the vertex i for all i = 1, . . . , n

Note that

u i = x i-1 x i x i+1 , where i = 2, . . . n -1 and u 1 = x n x 1 x 2 , u n = x n-1 x n x 1 .
To synchronize this notation for all i with i > n, we read i as i(mod n). In this way, we can write u i = x i-1 x i x i+1 for all i = 1, . . . n. Remark 6.5.4. Let i = j. Note that each variable x i appears in exactly three monomials in G(N I(C n )), and these monomials are

u i-1 = x i-2 x i-1 x i , u i = x i-1 x i x i+1 and u i+1 = x i x i+1 x i+2 .
From this observation, we conclude that {x i , x j } ∈ E(Γ) if and only if there exists a path of length three from i to j in C n . Note that here a path P of length n is defined on n + 1 vertices and n edges. 1. |V (Γ In )| = n. This can be easily verified because for every i, we can find another vertex j such that there is a path of length three from i to j in C n .

Γ In has one connected component if n = 3k

, for all k ≥ 2. Indeed, if n = 1(mod 3), that is, n = 3k + 1 for some k ≥ 1, then we have

E(Γ In ) = {{x 1 , x 4 }, {x 4 , x 7 }, . . . , {x 3k-2 , x 3k+1 }, {x 3k+1 , x 3 }, {x 3 , x 6 }, . . . , {x 3k-3 , x 3k }, {x 3k , x 2 }, {x 2 , x 5 }, . . . , {x 3k-2 , x 1 }}.
If n = 2(mod 3), that is, n = 3k + 2 for some k ≥ 1, then we have

E(Γ In ) = {{x 1 , x 4 }, {x 4 , x 7 }, . . . , {x 3k-2 , x 3k+1 }, {x 3k+1 , x 2 }, {x 2 , x 5 }, . . . , {x 3k-1 , x 3k+2 }, {x 3k+2 , x 3 }, {x 3 , x 6 }, . . . , {x 3k , x 1 }}.
3. Γ In has three connected components if n = 3k, for some k ≥ 2. In fact, set

V (Γ 1 ) = {x 1 , x 4 , . . . , x 3k-2 }
, and

E(Γ 1 ) = {{x 1 , x 4 }, {x 4 , x 7 }, . . . , {x 3k-2 , x 1 }}. Set V (Γ 2 ) = {x 2 , x 5 , . . . , x 3k-1 }
, and

E(Γ 2 ) = {{x 2 , x 5 }, {x 5 , x 8 }, . . . , {x 3k-1 , x 2 }}.
Set V (Γ 3 ) = {x 3 , x 6 , . . . , x 3k }, and

E(Γ 3 ) = {{x 3 , x 6 }, {x 6 , x 9 }, . . . , {x 3k , x 3 }}.
It can be easily verified that Γ In is the disjoint union of Γ 1 , Γ 2 , and Γ 3 .

Now, Theorem 6.5.3 together with Remark 6.5.5 leads us to the following corollary. ) and lim k→∞ depthR/I k n = 0.

We close this section by providing the fourth main result in the subsequent theorem, which is related to the dominating ideals of cycles. Theorem 6.5.7. The dominating ideals of cycles are nearly normally torsion-free.

Proof. Let C n denote a cycle graph of order n with V (C n ) = {x 1 , . . . , x n } and E(C n ) = {{x i , x i+1 } : i = 1, . . . , n -1} ∪ {{x n , x 1 }}. In the light of [122, Lemma 2.2], the dominating ideal of C n is given by DI(C n ) = n i=1 (x i , x i+1 , x i+2 ) ⊂ R = [x 1 , . . . , x n ],
where x n+1 (respectively, x n+2 ) represents x 1 (respectively, x 2 ). Set I := DI(C n ). Our strategy is to use Corollary 6.1.3. To do this, we must show that I(m \ {x i }) is normally torsion-free for all i = 1, . . . , n, where m = (x 1 , . . . , x n ). Without loss of generality, it is sufficient for us to prove that I(m \ {x 1 }) is normally torsionfree. To simplify our notation, set F := n-2 i=2 (x i , x i+1 , x i+2 ). By virtue of Corollary 189 6.1.3, one has to show that the ideal F = I(m \ {x 1 }) is normally torsion-free. To do this, let T = (V (T ), E(T )) be the rooted tree with the root 2, the vertex set V (T ) = {x 2 , . . . , x n }, and the edge set E(T ) = {(x i , x i+1 ) : i = 2, . . . , n -1}, where (x i , x i+1 ) denotes the directed edge from the vertex x i to the vertex x i+1 for all i = 2, . . . , n -1. It is not hard to check that F is the Alexander dual of the path ideal generated by all paths of length 2 in the rooted tree T . Now, one can deduce from Theorem 5.1.6 that F = I(m \ {x 1 }) is normally torsion-free. This finishes the proof.

Chapter 7

On the normality of monomial ideals

Let I be an ideal in a commutative Noetherian ring R. An element f ∈ R is integral over I if there exists an equation

f k + c 1 f k-1 + • • • + c k-1 f + c k = 0 with c i ∈ I i .
The set of elements I in R which are integral over I is the integral closure of I. The ideal I is integrally closed, if I = I, and I is normal if all powers of I are integrally closed. If I is a monomial ideal, then the notion integrality becomes simpler, namely, an element u ∈ R is integral over I if and only if there exists an integer k such that u k ∈ I k , see Section 1.1. The normality of square-free monomial ideals has been of interest for many authors, for instance see [START_REF] Faridi | Monomial Ideals via Square-free Monomial Ideals[END_REF][START_REF] Restuccia | On the normality of monomial ideals of mixed products[END_REF][START_REF] Simis | On the ideal theory of graphs[END_REF][START_REF] Villarreal | Rees cones and monomial rings of matroids[END_REF]. Such normality has connections to combinatorial commutative algebra as in [35,[START_REF] Martinez-Bernal | Associated primes of powers of edge ideals[END_REF][START_REF] Villarreal | Normality of subrings generated by square-free monomials[END_REF]. One aspect of these connections is that the square-free monomial ideals come out as the edge and cover ideals of graphs. For example, it is shown in [START_REF] Simis | On the ideal theory of graphs[END_REF] that the edge ideals of bipartite graphs are normal. Also, it has been shown in [START_REF] Villarreal | Monomial algebras[END_REF]Corollary 14.6.25] that the cover ideals of perfect graphs are normal. However, little is known for the cover ideals of imperfect graphs.

In this chapter, we first argue on the normality of monomial ideals under some monomial operations. After that, we employ these consequences in cover ideals of some imperfect graphs. Next, we turn our attention to exploring the normality and associated primes of powers of the cover ideals of theta graphs and Jahangir's graphs. Moreover, we handle the normality of closed neighborhood and dominating ideals of complete bipartite graphs. We finally deal with the normality of dominating ideals of h-wheel graphs.

Normality and Cover Ideals

In this section, we are interested in proving the normality of cover ideals of graphs. In Subsection 7.1.1, we develop a technique for proving such normality. In Subsection 7.1.2, we prove that the cover ideals of the cycle graphs of odd orders are normal, and thus, using the results of Subsection 7.1.1, we prove that the cover ideal of the wheel graph is normal as well. Motivated by the method of the proof of Theorem 7.1.3, we tailor this section with Subsection 7.1.3 that contains some results on the integral closedness. It should be noted that the results of this section can be found in [4].

Given a monomial ideal I in the polynomial ring

R = K[x 1 , . . . , x n ],
where K is a field. Let Γ(I) be the set of exponents of all monomials in I. It is well-known that finding the integral closure of the monomial ideal I is the same as finding all the integer lattice points in the convex hull in R n of Γ(I), see [START_REF] Swanson | Integral Closure of Ideals, Rings, and Modules[END_REF]Proposition 1.4.6] or [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.4.3]. According to the discussion before Proposition 1.4.2 of [START_REF] Swanson | Integral Closure of Ideals, Rings, and Modules[END_REF], the problem of finding an equation of integral dependence of a monomial m over a monomial ideal I is reduced to finding an integer k and monomials u

1 , . . . , u k in I such that m k = u 1 • • • u k . Thus, m ∈ I if and only if m k ∈ I k for some k.
The following remark is crucial for us to provide our results and can be directly proved by using [60, Theorem 1.4.2]. Remark 7.1.1. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] and let h ∈ R be a monomial. Then hI = hI, and also that I is normal if and only if hI is normal.

Studying the behavior of the integral closedness of powers of monomial ideals has been of interest for many researchers. Before we proceed to our results of this subsection, we would like to demonstrate some of these behaviors through certain monomial ideals. We do this in the following remark.

Remark 7.1.2. (i) We give an ideal I which is not integrally closed but

I k is integrally closed for every k ≥ 2. Let I = (x 6 , x 4 y 2 , x 3 y 3 , x 2 y 4 , y 6 ) ⊂ K[x, y]. Then I = (x 6-i y i ) 6
i=0 , and thus I is not integrally closed. However, simple computations show that I 2 = (x 12-i y i ) 12 i=0 and I 3 = JI 2 , where J = (x 6 , y 6 ). Hence, J is a reduction of I with reduction number 2. That is, I k+1 = JI k for k ≥ 2. Thus, I k is integrally closed for every k ≥ 2.

(ii) We give an ideal I which is integrally closed but I k is not integrally closed for every k ≥ 2. In an unpublished work of Jockusch and Swanson (see Exercise 1.14 of [START_REF] Swanson | Integral Closure of Ideals, Rings, and Modules[END_REF]) it is pointed out that if I = (x 2 , y 3 , z 7 ) ⊂ K[x, y, z], then I 2 is not integrally closed. Here, we show that α := xy 2 z 6 z 7(k-2) ∈ I k \ I k for any positive integer k ≥ 2; hence, I k is not integrally closed for every k ≥ 2. To show α ∈ I k it suffices to show that (1, 2, 6 + 7(k -2)) is in the convex hull of the set of exponents of the monomials x 2 , y 3 , and z 7 . But this is equivalent to finding nonnegative rational numbers λ 1 , λ 2 , and λ 3 with λ 1 + λ 2 + λ 3 = 1 and with 2kλ 1 ≤ 1, 3kλ 2 ≤ 2, and 7kλ 3 ≤ 6 + 7(k -2). It is routine to see that this is achieved if we choose λ 1 = 1/2k, λ 2 = 1/3k, and λ 3 = 1 -(λ 1 + λ 2 ). Now, we show that α / ∈ I k for any k ≥ 2. Using Normaliz [24] we find that the set of generators of I is given by G

(I) = {x 2 , y 3 , z 7 } ∪ {y 2 z 3 , yz 5 , xy 2 , xyz 2 , xz 4 }. If α ∈ I k for some k ≥ 2, then α = z 7(k-(i 1 +•••+i 5 )) (y 2 z 3 ) i 1 (yz 5 ) i 2 (xy 2 ) i 3 (xyz 2 ) i 4 (xz 4 ) i 5 with i j nonnegative. Now, the x-deg(α) forces that i 3 + i 4 + i 5 = 1. If i 3 = 1, then i 4 = i 5 = 0, also i 1 = i 2 = 0 by virtue of the y-deg(α); hence, z-deg(α) = 7(k -1) > 6 + 7(k -2), a contradiction.
The cases i 4 = 1 and i 5 = 1 are manipulated similarly.

(iii) We give an ideal L such that L 2 is not integrally closed but L 3 is integrally closed. This ideal is inspired to us while working with cover ideals of graphs.

Given the ideals I = (x 1 x 4 , x 1 x 5 , x 2 x 3 , x 2 x 5 , x 3 x 4 ) and

J = (x 1 x 2 x 4 , x 1 x 3 x 5 , x 1 x 2 x 3 , x 2 x 4 x 5 , x 3 x 4 x 5 ).
Let L = x 6 I + x 7 J. Then L is integrally closed since it is square-free (see Theorem 7.1.9). Note that α := x 1 x 2 x 3 x 4 x 5 x 6 x 7 / ∈ L 2 and

α 2 = (x 2 x 5 x 6 )(x 3 x 4 x 6 )(x 1 x 2 x 4 x 7 )(x 1 x 3 x 5 x 7 ) ∈ L 4 ,
hence α ∈ L 2 by virtue of [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.2]; therefore, L 2 is not integrally closed. On the other hand, computations with Normaliz [24] verify that L 3 is integrally closed. In fact, our computations also show that L k is integrally closed for k = 4, 5, 6, 7. The authors wonder whether L k is integrally closed for all k ≥ 3.

Constructing Normal Ideals with Application to Cover

Ideals

In the following theorem, we introduce a procedure for obtaining normal monomial ideals from other ideals that are assumed to be normal. This theorem enables us to prove the normality of the cover ideals of graphs upon the normality of cover ideals of other graphs. This is illustrated in Theorem 7.1.5 which in turn is used to prove that the cover ideals of complete graphs are normal (see Corollary 7.1.6), and also to prove that the cover ideals of the wheel graph is normal (see Theorem 7.1.8). A direct application of Theorem 7.1.3 to graphs is the following theorem which is needed to conclude Corollary 7.1.6 and also to prove Theorem 7.1.8.

Theorem 7.1.5. Let G = (V (G), E(G)) and H = (V (H), E(H)) be two finite simple graphs such that V (H) = V (G) ∪ {w} with w /

∈ V (G), and

E(H) = E(G) ∪ {{v, w} : for all v ∈ V (G)}.
Let J(G) (respectively, J(H)) denote the cover ideal of the graph G (respectively, graph H). If J(G) is normal, then J(H) is normal.

Proof. Suppose that J(G) is normal. Without loss of generality, one may assume that x i and G(I) ⊆ K[x 1 , . . . , x n ], one obtains g ∈ I. We next prove that L = x n+1 IR + gR. Let M ∈ R be a monomial. The condition that M will be in L is that for each edge {i, j} of H, we have at least one of x i |M , x j |M (in which case we will say that M covers {i, j}). First consider the case x n+1 ∤ M . In order to cover all the edges {i, n + 1}, 1 ≤ i ≤ n of H we must have

V (G) = [n] and V (H) = V (G) ∪ {n + 1}. Also, let R = K[x 1 , . . . ,
x i |M, 1 ≤ i ≤ n.
Thus, in this case M ∈ L ⇐⇒ g|M . Now, consider the case x n+1 |M . Then the edges {i, n + 1}, 1 ≤ i ≤ n are all covered, so M ∈ L if and only if M covers all the edges {i, j} ∈ E(G). This will be the case ⇐⇒ M = x n+1 M ′ with M ′ ∈ IR, from which it follows that L = x n+1 IR + gR. In the following, we show that

IR is normal in R = K[x 1 , . . . , x n+1 ].
To see this, let F be any monomial ideal in R ′ = K[x 1 , . . . , x n ] with minimal generating set G(F ) = {α 1 , . . . , α r }. Then by definition F R has minimal generating set G(F ), now regarded as a subset of R by the canonical inclusion R ′ ⊂ R. Similarly F t R = (F R) t because both have generating set all products of t elements of G(F ). We claim that the same sort of thing holds for integral closure, namely that F R = F R (where each integral closure operation takes place in the ring of the ideal). The inclusion F R ⊆ F R is obvious. Conversely, suppose that M is a monomial of R in F R. This means that there is a positive integer

d such that M d ∈ (F R) d . Write M = M ′ x a
n+1 with a a nonnegative integer and M ′ a monomial in R ′ . We now have M ′d x ad n+1 = r i=1 α a i i N , where r i=1 a i = d and N is a monomial in R. Setting x n+1 = 1 in this expression we obtain M ′ d ∈ F d , so that M ′ ∈ F and finally that M ∈ F R, proving the reverse inclusion. Since I is normal, applying the previous observations to I d and noting that I d = I d for all d we have

(IR) d = I d R = I d R = I d R = (IR) d , so (IR) d is integrally closed for all d, hence IR is normal. Now, assume that G(I) = {u 1 , . . . , u m }. This leads to G(L) = {x n+1 u 1 , . . . , x n+1 u m , g}.
In the light of L = x n+1 IR + gR, g ∈ I, gcd(x n+1 , u i ) = 1 for all i = 1, . . . , m, and because IR is normal, Theorem 7.1.3 yields that L is normal, as claimed.

The following corollary, which is an immediate consequence of Theorem 7.1.5, tells us that the cover ideals of complete graphs are normal.

Corollary 7.1.6. Let K n be the complete graph on n vertices with n ≥ 2. Let

J(K n ) denote the cover ideal of the graph K n . Then J(K n ) is a normal square-free monomial ideal. Proof. Let V (K n ) = {v 1 , . . . , v n }. We claim that one can inductively construct the graph K n . Since E(K n ) = {{v i , v j } : for all 1 ≤ i = j ≤ n}, we get {v 1 , v 2 } ∈ E(K n ). Consider a graph H 1 with V (H 1 ) = {v 1 , v 2 } and E(H 1 ) = {{v 1 , v 2 }}. Now, for each 2 ≤ k ≤ n -1, define a graph H k with vertices V (H k ) = {v 1 , . . . , v k+1 } and with edges E(H k ) = E(H k-1 ) ∪ {{v k+1 , w} : for all w ∈ V (H k-1 )}. It is routine to check that H n-1 = K n . Because J(H 1 ) = (x v 1 , x v 2 )
as a monomial prime ideal is normal, it follows from the iteration of Theorem 7.1.5 that, for each i = 2, . . . , n -1, J(H i ) is normal as well. In particular, we deduce that J(K n ) is normal, and the proof is complete.

Normality of Cover Ideals of Some Imperfect Graphs

We now explore the normality of the cover ideals of some imperfect graphs. It is well-known that the cover ideals of perfect graphs are normal, see [START_REF] Villarreal | Monomial algebras[END_REF]Corollary 14.6.25]. For example, cycle graphs of even orders are perfect, and so their cover ideals will be normal. However, little is known for the cover ideals of imperfect graphs. In what follows, we consider some classes of imperfect graphs, i.e., cycle graphs of odd orders and wheel graphs of even orders. Our aim is to show that the cover ideals of these graphs are normal.

Let R = K[x 1 , . . . , x 2n+1 ] be a polynomial ring over a field K. Let I = J(C 2n+1 ) ⊂ R be the cover ideal associated to the cycle graph C 2n+1 with the vertex set

[2n + 1]. That is, I = 2n+1 i=1 (x i , x i+1 ) = (x 1 , x 2 ) ∩ • • • ∩ (x 2n , x 2n+1 ) ∩ (x 2n+1 , x 1 )
, where subscripts are taken mod 2n + 1, for instance x 2n+1+i represents x i . Now, we are in a position to demonstrate that the cover ideals of cycle graphs of odd orders are normal.

= x α 1 1 • • • x α 2n+1 2n+1 ∈ I t with α = (α 1 , . . . , α 2n+1 ) ∈ N 2n+1 0 . Let {µ j } r j=1 ⊂ N 2n+1 0
be the set of exponents of the minimal generators of I t and let µ j,i be the i-th component of µ j . Since m ∈ I t , it follows by [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.4.3] that α belongs to the Newton polyhedron of I. That is, there exist λ 1 , . . . , λ r ∈ Q ≥0 with We now obtain that I d is integrally closed for all d. This gives that I is normal, as required.

r j=1 λ j = 1 such that α i ≥ r j=1 λ j µ j,i for each i = 1, . . . , 2n + 1. If i ∈ [2n + 1], then (with 2n + 2 = 1)
α i + α i+1 ≥ r j=1 λ j µ j,i + r j=1 λ j µ j,i+1 = r j=1 λ j (µ j,i + µ j,i+1 ) ≥ t
(α i + α i+1 ) -t(2n + 1) ≥ 2n+1 i=1 r j=1 λ j (µ j,i + µ j,i+1 ) -t(2n + 1) = r j=1 λ j 2n+1 i=1 (µ j,i + µ j,i+1 ) -t(2n + 1) ≥ t
We are now going to investigate the normality of the cover ideals of wheel graphs of even orders. To remember the definition of wheel graphs, refer to Definition 3.1.16.

Theorem 7.1.8. Suppose that W 2n is a wheel graph of order 2n on the vertex set

[2n]. Then J(W 2n ) is normal.
Proof. Label the vertices of the cycle graph C 2n-1 of order 2n-1 in counterclockwise order with 1, . . . , 2n -1, and the hub with 2n, as shown in the figure below.

1 2 3 4 W 2n 2n -1 2n -2 2n
Now, the desired conclusion follows readily from Theorems 7.1.7 and 7.1.5, as claimed.

Some Related Results in Normality

Motivated by the methods of proving the integral closedness in the previous subsections, we finish this section with some results on the integral closedness of monomial ideals.

The result below is well-known, but we re-prove it by a different method from the proof which has been introduced in Corollary 3.2.45. Theorem 7.1.9. Every square-free monomial ideal is integrally closed. [START_REF] Swanson | Integral Closure of Ideals, Rings, and Modules[END_REF] and the discussion before it, there exist λ 1 , . . . , λ r ∈ Q ≥0 with r j=1 λ j = 1 such that α k ≥ r j=1 λ j a jk for all k. As r j=1 λ j = 1, this yields that λ l > 0 for some l ∈ {1, . . . , r}, and thus α k ≥ λ l a lk for all k. Since a lk equals 0 or 1 and since α k is a nonnegative integer, we derive that α k ≥ a lk for all k. But this gives that α is a multiple of u l , that is, α ∈ I; hence I is integrally closed.

Proof. Let I ⊂ R = K[x 1 , . . . , x n ] be a square-free monomial ideal with G(I) = {u 1 , . . . , u r }. Write u i = x a i1 1 • • • x a in n with a ij ∈ {0, 1}. Let α = x α 1 1 • • • x αn n ∈ I. According to Proposition 1.4.6 of
In the following lemma we give the best possible hypotheses so that the integral closure commutes under addition. Lemma 7.1.10. Let J ⊆ I ⊆ R = K[x 1 , . . . , x n ] be monomial ideals. Let v be a square-free monomial in R such that gcd(v, f ) = 1 and gcd(v, g) = 1 for every f ∈ G(I) and g ∈ G(J). Let w be any monomial in R. Then vI + wJ = vI + wJ.

Proof. By Remark 7.1.1, we may assume that gcd(v, w) = 1. It is known that the integral closure operation on ideals preserves inclusion. Thus, and by Remark 7.1.1,

vI + wJ ⊆ vI + wJ. It remains to prove vI + wJ ⊆ vI + wJ. Write v = x 1 • • • x m ∈ K[x 1 , . . . , x m ] and let G(I) ⊆ K[x m+1 , . . . , x n ] and G(J) ⊆ K[x m+1 , . . . , x n ].
Assume α is a monomial in vI + wJ, thus α k ∈ (vI + wJ) k for some k. Hence, α k ∈ (vI) p (wJ) q for some p and q with p + q = k. Choose q to be the minimal according to this membership. If q = 0, then α k ∈ (vI) k ; thus, α ∈ vI = vI. Accordingly, assume q > 0. Let G(I) = {f 1 , . . . , f r } and G(J) = {g 1 , . . . , g s }. On account of α k ∈ (vI + wJ) k , we get the following equality (7.1.4)

α k = r i=1 f p i i v p s j=1 g q j j w q β, with r i=1 p i = p, s j=1 q j =
q, and β is some monomial in K[x 1 , . . . , x n ]. By our hypothesis, every g j ∈ G(J) is divisible by some f i j ∈ G(I), say g j = f i j g ′ j . Therefore, if v divides β, say β = vβ ′ , then (7.1.4) can be rewritten as (after some reordering)

α k = r-1 i=1 f p i i f pr+1 r v p+1 s-1 j=1 g q j j g qs-1 s w q-1 g ′ s wβ ′ ,
and hence α k ∈ (vI) p+1 (wJ) q-1 , contradicting the minimality of q. Hence, we may assume in (7.1.4)

that v ∤ β, say x 1 ∤ β. Write α = x b 1 1 • • • x bm m δ with each b i nonnegative and δ ∈ K[x m+1 , . . . , x n ]. Now, note that b 1 k = p because v is square-free with gcd(v, w) = 1, gcd(v, f i ) = 1,
and gcd(v, g j ) = 1 for all i and j. As p + q = k, we obtain q = (1 -b 1 )k. But q > 0, hence b 1 = 0; thus, p = 0 and q = k. Therefore, α k ∈ (wJ) k , and thus α ∈ wJ = wJ. Remark 7.1.11. We show that the hypotheses in the above lemma are best possible.

(i) The square-freeness of the monomial v is necessary. For instance, let I = (x 1 , x 2 ) and J = (x 2 1 ) in the polynomial ring K[x 1 , x 2 , x 3 , x 4 ] and let v = x 2 3 and w = x 4 . Then we have vI + wJ = vI + wJ = (x 1 x 2 3 , x 2 x 2 3 , x 2 1 x 4 ). Consider α := x 1 x 2 x 3 x 4 and also notice α 2 ∈ (vI + wJ) 2 , hence α ∈ vI + wJ\(vI + wJ).

(ii) The containment J ⊆ I is necessary. For instance, let

I = (x 2 1 , x 1 x 2 , x 2 2 ) and J = (x 2
3 ) in the polynomial ring K[x 1 , . . . , x 5 ] and let v = x 4 and w = x 5 . Then we have vI + wJ = vI + wJ = (x 2 1 x 4 , x 1 x 2 x 4 , x 2 2 x 4 , x 2 3 x 5 ). Consider α := x 1 x 3 x 4 x 5 and notice α 2 ∈ (vI + wJ) 2 , hence α ∈ vI + wJ\(vI + wJ). Remark 7.1.12. By Theorem 7.1.9, any square-free monomial ideal is integrally closed. Therefore, if I and J are square-free monomial ideals, and if v and w are square-free monomials with gcd(v, f ) = 1 and gcd(w, g) = 1 for every f ∈ G(I) and g ∈ G(J), then vI + wJ is always integrally closed since it is square-free as well, in particular, vI + wJ = vI + wJ, and so the containment J ⊆ I and the condition gcd(v, g) = 1 for every g ∈ G(J) are not necessary for the conclusion of Lemma 7.1.10 as long as the ideal vI +wJ is square-free. Though, to guarantee the normality of the ideal vI + wJ, even though it is square-free and the condition gcd(v, g) = 1 for every g ∈ G(J) holds, one cannot give up the containment condition of J ⊆ I, as the following example shows.

Example 7.1.13. Let I = (x 1 x 2 , x 2 x 3 , x 3 x 4 ) and J = (x 1 x 3 , x 2 x 3 , x 2 x 4 ) in the polynomial ring K[x 1 , . . . , x 6 ] and let v = x 6 and w = x 5 . Then we have 4 , so α ∈ (vI + wJ) 2 . Thus, (vI + wJ) 2 is not integrally closed; hence, vI + wJ is not normal.

vI + wJ = (x 1 x 2 x 6 , x 2 x 3 x 6 , x 3 x 4 x 6 , x 1 x 3 x 5 , x 2 x 3 x 5 , x 2 x 4 x 5 ) . Consider α := x 1 x 2 x 3 x 4 x 5 x 6 / ∈ (vI + wJ) 2 . Direct computations show that α 2 = (x 1 x 2 x 6 )(x 2 x 4 x 5 )(x 1 x 3 x 5 )(x 3 x 4 x 6 ) ∈ (vI + wJ)

Normality under some Operations

In this section, we investigate the normality under expansion, weighting, and polarization. In Subsection 7.2.1, we prove some results concerning the normality under expansion and then apply these results to obtain a generalization of the whole work of [1]. In Subsection 7.2.2, we show that if a weighted ideal is normal, then the ideal itself is normal. It should be noted that the results of this section can be found in [4].

Normality under Expansion

First, one should recall the definition of the expansion of a monomial ideal as has been stated in Definition 2.2.9.

The following theorem states that the normality of a monomial ideal is equivalent to the normality of its expansion. As an application of this theorem, we show in Proposition 7.2.2 that the integral closure commutes with expansion. Let R be an arbitrary Noetherian ring and I, J, K, L be ideals of R. By Remark 1.3.2(4) in [START_REF] Swanson | Integral Closure of Ideals, Rings, and Modules[END_REF], I J ⊆ IJ, hence I J ⊆ IJ + KL. Similarly, K L ⊆ IJ + KL. Since the latter is integrally closed, we have I J + K L ⊆ IJ + KL. The reverse inclusion is obvious, thus we obtain I J + K L = IJ + KL. This last equality helps us to prove a general relation between expansions and the integral closure as in the following lemma.

Lemma 7.2.3. Let

L = (u 1 , . . . , u s ) ⊂ K[x 1 , . . . , x n ] be a monomial ideal. Write u j = x a j 1 u ′ j with a j ≥ 0 and gcd(x 1 , u ′ j ) = 1.
Let L * be the expansion of L with respect to the n-tuple (m, 1, . . . , 1), that is,

L * = s j=1 (x 11 , . . . , x 1m ) a j u ′ j .
Define the ideal N in the ring

L * = s j=1 (x 11 , . . . , x 1m ) a j u ′ j = s j=1 (x a j 11 , . . . , x a j 1m )u ′ j = s j=1 (x a j 11 , . . . , x a j 1m )u ′ j = N ,
as required.

Let I = (x a 1 1 , . . . , x an n ) ⊂ R = K[x 1 , . . . ,
x n ] and let J := I. The normality of the ideal J has been of interest for many authors, see [1,2,33,[START_REF] Reid | Some results on normal monomial ideals[END_REF]. A result of Reid, Roberts, and Vitulli [START_REF] Reid | Some results on normal monomial ideals[END_REF] relates the normality of both of the ideals J and L, where L = (x a 1 1 , . . . , x an+l n

) and l = lcm(a 1 , . . . , a n- 1). Yet, it is still a challenging problem to give necessary and sufficient conditions, on the integers a i , for the normality of the ideal J. Al-Ayyoub [1] proved that the integral closure of the ideal I = (x a 1 1 , . . . , x an n ) in R is normal provided that a i ∈ {s, l}, where s and l are positive integers. Generalizing the whole work of [1], it is proved in [2] that the integral closure of x a 1 1 , . . . , x [2] is long since its accomplished after developing an algorithm and some consequences. As an application of Theorem 7.2.1, we give a direct proof of this result.

a i-1 i-1 , x a i i1 , x a i i2 , x a i+1 i+1 , . . . , x an n is normal in R[x i1 , x i2 ]/(x i ) if and only if the integral closure of (x a 1 1 , . . . , x an n ) is normal in R, where 1 ≤ i ≤ n. The proof of
L 2 := x a 1 1 , . . . , x a i-1 i-1 , x a i i1 , x a i i2 , x a i+1 i+1 , . . . , x an n , in R[x i1 , x i2 ]/(x i ). Then I is normal if and only if J is normal.
Proof. By virtue of Theorem 7.2.1, it suffices to show that J = I * , the expansion of I with respect to the n-tuple (1, . . . , 1, 2, 1, . . . , 1), where 2 is in the i-th place. By Lemma 7.2.3,

L * 1 = L 2 . But J = L 2 and L * 1 = L 1 * = I * by Proposition 7.2.2.
This leads to J = I * , as desired.

Normality under Weighting

u k W ∈ (I t W ) k . Now, [60, Theorem 1.4.2] gives that u W ∈ (I W ) t .
The assumption deduces that (I W ) t = (I W ) t , and so u W ∈ (I W ) t . By virtue of (I W ) t = (I t ) W , we get u W ∈ (I t ) W . Hence, there exists a monomial v ∈ G(I t ) such that v W |u W . Accordingly, one has v|u, and so u ∈ I t . This means that I is a normal ideal, and the proof is complete.

Normality under More Operations

In this section, we investigate the normality under localization, deletion, and contraction. In Subsection 7.3.1, we show that monomial localization of a normally torsion-free square-free monomial ideal I, in a commutative Noetherian ring S, with respect to p is normal, where p ∈ Ass S (S/I), see Theorem 7.3.2. We also show that the converse of this theorem is not true, see Remark 7.3.4. Upon our results on normality of cover ideals of cycle graphs of odd order, and also upon [START_REF] Villarreal | Monomial algebras[END_REF]Corollary 10.5.9], we construct a square-free monomial ideal L ⊂ R = K[x 1 , . . . , x n ] which is not normal, while the contraction L/x j = (L : R x i ) is normal for some 1 ≤ i ≤ n, see Subsection 7.3.2. In Subsection 7.3.3, we show that if I is a normally torsion-free square-free monomial ideal, then every deletion I \ x j is normal, see Theorem 7.3.9. We also show that the converse of this theorem is not true, see Example 7.3.11 in which proving normality depends on Theorem 7.1.3. It should be noted that the results of this section can be found in [3].

The subsequent theorem expresses that under certain conditions, the sum of two square-free monomial ideals is normal, see also [START_REF] Vasconcelos | Arithmetic of Blowup Algebras[END_REF]Corollary 5.5.13]. This theorem is needed (along with Theorem 7.1.3 and Remark 7.1.1) to produce an interesting example about the normality under deletion, see Example 7.3.11.

Theorem 7.3.1. Suppose that I is a square-free monomial ideal in R = K[x 1 , . . . , x n ] such that I = I 1 R + I 2 R, where G(I 1 ) ⊂ R 1 = K[x 1 , . . . , x m ] and G(I 2 ) ⊂ R 2 = 203 K[x m+1 , . . . ,
x n ] for some positive integer 1 ≤ m < n. If I 1 and I 2 are normally torsion-free square-free monomial ideals, then I is a square-free normally torsion-free monomial ideal; hence, normal.

Proof. Theorem 5.3.4 together with [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.6] yield the claim.

Normality under Localization

First, we need to recall the definition of the monomial localization of a monomial ideal with respect to a monomial prime ideal as has been stated in Definition 3.1.14.

In the following theorem, we prove that if I is a normally torsion-free square-free monomial ideal, then every monomial localization of I is normal. However, one should note that if I is a normal (not necessarily square-free) monomial ideal, then

I(p)
G(I) = {u 1 , . . . , u m }, m
i=1 supp(u i ) = {x 1 , . . . , x r }, and J 1 , . . . , J m be some nonzero square-free monomial ideals in

R 2 = K[x r+1 , . . . , x n ]. If L := u 1 J 1 R + • • • + u m J m R
is normally torsion-free, where R = K[x 1 , . . . , x n ], then I is square-free normally torsion-free, hence normal.

Proof. Set p := ( m i=1 supp(u i ))R = (x 1 , . . . , x r )R. It is routine to check that p is a monomial prime ideal of R, and L ⊆ p. It follows now from Theorem 7.3.2 that L(p) is normally torsion-free and normal as well. Since L(p) = I, one can conclude that I is normally torsion-free and normal as well, as required.

The opposite of Theorem 7.3.2 is not true as we show in the following remark. Recall from Theorem 3.2.34 that any normal monomial ideal has the strong persistence property. In particular, it has been proved in [START_REF] Villarreal | Monomial algebras[END_REF]Theorem 7.7.3] that every normal monomial ideal has the persistence property. Remark 7.3.4. Let I be a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] such that I(p) is normal for all p ∈ Ass R (R/I). Then we show that, in general, we cannot deduce that I is normal. To see this, consider the graph G = H 4 , which has been presented in Proposition 3.2.5.

To simplify the notation, one may assume that V (G) = [12]. Let I be the cover ideal associated to the graph G in the polynomial ring R = K[x 1 , . . . , x 12 ], that is, I = {i,j}∈E(G) (x i , x j ), and so Ass R (R/I) = {(x i , x j )|{i, j} ∈ E(G)}. Since I(p) = p for each p ∈ Ass R (R/I), one can readily conclude that I(p) is normal for each p ∈ Ass R (R/I). Now, suppose, on the contrary, that I is normal. It follows now from [START_REF] Villarreal | Monomial algebras[END_REF]Theorem 7.7.3] that I has the persistence property. On the other hand, using the commutative algebra program Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], we get m = (x 1 , . . . , x 12 ) ∈ Ass R (R/I 3 ) \ Ass R (R/I 4 ). Consequently, I does not satisfy the persistence property. This leads to a contradiction. Therefore, I is not normal, as required.

As we have mentioned before the above remark, it has been proved in Theorem 3.2.34 that any normal monomial ideal has the strong persistence property. Therefore, in the light of Theorems 7.1.7 and 7.1.8, the following two corollaries follow immediately.

Normality under Contraction

To state the subsequent result, we require to remember the definition of the contraction operator as has been given in Definition 3. It is well-known that if I is a square-free monomial ideal in a polynomial ring K[x 1 , . . . , x n ], then I/x j = (I : R x j ) for each j = 1, . . . , n. Hence, one should be aware that it is possible a square-free monomial ideal such as

I ⊂ R = K[x 1 , . . . , x n ]
is not normal, while (I : R x i ) is normal for some 1 ≤ i ≤ n. To see this, consider the following example.

Example 7.3.8. Consider two monomial ideals I := (x 1 x 4 , x 1 x 5 , x 2 x 3 , x 2 x 5 , x 3 x 4 ) and J := (x

1 x 2 x 4 , x 1 x 3 x 5 , x 1 x 2 x 3 , x 2 x 4 x 5 , x 3 x 4 x 5 ) in K[x 1 , . . . , x 7 ]. Set L := x 6 I + x 7 J.
One can rapidly conclude that I and J have the following minimal primary decomposition

I = (x 3 , x 2 , x 1 ) ∩ (x 5 , x 4 , x 3 ) ∩ (x 5 , x 3 , x 1 ) ∩ (x 5 , x 4 , x 2 ) ∩ (x 4 , x 2 , x 1
), and

J = (x 1 , x 4 ) ∩ (x 4 , x 3 ) ∩ (x 3 , x 2 ) ∩ (x 2 , x 5 ) ∩ (x 5 ,
α 2 = (x 1 x 2 x 3 x 7 )(x 1 x 5 x 6 )(x 2 x 4 x 5 x 7 )(x 3 x 4 x 6 ) ∈ (L 2 ) 2 ,
and so α ∈ L 2 , while α / ∈ L 2 . This yields that L is not normal.

Normality under Deletion

To express Theorem 7.3.9, we must recollect the definition of the deletion operator as has been stated in Definition 3.2.19.

In the next theorem, we show that if I is a normally torsion-free square-free monomial ideal, then every deletion I \ x j is normal. Remark 7.3.10. We finally emphasize that if I is a normal (not necessarily squarefree) monomial ideal, then both I/x j and I \ x j are normal by [START_REF] Villarreal | Monomial algebras[END_REF]Proposition 12.2.3].

As an application of Theorem 7.3.1, Remark 7.1.1, and Theorem 7.1.3, we show that if I is a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] such that I \ x j is normal for all i = 1, . . . , n, then we cannot, in general, conclude that I is normal. This is illustrated in the following example.

I = (x 3 x 5 x 6 , x 3 x 4 x 6 , x 2 x 5 x 6 , x 2 x 4 x 5 , x 1 x 3 x 5 , x 1 x 2 x 6 ), in the polynomial ring R = K[x 1 , . . . , x 6 ]
. We demonstrate that I \ x i is normal for all i = 1, . . . , 6. One can rapidly gain the following results: We claim that I is not normal. To show this, put α := x 1 x 2 x 3 x 4 x 5 x 6 . Since

(i) I \ x 1 = x 6 (x 3 (x 5 , x 4 ) + (x 2 x 5 )) + (x 2 x 4 x 5 ). (ii) I \ x 2 = x 3 (x 5 (x 1 , x 6 ) + (x 4 x 6 )). (iii) I \ x 3 = x 2 (x 6 (x 5 , x 1 ) + (x 4 x 5 )). (iv) I \ x 4 = x 6 (x 5 (x 3 , x 2 ) + (x 1 x 2 )) + (x 1 x 3 x 5 ). (v) I \ x 5 = x 6 (x 1 x 2 , x 3 x 4 ). (vi) I \ x 6 = x 5 (x 1 x 3 , x 2 x 4 ).
α 2 = (x 1 x 3 x 5 )(x 1 x 2 x 6 )(x 2 x 4 x 5 )(x 3 x 4 x 6 ) ∈ (I 2 ) 2 ,
we get α ∈ I 2 . On the other hand, it is easy to see that α / ∈ I 2 , and so α ∈ I 2 \ I 2 . Accordingly, one derives that I is not normal.

More results on the normality of square-free monomial ideals

In this section, we express several outcomes on the normality of square-free monomial ideals. It should be noted that the results of this section can be found in [3].

We begin with the following theorem, which is essential for us to show Corollary 7.4.2.

Theorem 7.4.1. Let I be a normal monomial ideal in R and h

∈ R a monomial. Assume v ∈ R is a square-free monomial with gcd(u, v) = 1 for all u ∈ G(I) ∪ {h}. Then L := I + vhR is normal if and only if J := I + hR is normal. Proof. Because gcd(u, v) = 1 for all u ∈ G(I) ∪ {h}, one may assume that v = x 1 • • • x m ∈ K[x 1 , . . . , x m ] and G(I + hR) := {u 1 , . . . , u s , h} ⊆ K[x m+1 , . . . ,
x n ] for some positive integer 1 ≤ m < n. The ideal J is obtained from L by making every variable x i (for i = 1, . . . , m) is equal to 1. Proposition 12.2.3 of Villarreal [START_REF] Villarreal | Monomial algebras[END_REF] asserts that a normal monomial ideal stays normal if we make any variable equals 1. This proves the necessary part. Conversely, we prove that L t = L t for all integers t ≥ 1, provided that J is normal. It suffices to show that L t ⊆ L t . Let α be a monomial in L t and write α = v b δ with v ∤ δ and δ ∈ R. By [60, Theorem 1.4.2],

α k ∈ L tk for some integer k ≥ 1. Write (7.4.1) v bk δ k = s i=1 u p i i v q+ε h q β,
with s i=1 p i + q = tk, ε ≥ 0, and β is some monomial in R such that v ∤ β. Let x l ∈ supp(v) \ supp(β) for some l ∈ {1, . . . , m}. Now, if x l ∈ supp(δ), then, since v is square-free, the x l -degree in both sides of (7.4.1) gives that bk + jk = q + ε, where j = x l -deg(δ). Hence, cancelling v bk from both sides of (7.4.1) gives that

δ k = s i=1 u p i i v q+ε-bk h q β.
Since v is square-free, this yields that v must divide δ, a contradiction. Therefore, we proceed with the assumption that there exists l ∈ {1, . . . , m} with x l ∈ supp(v) \ (supp(β) ∪ supp(δ)). This assumption, along with the hypothesis gcd(u, v) = 1 for all u ∈ G(I) ∪ {h}, forces that bk = q + ε. Therefore, by (7.4.1) we obtain δ k = s i=1 u p i i h q β ∈ J tk . This implies that δ ∈ J t . Since J is normal, we obtain J t = J t , and thus, δ ∈ J t ; hence, write

(7.4.2) δ = s i=1 u l i i h z γ, with s i=1 l i + z = t, z ≥ 0, and γ is some monomial in R. Note v ∤ γ as v ∤ δ. Now, since v bk δ k ∈ L tk , by virtue of (7.4.2), we get s i=1 u l i k i v bk h zk γ k ∈ L tk = (I + vhR) tk . We therefore have bk ≥ zk, that is, b ≥ z. Thus, v b δ = s i=1 u l i i v b h z γ ∈ (I + vhR) t
, and the proof is done.

Setting h = 1 in Theorem 7.4.1, we obtain the following corollary.

Corollary 7.4.2. Let I be a normal monomial ideal in R and v ∈ R a square-free monomial with gcd(u, v) = 1 for all u ∈ G(I). Then I + vR is normal.

Inductively and in view of [START_REF] Villarreal | Monomial algebras[END_REF]Proposition 12.2.3], the following proposition follows directly from Corollary 7.4.2.

Proposition 7.4.3. Any square-free monomial ideal of R with a set of pairwise relatively prime generators is normal. In particular, assume G(I) ⊂ K[x 1 , . . . , x m ] and J is square-free with G(J) ⊂ K[x m+1 , . . . , x n ], where 1 ≤ m < n, and the generators of J are pairwise relatively prime. Then I + J is normal if and only if I is normal.

Remark 7.4.4. It should be noted that Proposition 7.4.3 may be false if we drop the condition that the generators of J are pairwise relatively prime. To see a counterexample, let I := (x 1 x 2 x 4 , x 1 x 3 x 4 , x 1 x 3 x 5 , x 2 x 3 x 5 , x 2 x 4 x 5 ) ⊂ K[x 1 , . . . , x 5 ] and J := (y 1 y 2 y 4 , y 1 y 3 y 4 , y 1 y 3 y 5 , y 2 y 3 y 5 , y 2 y 4 y 5 ) ⊂ K[y 1 , . . . , y 5 ] be the cover ideals of the odd cycle graphs G and H, respectively, where Because α 2 = (x 1 x 2 x 4 )(x 1 x 3 x 5 )(x 2 x 3 x 5 )x 4 (y 1 y 2 y 4 )(y 1 y 3 y 5 )(y 2 y 3 y 5 )y 4 ∈ Q 6 , one can deduce that α ∈ Q 3 . On the other hand, it is easy to check that α / ∈ Q 3 . This means that Q is non-normal.

V (G) = {x 1 , x 2 , x 3 , x 4 , x 5 } with E(G) = {{x i , x i+1 }} 5 i=1 ,
Assume a graph H is obtained from a graph G by connecting all the vertices of G with a new vertex. In Theorem 7.1.5 it is proved that the cover ideal of the graph H is normal provided that the cover ideal of the graph G is normal. The proof relies on Theorem 7.1.3 in which it is proved that vI + hR is normal provided that I is a normal monomial ideal, h ∈ I, and v is a square-free monomial which is relatively prime to every generator of I. In Theorem 7.4.6, we give a generalization of Theorem 7.1.3. The following lemma is needed in the proofs of Theorems 7. Proof. Assume that G(I) := {u 1 , . . . , u r } and G(J) := {v 1 , . . . , v s }. As before, it suffices to show that L t ⊆ L t for all t ≥ 1. Let α be a monomial in L t . By [60, Theorem 1.4.2], we have α k ∈ (JI + hR) tk for some positive integer k. The binomial expansion implies that α k ∈ (JI) p (hR) q for some nonnegative integers p and q with p + q = tk. Choose q to be the minimal according to this membership. If q = 0, then p = tk; hence, α k ∈ (JI) tk . Accordingly, α ∈ (JI) t . In the light of I and J being normal and gcd(u, v) = 1 for all u ∈ G(I) and v ∈ G(J), Lemma 7.4.5 yields that JI is normal, and so α ∈ (JI) t ; thus, we have α ∈ L t and the proof is over. Therefore, let q ≥ 1. Since α k ∈ (JI) p (hR) q , we get the following equality (7.4.3)

α k = v z 1 1 • • • v zs s u p 1 1 • • • u pr r h q β,
with s i=1 z i = p, r i=1 p i = p, and β some monomial in R. Since h ∈ I, this implies that h = h ′ u λ for some 1 ≤ λ ≤ r and monomial h ′ in R. If β ∈ J, then β = v θ β ′ for some 1 ≤ θ ≤ s and monomial β ′ in R. Hence, (7.4.3) can be rewritten as follows

α k = v z 1 1 • • • v zs s v θ u p 1 1 • • • u pr r u λ h q-1 h ′ β ′ . 209
This leads to α k ∈ (JI) p+1 (hR) q-1 , which contradicts the minimality of q. Accordingly, one can assume in (7.4.3) that v i ∤ β for each i = 1, . . . , s. Let x j i |v i but

x j i ∤ β for each i = 1, . . . , s. Note that x j 1 , . . . , x js are distinct since gcd(v i , v j ) = 1 for any 1 ≤ i = j ≤ s. Write α = x b 1 j 1 • • •
x bs js δ with x j d ∤ δ for each d = 1, . . . , s. This gives rise to the following equality (7.4.4)

x b 1 k j 1 • • • x bsk js δ k = v z 1 1 • • • v zs s u p 1 1 • • • u pr r h q β.
In view of (7.4.4), we have

b i k = z i for each i = 1, . . . , s. Set b := b 1 + • • • + b s .
Thanks to s i=1 z i = p, one has bk = p, and so q = (t-b)k. We therefore obtain

α k ∈ (JI) bk (hR) (t-b)k . Consequently, α ∈ (JI) b (hR) t-b . It follows from Lemma 7.4.5 that (JI) b (hR) t-b = (JI) b (hR) t-b . This yields that α ∈ (JI) b (hR) t-b ⊆ (JI +hR) t = L t , as required.
The subsequent theorem is one of the main results in this section which is used in proving Theorem 7.4.12.

Theorem 7.4.7. Let I be a normal square-free monomial ideal in

R = K[x 1 , . . . , x n ] with G(I) ⊂ R. Then the ideal L := IS ∩ (x n , x n+1 ) ⊂ S = R[x n+1 ] is normal. Proof. Since I ∩ (x n ) = x n (I : S x n ) and I ∩ (x n+1 ) = x n+1 I, one can conclude that L = x n (I : S x n ) + x n+1 I.
To simplify our notation, set F := (I : S x n ) and L = x n F + x n+1 I. Since I is square-free, then the ideal F is obtained from I by making the variable x n is equal to 1. Therefore, F is normal by virtue of Proposition 12.2.3 in [START_REF] Villarreal | Monomial algebras[END_REF], and hence x n F is also normal. Let G(I) = {g 1 , . . . , g d , g d+1 , . . . , g s } with x n | g j for j = 1, . . . , d and x n ∤ g j for j = d + 1, . . . , s. Then a generating set (not necessarily minimal) of F is given by {f 1 , . . . , f d , f d+1 , . . . , f s } with f i x n = g i for i = 1, . . . , d and f i = g i for i = d + 1, . . . , s. Note that x n F + I = I; therefore,

x n F + I is normal, that is, (x n F + I) t = (x n F + I) t for all t ≥ 1. Our goal is to show that L t = L t for all t ≥ 1.
Let α be a monomial in L t and write α = x b n+1 δ for some integer b and some monomial δ ∈ R with x n+1 ∤ δ. As α ∈ L t , [60, Theorem 1.4.2] implies that α k ∈ L tk = (x n F + x n+1 I) tk for some integer k; therefore, α k ∈ (x n F ) p (x n+1 I) q for some integers p and q with p+q = tk. Assume q is maximal according to this membership. Note that if p = 0, then α k ∈ (x n+1 I) tk , and hence α ∈ (x n+1 I) t = (x n+1 I) t ⊂ L t . Henceforth, assume p > 0. Similarly, and since x n F is normal, we may also assume q > 0. Write (7.4.5)

α k = x bk n+1 δ k = s i=1 f p i i x p n s j=1 g q j j x q n+1 β,
with s i=1 p i = p, s j=1 q j = q, and β some monomial in S. If x n+1 | β, then we get a contradiction to the maximality of q since either f i x n = g i or f i = g i . Therefore, we may assume in (7.4.5) that x n+1 ∤ β, and thus we can conclude that q = bk and also that (7.4.6)

δ k = s i=1 f p i i x p n s j=1 g q j j β ∈ (x n F + I) tk .
Therefore, δ ∈ (x n F + I) t , and hence δ ∈ (x n F + I) t ; thus, let δ ∈ (x n F ) l I h with l + h = t and l being maximal with respect to this membership. Note that if h = 0,

then δ ∈ (x n F ) t ⊂ L t . Henceforth, assume h > 0. Write δ = s i=1 f l i i x l n s j=1 g h j j γ,
with s i=1 l i = l, s j=1 h j = h, and γ is some monomial in S. Note that if d = 0, then L = I ∩ (x n , x n+1 ) is normal by Lemma 7.4.5. In addition, note that if d = s, then L = I, which is normal. Henceforth, assume that s > d > 0. The maximality of l implies that h j = 0 for j = 1, . . . , d and also that x n ∤ γ since g j = f j for

j = d + 1, . . . , s. Hence, x n -deg(δ) = l. Since p = (t -b)k, then (7.4.6) gives that x n -deg(δ) ≥ t -b, that is, l ≥ t -b; thus, b ≥ h. Therefore, since δ ∈ (x n F ) l I h , one can deduce that x b n+1 δ ∈ (x n F ) l (x n+1 I) h ⊆ (x n F + x n+1 I) l+h = L t ,
which finishes the proof.

Remark 7.4.8. As the reader may notice, in the proof of Theorem 7.4.7 we proved that the linear combination x n F + x n+1 I is normal, where I ⊆ F . Initial endeavors of the authors were to prove a more general result, that is, investigating the normality of square-free monomial ideals resulting from linear combinations vF + wI, where F and I are normal ideals, v and w are square-free monomials with gcd(v, f ) = 1 for all f ∈ G(F ) and gcd(w, g) = 1 for all g ∈ G(I). Since vF +wI is square-free, one gains it is integrally closed. However, we found that one cannot guarantee the normality of vF +wI, even though one has I ⊆ F or G(I) ⊂ G(F ). In this remark we demonstrate an example supporting this assertion. Let F = (x 1 x 2 x 4 , x 1 x 3 x 5 , x 2 x 3 , x 2 x 5 , x 3 x 4 ) and

I = (x 1 x 2 x 4 , x 1 x 3 x 5 ) in R = K[x 1 , . . . , x 7 ]
, and let v = x 6 and w = x 7 . Then we get

T := vF + wI = (x 1 x 2 x 4 x 6 , x 1 x 3 x 5 x 6 , x 2 x 3 x 6 , x 2 x 5 x 6 , x 3 x 4 x 6 , x 1 x 2 x 4 x 7 , x 1 x 3 x 5 x 7 ).
The normality of I can be deduced from Proposition 7.4.3. In addition, using Normaliz [24] yields that F is a normal monomial ideal. Now, put α := x 1 x 2 x 3 x 4 x 5 x 6 x 7 . Direct computations show that α / ∈ T 2 . Since

α 2 = (x 2 x 5 x 6 ) (x 3 x 4 x 6 ) (x 1 x 2 x 4 x 7 ) (x 1 x 3 x 5 x 7 ) ∈ T 4 ,
we conclude that α ∈ T 2 \T 2 , that is, vF + wI is not normal.

Corollary 7.4.9. Let I be a normal square-free monomial ideal in

R = K[x 1 , . . . , x n ] with G(I) ⊂ R. Then the ideal L := IS ∩ (x n , x n+1 • • • x m ) ⊂ S = R[x n+1 , . . . , x m ] is normal. Proof. Since I ∩ (x n , x n+1 • • • x m ) = I ∩ (x n , x n+1 ) ∩ (x n , x n+2 ) ∩ • • • ∩ (x n , x m ),
this claim follows at once from Theorem 7.4.7.

The above corollary motivates for the following two questions.

Question 7.4.10. Let I be a normal square-free monomial ideal in R = K[x 1 , . . . , x n ] with G(I) ⊂ R, and {i 1 , . . . , i r } ⊆ {1, . . . , n} with r > 1. Then, in general, can we deduce that the ideal

L := IS ∩ (x i 1 . . . x ir , x n+1 ) ⊂ S = R[x n+1 ] is normal? Question 7.4.11. Let I be a normal square-free monomial ideal in R = K[x 1 , . . . , x n ] G(I) ⊂ R. Then, in general, can one conclude L := IS ∩ (x n , x ℓ n+1 ) ⊂ S = R[x n+1 ] with ℓ > 1, is normal?
We show that Question 7.4.10 has a negative answer, while we will show that Question 7.4.11 has an affirmative answer, see Proposition 7.7.6. For this purpose, we provide a counterexample. Firstly, one should recall that, given a graph G, if v is a vertex of G, we may obtain a graph on n -1 vertices by deleting from G the vertex v together with all the edges incident with v. The resulting graph is denoted by G \ v. In the following, one should consider a graph which has been stated in [START_REF] Kaiser | Replication in critical graphs and the persistence of monomial ideals[END_REF]. In fact, for a positive integer n, let [n] denote the set {0, . . . , n -1}. Assume that P n is a path with vertex set [n], with vertices in the increasing order along P n . Let also K 3 be the complete graph whose vertex set is the group Z 3 . For n ≥ 4, we define H n as the graph obtained from the Cartesian product P n 2K 3 by adding the three edges joining (0, j) to (n -1, -j) for j ∈ Z 3 . Figure below is the graph of H 4 in [START_REF] Kaiser | Replication in critical graphs and the persistence of monomial ideals[END_REF].

v 0,2 v 0,0 v 0,1 v 1,2 v 1,0 v 1,1 v 2,2 v 2,0 v 2,1 v 3,2 v 3,0 v 3,1
= K[x α : α ∈ V (H 4 )] over a field K. Now, put G := H 4 \ v 0,1
, where the graph G is obtained from H 4 by deleting from H 4 the vertex v 0,1 together with all the edges incident with v 0,1 . Let J(G) denote the cover ideal of the graph G. It should be noted that

J(H 4 ) =J(G) ∩ (x v 0,1 , x v 1,1 ) ∩ (x v 0,1 , x v 0,0 ) ∩ ((x v 0,1 , x v 0,2 ) ∩ (x v 0,1 , x v 3,2 ) =J(G) ∩ (x v 0,1 , x v 1,1 x v 0,0 x v 0,2 x v 3,2 ).
It follows from Normaliz [24] that J(G) is a normal monomial ideal, while based on Remark 7.3.4, J(H 4 ) is non-normal.

An application to the cover ideals of imperfect graphs

The point of this subsection is to hire some of the consequences of the previous section to cover ideals of imperfect graphs. Indeed, we first prove Theorem 7.4.12, which says that if we take any graph G with normal cover ideal, and add a leaf to G, then the cover ideal of the new graph is also normal. Note that, in this section,

[n] = {1, . . . , n}. Theorem 7.4.12.

Let G = (V (G), E(G)) and H = (V (H), E(H)) be finite simple graphs such that V (H) = V (G) ∪ {w} with w / ∈ V (G)
, and E(H) = E(G) ∪ {{v, w}} for some vertex v ∈ V (G). Let J(G) and J(H) be the cover ideals of the graphs G and H, respectively. If J(G) is normal, then J(H) is normal.

Proof. Suppose that J(G) is normal. Without loss of generality, one may assume that

V (G) = [n], V (H) = V (G) ∪ {n + 1}, and E(H) = E(G) ∪ {{n, n + 1}}. Since J(H) = J(G) ∩ (x n , x n+1
), the claim is a straightforward consequence of Theorem 7.4.7, that is, J(H) is normal.

Here, we want to explore the normality of the cover ideals of helm graphs H n for n ≥ 5 odd. For this purpose, one requires to recall the definition of helm graphs, see Definition 3.1. 19.

As an application of Theorem 7.4.12, we illustrate that every cover ideal of helm graphs H n for n ≥ 5 odd, is normal. Since H 2n+1 with n ≥ 2, contains an induced odd cycle of length n ≥ 5, this graph is imperfect, and so the normality of its cover ideal is of special interest. Proof. Label the vertices of C 2n+1 in counterclockwise order with 1, . . . , 2n + 1, and the hub with 2n + 2, as shown in the figure below, such that we have We can now combine together Theorem 7.1.8 and the iteration of Theorem 7.4.12 to obtain the normality of J(H 2n+1 ). The last assertion follows readily from the normality of J(H 2n+1 ), Theorem 3.2.34, and Proposition 3.1.12.

E(H 2n+1 ) = E(W 2n+2 ) ∪ {{i, 2n + 2 + i} : i = 1, . . . , 2n + 1}.

On the normality of linear combinations of normal ideals

Let R = Q[x 1 , . . . , x n+2 ], let I and J be two square-free monomial ideals in the ring Q[x 1 , . . . , x n ], and let L = x n+1 IR + x n+2 JR. As we have mentioned in Remark 7.4.8, this construction is a fruitful source of interesting square-free monomial ideals

L which are not normal.
In what follows, we investigate the case where I is the edge ideal

I n = (x 1 x 2 , x 2 x 3 , . . . , x n-1 x n , x n x 1 ),
of an n-cycle C n and J is the cover ideal J n = (x n , x 1 ) ∩ n-1 i=1 (x i , x i+1 ) of I n . We will write L n instead of L, so that L n = x n+1 I n R+x n+2 J n R. By abuse of notation, we will write more simply L n = x n+1 I n + x n+2 J n . Also, all statements about L n , I n , J n will take place in the ring R = Q[x 1 , . . . , x n+2 ], which for simplicity of notation, we might not explicitly mention. In a similar vein, L i n \L i n literally means the complement of L i n in L i n , i.e., those elements of L i n not in L i n . But computationally we like to think of it as all the monomials in L i n that are not in L i n . We need n ≥ 3 for cycles to exist at all (as simple undirected graphs). Macaulay2 calculation shows that L 3 is normal. In the following, we will always have n ≥ 4.

Our first conclusion is that L n is not normal for n ≥ 4. More specifically, L 2 n is not integrally closed for any n ≥ 4 (Theorem 7.4.14). If i ≥ 3, then L i n is not integrally closed for even n ≥ 4, and for odd n > 5 (Theorem 7.4.16). If n = 5, then L i 5 is integrally closed for i ≥ 3 (Theorem 7.4.17). We used the Hilbert basis of the Rees cone to prove the last result. Thus, we characterize all cases when L i n is integrally closed, i.e., when L i n /L i n = 0. 214

We finally show that L 2 n /L 2 n is not a finite dimensional vector space for all odd n, n ≥ 5 (Theorem 7.4.18).

The case of L 2 n

We start by proving our first claim in the following theorem. Proof.

Let f = x 1 x 2 • • • x n x n+1 x n+2 . Then we claim that f / ∈ L 2 n .
For if it were, then f would be the product of one minimal generator of x n+1 I n and one minimal generator of x n+2 J n . By cyclic symmetry of C n , we can assume that the element of

x n+1 I n is x 1 x 2 x n+1 . What is left is x 3 x 4 • • • x n x n+2 which is not in x n+2 J n because x 3 x 4 • • • x n
does not contain one of the variables {x 1 , x 2 } (according to the definition of the cover ideal). But now consider

f 2 = x 2 1 x 2 2 • • • x 2 n x 2 n+1 x 2 n+2 .
If n is even, this can be written as

f 2 = (x 1 x 3 x 5 • • • x n-1 x n+2 )(x 2 x 4 x 6 • • • x n x n+2 )(x 1 x 2 x n+1 )(x 3 x 4 x n+1 )(x 5 • • • x n ) ∈ L 4 n , and so f ∈ L 2 n \L 2 n .
If n is odd, then we have

f 2 = (x 1 x 2 x n+1 )(x 1 x 3 • • • x n x n+2 )(x 2 x 4 • • • x n-1 x n x n+2 )(x 3 x 4 x n+1 )(x 5 • • • x n-1 ) ∈ L 4 n ,
and so again f ∈ L 2 n \L 2 n . This finishes our proof.

In view of Theorem 7.4.14 and remembering the fact that a monomial ideal is normal if and only if all its powers are integrally closed, we can deduce the following corollary.

Corollary 7.4.15. If n ≥ 4, then L n is not normal.

The case of L i n , i ≥ 3 We are in a position to establish our second claim in the following theorem. Proof. We produce an explicit element in L i n \L i n for i ≥ 3 in these ranges of n. For n even, n ≥ 4, consider

h i = x 1 x i-1 2 x 3 x i-1 4 • • • x n-1 x i-1 n x n+1 x i-1 n+2 . Because of the x n+1 x i-1 n+2 portion, h i is potentially in (x n+1 I n )(x i-1 n+2 J i-1 n ) ⊂ L i n .
By cyclic symmetry, we can assume that the generator in x n+1 I n is x 1 x 2 x n+1 . Dividing by this, we get

x i-2 2 x 3 x i-1 4 • • • x n-1 x i-1 n x i-1 n+2
. We now need i -1 minimal generators of x n+2 J n each of which has to be divisible by a variable from {x 1 , x 2 }, which must be x 2 since x 1 is now absent. But there are only i -2 x 2 factors left, so this is not possible. Thus, h i / ∈ L i n . But we can write

h 2 i = x 2 1 x 2i-2 2 x 2 3 x 2i-2 4 • • • x 2 n-1 x 2i-2 n x 2 n+1 x 2i-2 n+2 = (x 2 x 4 • • • x n x n+2 ) 2i-4 f 2 ∈ L 2i n ,
since we saw in the proof of Theorem 7.4.14 that f 2 ∈ L 4 n . Thus, h i ∈ L i n \L i n , as desired.

For n odd, n > 5, let

g i = x 2 x 3 x 4 • • • x n-2 x i-1 n-1 x i-1 n x i-1 n+1 x n+2 . Because of the x i-1 n+1 x n+2 portion, if g i ∈ L i n , then g i ∈ (x n+1 I n ) i-1 (x n+2 J n ).
If p is any minimal generator of x n+2 J n , then p must be divisible by at least one generator from each of the pairs

{x 1 , x 2 }, {x 2 , x 3 }, . . . , {x n-1 , x n }, {x n , x 1 }. Because x 1 is absent from g i ,
p must be divisible by x 2 and x n . Furthermore, g i /p cannot be divisible by any of the products x 1 x 2 , x 2 x 3 , . . . , x n-3 x n-2 . The only part of g i /p left to give an element of (

x n+1 I n ) i-1 is possibly x n-2 x i-1 n-1 x i-2 n x i-1 n+1 .
Here, the exponents of x n and x n+1 are certain but those of x n-2 and x n-1 are possibly one lower. Thus, g i /p is always divisible by (x n-1 x n ) i-2 . Dividing by this, we are left at most with x n-2 x n-1 . But p has to be divisible by at least one of these variables, so g i /p cannot be in (x n+1 I n ) i-1 , and g i cannot be in (x n+1 I n ) i-1 (x n+2 J n ), and hence is not in L i n . Note that g i is of degree n -2 + 3(i -1) and that x n+2 J n has minimal generators of degree 2 + (n -1)/2 (e.g., x 1 x 3 x 5 x 7 x 9 for C 7 ). Thus, potentially g 2 i is the product of 2 minimal generators of x n+2 J n and 6(i -1) minimal generators of x i+1 I n , with 2n -4 + 6(i -1) -(4 + (n -1) + 6(i -1)) = n -7 extra variables. Indeed

g 2 i =(x 2 x 3 x 5 x 7 • • • x n x n+2 )(x 2 x 4 x 6 • • • x n-1 x n x n+2 )(x 3 x 4 x n+1 )(x 5 x 6 x n+1 ) (x n-1 x n x n+1 ) 2i-4 (x 7 • • • x n-1 ).
Notice that the last bit

x 7 • • • x n-1 being the "extra variables". Thus, g 2 i ∈ L 2i n , and so g i ∈ L i n . The final conclusion is that g i ∈ L n i \L i n .
It should be noted that the proof of Theorem 7.4.16 fails for n = 5 because 5 -7 is negative. The part of the proof showing that g i / ∈ L i 5 is valid, but the expression for g 2 i ∈ L 2 n does not work for n = 5. In order to tackle n = 5, we found the Rees cone helpful. This is reviewed in the next subsection.

The Rees Cone, illustrated by application to L 5

The Hilbert basis of the Rees cone can be computed with Normaliz [24] (as in Example 2.16 of version 3.8.4). A definition of the Rees cone and its properties can be found in [START_REF] Villarreal | Monomial algebras[END_REF]Chapters 13 and 14]. Let I be a monomial ideal in a polynomial ring Q[x 1 , . . . , x n+2 ] whose minimal generators have exponent vectors {v 1 , . . . , v m }. Let {e i , 1 ≤ i ≤ n + 3} be the unit vectors in Q n+3 (dimension 1 more than the number of variables). Then the Rees cone C(I) is the rational cone in Q n+3 spanned by the vectors (v i , 1), 1 ≤ i ≤ m and e i , 1 ≤ i ≤ n + 2 (leave out e n+3 ). This has the property that (v, d) ∈ C(I) if and only if x v ∈ I d . The next example is motivated by Remark 7.1.2(iii).

Using Normaliz directly we find that the Hilbert basis of the Rees cone of I 5 is the rows of the matrix 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 0 1 0 0 1 0 0 0 0 0 0 0 1 1 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 1 0 0 1 0 1 0 1 1 0 1 0 1 1 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 1 1 0 1 0 1 0 1 1 1 0 1 1 0 0 1 1

1 1 0 0 0 1 0 1 1 1 0 1 0 0 1 1 1 1 1 1 1 1 1 2
The rows of this matrix are the e i , 1 ≤ i ≤ 7, (v i , 1) where v i is the exponent vector of a minimal generator of x 6 I 5 (e.g., (0, 0, 0, 1, 1, 1, 0, 1) corresponds to x 4 x 5 x 6 ), there are five of these, and (0, 1, 0, 1, 1, 0, 1, 1) corresponds to x 2 x 4 x 5 x 7 , a minimal generator of x 7 J 5 (there are also five of these). The final row corresponds to x 1 x 2 x 3 x 4 x 5 x 6 x 7 , which according to the theory of the Rees cone (and Theorem 7.4.14 above) is in L 2 5 \L 2 5 .

Theorem 7.4.17.

If i ≥ 3, then L i 5 is integrally closed.
Proof. What we must show is that if (v, i), i ≥ 3 is in the Rees cone of I 5 , then (v, i) is the sum of i rows of the above matrix that end in 1 (possibly with some rows ending in 0). If there is no row (

, then we are done. At most one (

is required in this summation, by the proof of Theorem 7.4.14. Thus, it suffices to show that (1, 1, 1, 1, 1, 1, 1, 2) plus any row ending in 1 can be rewritten without the (

. By cyclic symmetry in the variables {x 1 , . . . , x 5 }, it suffices to consider one minimal generator of x 6 I 5 and one of x 7 J 5 . Thus, we have the following equalities

+ (0, 0, 0,

= (0, 1, 0, 1, 1, 0, 1, 1) + (1, 0, 0, 0,

+ (0, 0,

+ (0, 0, 0, 1, 1, 1, 0, 1).

We have already seen that f = x 1 x 2 x 3 x 4 x 5 x 6 x 7 is not in L 2 5 . In fact, we have the following stronger result. The proof works for any odd n ≥ 5, so we state it in that generality.

Theorem 7.4.18. Let f = x 1 x 2 • • • x n x n+1 x n+2 , where n is an odd integer ≥ 5. For any i ≥ 1, x i n+2 f / ∈ L 2 n , so that L 2 n /L 2
n is an infinite dimensional vector space. Proof. We have

x i n+2 f = x 1 x 2 • • • x n+1 x i+1 n+2 . If this is in L 2 n , then from the x n+1 x i+1 n+2
portion, it must be in either (x n+2 J n ) 2 or x n+1 I n x n+2 J n . It cannot be in the former because all minimal generators of J n are of degree ≥ (n + 1)/2, and x i n+2 f is only of degree n in {x 1 , . . . , x n }. It cannot be in x n+1 I n x n+2 J n by the same argument that f / ∈ L 2 5 (based on the proof of Theorem 7.4.14). In the proof of Theorem 7.4.14, it was seen that f ∈ L 2 n and of course x i n+2 f is still in L 2 n . Finally, note that all ideals involved here are monomial ideals, and those monomials in

L 2 n , but not in L 2 n form a Q-basis of L 2 n /L 2 n .
We have found an infinite number of these, and so the conclusion follows.

We are ready to verify our last outcome in the subsequent theorem. Proof. We have

x 1 f = x 2 1 x 2 x 3 x 4 x 5 x 6 x 7 = (x 1 x 2 x 4 x 7 )(x 1 x 5 x 6 )x 3 ∈ L 2 5
. By cyclic symmetry in {x 1 , . . . , x 5 }, we also obtain x 2 f, x 3 f, x 4 f, x 5 f ∈ L 2 5 . We also deduce that

x 6 f = x 1 x 2 x 3 x 4 x 5 x 2 6 x 7 = (x 1 x 5 x 6 )(x 2 x 3 x 6 )x 4 x 7 ∈ L 2 5 .
As in the proof of Theorem 7.4.17, any monomial in L 2 5 can be written with at most one factor f . It follows now that (x 1 , . . . , x 6 ) ∈ L 2 5 : L 2 5 . The only monomial ideals between (x 1 , . . . , x 6 ) and (1) are of the form (x 1 , . . . , x 6 , x i 7 ), i ≥ 1. By virtue of Theorem 7.4.18, we get x i 7 / ∈ L 2 5 : L 2 5 . This finishes the proof.

It should be noted that, by using similar techniques, we have been able to prove that L i 4 : L i 4 = (x 1 , . . . , x 6 ) for all i ≥ 2 and L i 4 \ L i 4 is a finite dimensional vector space for all i ≥ 1.

Powers of cover ideals of Jahangir's graphs

The Jahangir's graph J n,m , for m ≥ 3, is a graph on nm + 1 vertices, i.e., a graph consisting of a cycle C nm with one additional vertex which is adjacent to m vertices of C nm at distance n to each other on C nm . More explicitly, it consists of m consecutive cycles of equal length n+2 such that all these cycles have one vertex in common and every pair of consecutive cycles has exactly one edge in common. Also, a vertex v in the vertex set of C nm is called radial if {v, w} is an edge of J n,m , where w denotes the additional vertex of J n,m . Moreover, each edge {v, w} is called a spoke. In fact, J n,m has exactly m spokes.

As an example, consider the Figure below. It presents the Jahangir's graphs J 2,8 , J 5,4 , and J 3,5 . The Jahangir's graph J 2,8 appears on Jahangir's tomb in his mausoleum, which lies 5 kilometers north-west of Lahore, Pakistan across the River Ravi ( [7]).

(a) J 2,8 (b) J 5,4 (c) J 3,5 In general, Jahangir's graphs have been explored by many people. In [START_REF] Imbesi | Algorithmic releases on the spanning trees of Jahangir graphs[END_REF], algebraic and combinatorial properties and a computation of the number of the spanning trees are developed for a Jahangir's graph. Moreover, some algebraic and combinatorial characterizations of the spanning simplicial complex s (J n,m ) of the Jahangir's graph J n,m are examined by Raza et al. in [START_REF] Raza | On algebraic characterization of SSC of the Jahangir's graph[END_REF]. In addition, it follows from [START_REF] Villarreal | Monomial algebras[END_REF]Theorem 7.7.14] that the edge ideals of Jahangir's graphs satisfy the strong persis-tence property. It should be noted that, based on [133, Corollary 10.5.9], the edge ideals of Jahangir's graphs are normal.

In this section, we concentrate on the cover ideals of Jahangir's graphs. Our first main aim is to study the normality of the cover ideals of Jahangir's graphs. In fact, we show that the cover ideals of Jahangir's graphs are normal, see Theorem 7.5.2. The second main goal of us is to study the associated primes of the powers of cover ideals of Jahangir's graphs. Indeed, we demonstrate that the cover ideals of Jahangir's graphs satisfy the (symbolic) (strong) persistence property, see Corollary 7.5.3. In the third main result of this section, we determine when the unique homogeneous maximal ideal appears in the set of associated primes of powers of the cover ideals of Jahangir's graphs, refer to Theorem 7.5.7. It should be noted that the results of this section can be found in [5].

We continue with the following auxiliary result, which we require it to establish Theorem 7.5.2. In addition, throughout the following text, if G is a finite simple graph, then J(G) stands for the cover ideal of G. We are now in a position to state the first main result of this section in the following theorem. Theorem 7.5.2. The cover ideals of the Jahangir's graphs are normal.

Proof. Assume that J n,m is the Jahangir's graph with the vertex set V (J n,m ) = {0, 1, 2, . . . , nm}, 0 is the additional vertex, and the following edge set

E(J n,m ) ={{i, i + 1} : i = 1, . . . , nm -1} ∪ {{nm, 1}} ∪ {{0, 1 + rn} : r = 0, 1, . . . , m -1}. Let R = K[x 1 , . . . , x nm ], u := m-1 r=0 x 1+rn , and S = R[x 0 ]. Let J(J n,m
) (respectively, J(C nm )) denote the cover ideal of J n,m (respectively, C nm ). To simplify our notation, set L := J(J n,m ), I := J(C nm ), and F := (I : S u). Clearly, L = I ∩ (u, x 0 ) ⊂ S. Because I ∩ (u) = u (I : S u) and I ∩ (x 0 ) = x 0 I, we derive that L = u (I : S u) + x 0 I; thus, L = uF + x 0 I. Moreover, based on Proposition 7.5.1, the ideal I is normal. On account of I is square-free, then the ideal F is obtained from I by putting x 1+rn = 1 for all r = 0, 1, . . . , m -1. Hence, F is normal by [133, Proposition 12.2.3], and so uF is normal according to Remark 7.1.1. Due to F = (I : S u), one gains uF + I = I; therefore, uF + I is normal. Since gcd(v, x 0 ) = 1 for all v ∈ G(I) ∪ G(uF ), it follows from Theorem 6.5.1 that L is normal. This completes the proof.

As an application of Theorem 7.5.2, we can present the second main result in the next corollary.

Corollary 7.5.3. Let J n,m be the Jahangir's graph, and J(J n,m ) its cover ideal. Then the following statements hold.

(i) J(J n,m ) has the strong persistence property.

(ii) J(J n,m ) has the persistence property.

(iii) J(J n,m ) has the symbolic strong persistence property.

Proof. (i) -(iii) can be shown similar to the proof of Corollary 6.2.3.

Here, we give the following proposition, which helps us to demonstrate Lemma 7.5.5. Proposition 7.5.4. Let J n,m be the Jahangir's graph, n odd, and W a vertex cover set of J n,m . If W does not contain the additional vertex, then

|W | ≥ m(n + 1)/2. If W contains the additional vertex, then |W | ≥ (nm + 3)/2 if m is odd, or |W | ≥ (nm + 2)/2 if m is even.
Proof. Let w denote the additional vertex of J n,m . First, assume that w / ∈ W . Let v 1 , . . . , v m be the radial vertices of J n,m . It follows from w / ∈ W that v i ∈ W for i = 1, . . . , m. Since W ∩V (C nm ) is a vertex cover of C nm , and v i , v i+1 ∈ W ∩V (C nm ), we deduce that in the path from v i to v i+1 , one needs at least (n -1)/2 vertices for i = 1, . . . , m, where v m+1 represents v 1 . We therefore gain |W | ≥ m + m(n -1)/2, and so |W | ≥ m(n + 1)/2. Now, let w ∈ W . Then W ∩ V (C nm ) must be a vertex cover of C nm . If m is odd, then C nm is an odd cycle, and so W ∩ V (C nm ) has at least (nm+1)/2 elements; thus, we get |W | ≥ 1+(nm+1)/2, and hence |W | ≥ (nm+3)/2. If m is even, then C nm is an even cycle. Accordingly, W ∩ V (C nm ) has at least nm/2 elements, and so |W | ≥ (nm + 2)/2. This completes the proof.

In order to prove Theorem 7.5.7, we require the subsequent lemma. Lemma 7.5.5. Let J n,m be the Jahangir's graph with the vertex set V (J n,m ) = {0, 1, 2, . . . , nm}, 0 be the additional vertex, and the following edge set

E(J n,m ) ={{i, i + 1} : i = 1, . . . , nm -1} ∪ {{nm, 1}} ∪ {{0, 1 + kn} : k = 0, 1, . . . , m -1}.
Let C 2e 1 +1 , . . . , C 2ez+1 be all induced odd cycle subgraphs of J n,m . Then, for each positive integer s, if

x ℓ 0 0 x ℓ 1 1 • • • x ℓnm nm ∈ J(J n,m
) s , then the following conditions hold: 221

(i) ℓ α + ℓ β ≥ s for each {α, β} ∈ E(J n,m ); (ii) {α,β}∈E(C 2e j +1 ) ((ℓ α + ℓ β ) -s) ≥ s for each j = 1, . . . , z; (iii) If n and m are odd, then ((m -1)/2)ℓ 0 + nm i=1 ℓ i ≥ sm(n + 1)/2.
Proof. Fix s ≥ 1. To verify the claim, consider a monomial

x ℓ 0 0 x ℓ 1 1 • • • x ℓnm nm ∈ J(J n,m ) s . Take an arbitrary edge {α, β} in E(J n,m ). Since J(J n,m ) ⊆ (x α , x β ), this implies that x ℓ 0 0 x ℓ 1 1 • • • x ℓnm nm ∈ (x α , x β ) s
, and so ℓ α + ℓ β ≥ s. This proves condition (i). By virtue of J(J n,m ) s ⊆ J(C 2e j +1 ) s for all j, we have i∈V (C 2e j +1 ) x ℓ i i ∈ J(C 2e j +1 ) s for all j, where J(C 2e j +1 ) is the cover ideal of C 2e j +1 . Therefore, condition (ii) follows directly from Lemma 3.1.10. It follows from

x ℓ 0 0 x ℓ 1 1 • • • x ℓnm nm ∈ J(J n,m ) s that there exist monomials g 1 , . . . , g s ∈ G(J(J n,m )) and some monomial h in K[x 0 , x 1 , . . . , x nm ] such that x ℓ 0 0 x ℓ 1 1 • • • x ℓnm nm = g 1 • • • g s h.
Note that each g i corresponds to a minimal vertex cover set of J n,m . After reordering, there exists some integer λ ≤ ℓ 0 such that x 0 | g i for i = 1, . . . , λ, and x 0 ∤ g i for i = λ + 1, . . . , s. If n and m are odd, then one can derive from Proposition 7.5.4 that deg(g i ) ≥ (nm + 3)/2 for i = 1, . . . , λ, and also deg(g i ) ≥ m(n + 1)/2 for i = λ + 1, . . . , s. Note that m ≥ 3 and λ ≤ ℓ 0 . Hence, we get the following

2 nm i=0 ℓ i ≥ λ(nm + 3) + (s -λ)m(n + 1) = smn + sm -λ(m -3) ≥ smn + sm -ℓ 0 (m -3).
Consequently, we deduce that ((m -1)/2)ℓ 0 + nm i=1 ℓ i ≥ sm(n + 1)/2. Therefore, one concludes that condition (iii) holds. This finishes the proof. Remark 7.5.6. Assume that J n,m is the Jahangir's graph with the vertex set V (J n,m ) = {0, 1, 2, . . . , nm} with labeling counterclockwise, 0 is the additional vertex, m = (x 0 , x 1 , . . . , x nm ) the unique homogeneous maximal ideal, and J(J n,m ) its cover ideal. If n is an even number, then J n,m has no any odd cycle, and so J n,m is a bipartite graph. This implies that J(J n,m ) is normally torsion-free, and so the unique homogeneous maximal ideal does not appear in the set of associated primes of powers of the cover ideal of the Jahangir's graph, that is, m / ∈ Ass(J(J n,m ) s ) for all s ≥ 1. Now, assume that n is an odd number and m is an even number. Our computations by Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF] show that the unique homogeneous maximal ideal m does not appear in the set of associated primes of powers of the cover ideal of the Jahangir's graph, that is, m / ∈ Ass(J(J n,m ) s ) for all s ≥ 1. In particular, it should be noted that in this case, J(J n,m )/{x 0 } is exactly the cover ideal of the rim, and since m is even, this yields that the rim is an even cycle graph, and hence J(J n,m )/{x 0 } is normally torsion-free. Nevertheless, we need to discuss the case in which n is an odd number and m is an even number separately. In what follows, we explore the case in which m, n are odd numbers.

We are ready to give the third main result of this section in the next theorem. Indeed, we determine when the unique homogeneous maximal ideal appears in the set of associated primes of powers of the cover ideals of Jahangir's graphs. Theorem 7.5.7. Let J n,m be the Jahangir's graph, J(J n,m ) its cover ideal, and m, n odd numbers. Then, for all s ≥ 3,

(x i : i ∈ V (J n,m )) ∈ Ass(J(J n,m ) s ) \ (Ass(J(J n,m )) ∪ Ass(J(J n,m ) 2 )).
Proof. Without loss of generality, assume that J n,m is the Jahangir's graph with the vertex set V (J n,m ) = {0, 1, 2, . . . , nm} with labeling counterclockwise, 0 is the additional vertex, m = (x 0 , x 1 , . . . , x nm ), and the following edge set

E(J n,m ) ={{i, i + 1} : i = 1, . . . , nm -1} ∪ {{nm, 1}} ∪ {{0, 1 + kn} : k = 0, 1, . . . , m -1}.
Set L := J(J n,m ). Since L is a square-free monomial ideal, we have Ass(L) = Min(L), and so m / ∈ Ass(L). In addition, it follows from [START_REF] Francisco | Associated primes of monomial ideals and odd holes in graphs[END_REF]Corollary 3.4] that m / ∈ Ass(L 2 ). According to Corollary 7.5.3, the ideal L satisfies the persistence property. Hence, it is sufficent for us to show that m ∈ Ass(J(J n,m ) 3 ). To accomplish this, set

h := x 2 0 m-1 k=0 n-1 j=1 x 2 1+kn x λ 1+kn+j , where λ =    1 if j is odd, 2 if j is even.
In what follows, our aim is to show that h / ∈ L 3 and x i h ∈ L 3 for each i = 0, 1, . . . , nm. Suppose, on the contrary, that h ∈ L 3 . One can promptly deduce from Lemma 7.5.5(iii) that (7.5.1)

((m -1)/2)(2) + nm i=1 deg x i h ≥ 3m(n + 1)/2.
Note that there exist exactly n -1 numbers between any two consecutive radial vertices, and by virtue of n is odd, this implies that there exist (n-1)/2 odd numbers and also (n -1)/2 even numbers between any two consecutive radial vertices. We thus get

((m -1)/2)(2) + nm i=1 deg x i h =m -1 + 2m + 2m(n -1)/2 + m(n -1)/2 =(3m(n + 1) -2)/2,
which contradicts (7.5.1). Accordingly, we obtain h / ∈ L 3 . To complete the proof, one has to verify that x i h ∈ L 3 for each i = 0, 1, . . . , nm. To achieve this, fix 1 ≤ t ≤ nm. We relabel the vertices of the odd cycle C nm starting from t = y 1 counterclockwise y 2 , y 3 , . . . , y nm . Now, set

u 1 := x 0 i is odd y i , u 2 := x 0 y 1 i is even y i , u 3 := m-1 k=0 j=1,...,n-1,
where j is even

x 1+kn x 1+kn+j .
It is not hard to check that x t h = u 1 u 2 u 3 , and also u 1 , u 2 , and u 3 correspond to vertex cover sets of J n,m . This yields that u 1 , u 2 , u 3 ∈ L, and so x t h ∈ L 3 . It remains to show that x 0 h ∈ L 3 . To do this, put

v 1 := x 0 i is odd x i , v 2 := x 0 i=n+1,...,nm,
where i is even

x i i=1,...,n+1, where i is odd

x i , and

v 3 := x 0 i=1,...,n, where i is even x i m-1 k=1 j=1,...,n-1,
where j is even

x 1+kn x 1+kn+j .
It is routine to see that x 0 h = v 1 v 2 v 3 , and also v 1 , v 2 , and v 3 correspond to vertex cover sets of J n,m . We thus gain v 1 , v 2 , v 3 ∈ L, and hence x 0 h ∈ L 3 . This terminates the argument.

As an immediate consequence of Theorem 7.5.7, we give the following result:

Corollary 7.5.8. Let G = (V (G), E(G)) be a finite simple graph with the vertex set V (G) = {x 1 , . . . , x n }, and let G have a subgraph J n,m as the Jahangir's graph, where m, n be odd numbers. Let R = K[x 1 , . . . , x n ] and J denote the cover ideal of G. Then, for all s ≥ 3,

(x i : i ∈ V (J n,m )) ∈ Ass(R/J s ) \ (Ass(R/J) ∪ Ass(R/J 2 )).
Proof. This claim can be deduced from [45, Lemma 2.11] and Theorem 7.5.7.

Powers of cover ideals of theta graphs

Studying the strong persistence property and normality of the cover ideals of theta graphs are the major points of this section. It should be noted that the results of this section can be found in [6].

In order to formulate the outcomes of this section, one has to recall the definition of simple graphs of the form θ n 1 ,...,n k which were introduced in [START_REF] Mohammadi | Sequentially Cohen-Macaulay graphs of form θ n 1[END_REF]. Definition 7.6.1. Let k > 1 be an integer, and n 1 , . . . , n k be a sequence of positive integers. Then θ n 1 ,...,n k is the graph constructed by k paths of length n 1 , . . . , n k such that only their endpoints being in common. By length of a path, we mean the number of edges in the path. We call θ n 1 ,...,n k the theta graph of order n 1 , . . . , n k .

As an example, consider the figure below. It presents the theta graph θ 3,1,4,2 of order 3, 1, 4, 2, the left-hand side graph in Figure 1, and the theta graph θ 3,4,2,3 of order 3, 4, 2, 3, the right-hand side graph in Figure 1.

θ 3,1,4,2 θ 3,4,2,3 Figure 1
Remark 7.6.2. If k = 2, then θ n 1 ,...,n k is a cycle of length n 1 +n 2 ; thus, theta graphs are a generalization of cycle graphs. Furthermore, it should be noted that since we assume θ n 1 ,...,n k is a simple graph, there exists at most one i with 1 ≤ i ≤ k such that n i = 1.

Strong persistence property of cover ideals of theta graphs

In this subsection, our aim is to explore the strong persistence property of the cover ideals of theta graphs. To prove Theorem 7.6.4, we have to employ the following auxiliary theorem.

Theorem 7.6.3. With the notation of Definition 7.6.1, let θ n 1 ,...,n k be a theta graph of order n 1 , . . . , n k with V (θ n 1 ,...,n k ) = {1, . . . , m} and cover ideal

J(θ n 1 ,...,n k ) ⊂ R = K[x 1 , . . . , x m ] over a field K. Let C 2e 1 +1 , . . . , C 2ez+1 be all induced odd cycle subgraphs of θ n 1 ,...,n k . Then, for each positive integer s, x ℓ 1 1 • • • x ℓm m ∈ (J(θ n 1 ,...,n k ))
s if and only if the following conditions hold:

(i) ℓ α + ℓ β ≥ s for each {α, β} ∈ E(θ n 1 ,...,n k ); (ii) {α,β}∈E(C 2e j +1 ) [(ℓ α + ℓ β ) -s] ≥ s for each j = 1, . . . , z.
Proof. To simplify the notation, set G := θ n 1 ,...,n k . Fix s ≥ 1. The forward implication can be deduced from Lemma 3.1.10. Conversely, assume that a monomial

x ℓ 1 1 • • • x ℓm m satisfies the conditions (i) and (ii). We show that x ℓ 1 1 • • • x ℓm m ∈ J(G) s . Let z = 0.
This means that there is no induced odd cycle subgraph, and hence G is bipartite. By virtue of [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6], J(G) is normally torsion-free; thus, [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.6] implies that J(G) s = J(G) (s) , where J(G) (s) denotes the sth symbolic power of J(G). Because J(G) (s) = p∈Min(J(G)) p s , this yields that J(G) s = p∈Min(J(G)) p s . Thanks to the monomial x ℓ 1 1 • • • x ℓm m satisfies the condition (i), one obtains ℓ α + ℓ β ≥ s for each {α, β} ∈ E(G), and so

x ℓ 1 1 • • • x ℓm m ∈ (x α , x β ) s for each {α, β} ∈ E(G). In the light of Min(J(G)) = {(x α , x β ) : for all {α, β} ∈ E(G)}, this gives rise to x ℓ 1 1 • • • x ℓm m ∈ p∈Min(J(G)) p s , and hence x ℓ 1 1 • • • x ℓm m ∈ J(G) s .
We therefore assume that z ≥ 1. Let C be an induced odd cycle subgraph of G. Without loss of generality, one may assume that V (C) ∩ {1, . . . , m} = {1, . . . , t}. Since x ℓ 1 1 • • • x ℓm m satisfies the conditions (i) and (ii), Lemma 3.1.10 implies that

x ℓ 1 1 • • • x ℓt t ∈ J(C) s . Consequently, there exist monomials g 1 , . . . , g s ∈ G(J(C)) and some monomial h in R such that x ℓ 1 1 • • • x ℓt t = g 1 • • • g s h.
Note that any minimal generator of J(G) (respectively, J(C)) corresponds to a minimal vertex cover set of G (respectively, C). Hence, to finish the proof, one has to verify that

g 1 • • • g s hx ℓ t+1 t+1 • • • x ℓm m can be written as f 1 • • • f s h ′ such
that each f i corresponds to a vertex cover set of G. To accomplish this, let e = {α, β} be an edge in E(G) \ E(C). By the condition (i), we have ℓ α + ℓ β ≥ s. This enables us to cover the edge e = {α, β} at least s times. Accordingly, one can derive that x ℓ 1 1 • • • x ℓm m ∈ J(G) s , and the proof is done.

We are now in a position to express and prove Theorem 7.6.4. Theorem 7.6.4. With the notation of Definition 7.6.1, let θ n 1 ,...,n k be a theta graph of order n 1 , . . . , n k with V (θ n 1 ,...,n k ) = {1, . . . , m} and the cover ideal

J(θ n 1 ,...,n k ) in R = K[x 1 , . . . , x m ] over a field K. Let C 2e 1 +1 , . . . , C 2ez+1 be all induced odd cycle subgraphs of θ n 1 ,...,n k . Then, for all s ≥ 2, J(θ n 1 ,...,n k ) s = {α,β}∈E(θn 1 ,...,n k ) (x α , x β ) s ∩ z j=1 (x i : i ∈ V (C 2e j +1 )) s(e j +1)
, is a minimal primary decomposition of J(θ n 1 ,...,n k ) s . Proof. Fix s ≥ 2, and put G := θ n 1 ,...,n k . Because J(G) = {α,β}∈E(G) (x α , x β ), we need only prove the following equality (7.6.1)

( {α,β}∈E(G) (x α , x β )) s = {α,β}∈E(G) (x α , x β ) s ∩ z j=1 (x i : i ∈ V (C 2e j +1 )) s(e j +1) .
Assume A (respectively, B) denotes the ideal on the left-hand (respectively, righthand) side of (7.6.1). We first establish A ⊆ B. Since, for any {α, β} ∈ E(G), {α,β}∈E(G) (x α , x β ) ⊆ (x α , x β ), this yields the following containment

( {α,β}∈E(G) (x α , x β )) s ⊆ {α,β}∈E(G) (x α , x β ) s .
Fix 1 ≤ j ≤ z. In order to complete the proof, we have to verify that

( {α,β}∈E(G) (x α , x β )) s ⊆ (x i : i ∈ V (C 2e j +1 )) s(e j +1) . Thanks to {α,β}∈E(G) (x α , x β ) ⊆ {α,β}∈E(C 2e j +1 ) (x α , x β ), it is enough for us to show that ( {α,β}∈E(C 2e j +1 ) (x α , x β )) s ⊆ (x i : i ∈ V (C 2e j +1 )) s(e j +1
) . To do this, pick a minimal generator u in ( {α,β}∈E(C 2e j +1 ) (x α , x β )) s . This implies that u = s i=1 g i , where each g i is a minimal generator of {α,β}∈E(C 2e j +1 ) (x α , x β ). Due to {α,β}∈E(C 2e j +1 ) (x α , x β ) is exactly the cover ideal of the odd cycle C 2e j +1 and since any minimal generator of J(C 2e j +1 ) corresponds to a minimal vertex cover of C 2e j +1 , and also by observing the fact that any minimal vertex cover of C 2e j +1 has at least e j + 1 elements, one derives that degu = s i=1 degg i ≥ s(e j + 1). This gives that u ∈ (x i : i ∈ V (C 2e j +1 )) s(e j +1) . Accordingly, we get A ⊆ B.

We now demonstrate that B ⊆ A. Take a minimal generator u in B. Let

u := x ℓ 1 1 • • • x ℓm m with ℓ i ≥ 0 for each i = 1, . . . , m.
Our strategy for completing the proof is to use Theorem 7.6.3. It follows at once from u ∈ B that u ∈ (x α , x β ) s for each (α, β) ∈ E(G). We thus have x λ α x s-λ β |u for some 0 ≤ λ ≤ s. This leads to ℓ α + ℓ β ≥ s. To finish the argument, we must establish that, for each j = 1, . . . , z,

{α,β}∈E(C 2e j +1 ) [(ℓ α + ℓ β ) -s] ≥ s.
To achieve this, fix 1 ≤ j ≤ z. In the light of u ∈ B, we can deduce that u ∈ (x i : i ∈ V (C 2e j +1 )) s(e j +1) . Without loss of generality, one may assume that V (C 2e j +1 ) = {1, . . . , 2e j + 1} and E(C 2e j +1 ) = {{i, i + 1} : i = 1, . . . , 2e j + 1} with 2e j + 2 = 1. Consequently, there exist nonnegative integers θ 1 , . . . , θ 2e j +1 with

2e j +1 i=1 θ i = s(e j + 1) such that x θ 1 1 • • • x θ 2e j +1 2e j +1 | u. This gives rise to {α,β}∈E(C 2e j +1 ) [(ℓ α + ℓ β ) -s] = 2e j +1 i=1 [(ℓ i + ℓ i+1 )] -s(2e j + 1) =2 2e j +1 i=1 ℓ i -s(2e j + 1) ≥2 2e j +1 i=1 θ i -s(2e j + 1)
=s.

One can deduce from Theorem 7.6.3 that u ∈ A; thus, B ⊆ A, as required.

The corollaries below are some immediate consequences of Theorem 7.6.4.

Corollary 7.6.5. With the notation of Definition 7.6.1, let θ n 1 ,...,n k be a theta graph of order n 1 , . . . , n k . Then, for all positive integers s ≥ 2, we have

Ass(J(θ n 1 ,...,n k ) s ) = Min(J(θ n 1 ,...,n k ))
∪ {(x j 1 , . . . , x jc ) : induced graph on {j 1 , . . . , j c } is an odd cycle}.

Corollary 7.6.6. With the notation of Definition 7.6.1, let θ n 1 ,...,n k be a theta graph of order n 1 , . . . , n k , k > 2, and let m be the unique homogeneous maximal ideal of

R = K[x α : α ∈ V (θ n 1 ,...,n k )]. Then m / ∈ Ass R (R/J(θ n 1 ,...,n k ) s ) for all s ≥ 1.
Corollary 7.6.7. Let C be a cycle graph with a path between two vertices of the cycle C. Then, for all positive integers s ≥ 2,

Ass(J(C) s ) = Min(J(C))
∪ {(x i 1 , . . . , x ir ) : induced graph on {i 1 , . . . , i r } is an odd cycle}.

To illustrate Corollary 7.6.5, we give the following example.

Example 7.6.8. We now number the vertices of the graphs in Figure 1, as are illustrated by Figure 2 below. It is routine to check that there are only two induced odd cycle subgraphs of θ 3,1,4,2 . In fact, the first odd cycle is 1, 5, 6, 7, 4, 1 and the second odd cycle is 1, 8, 4, 1. Note that the induced subgraph of θ 3,1,4,2 on {1, 2, 3, 4, 5, 6, 7} (respectively, {1, 2, 3, 4, 8}) is not an odd cycle since it has the chord {1, 4}, while any cycle graph has no chord. In the light of Corollary 7.6.5, for all s ≥ 2, one can derive that Ass(J(θ 3,1,4,2 ) s ) = {(x 1 , x 2 ), (x 2 , x 3 ), (x 3 , x 4 ), (x 1 , x 4 ), (x 1 , x 5 ), (x 5 , x 6 ), (x 6 , x 7 ), (x 7 , x 4 ), (x 1 , x 8 ), (x 4 , x 8 ), (x 1 , x 8 , x 4 ), (x 1 , x 5 , x 6 , x 7 , x 4 )}.

In addition, one can easily see that there are exactly four induced odd cycle subgraphs C 1 , C 2 , C 3 , and C 4 such that V (C 1 ) = {1, 2, 3, 4, 5, 6, 7}, V (C 2 ) = {1, 2, 3, 4, 8}, V (C 3 ) = {1, 5, 6, 7, 4, 9, 10}, and V (C 4 ) = {1, 8, 4, 9, 10}. In view of Corollary 7.6.5, for all s ≥ 2, we deduce that Ass(J(θ 3,4,2,3 ) s ) = {(x 1 , x 2 ), (x 2 , x 3 ), (x 3 , x 4 ), (x 1 , x 5 ), (x 5 , x 6 ), (x 6 , x 7 ), (x 7 , x 4 ), (x 1 , x 8 ), (x 8 , x 4 ), (x 1 , x 9 ), (x 9 , x 10 ), (x 10 , x 4 ), (x 1 , x 2 , x 3 , x 4 , x 8 ), (x 1 , x 5 , x 6 , x 7 , x 4 , x 9 , x 10 ), (x 1 , x 8 , x 4 , x 9 , x 10 ), (x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 )}.

We are ready to state and prove Theorem 7.6.9 as the main theorem of this section.

Theorem 7.6.9. With the notation of Definition 7.6.1, let θ n 1 ,...,n k be a theta graph of order n 1 , . . . , n k with the cover ideal J(θ n 1 ,...,n k ). Then J(θ n 1 ,...,n k ) has the strong persistence property.

Proof. For convenience of notation, put L := J(θ n 1 ,...,n k ). In addition, let a and b be the endpoint vertices of the path graphs. Want to show that (L m+1 : R L) = L m for all positive integers m, where

R = K[x α : α ∈ V (θ n 1 ,...,n k )]. Fix m ≥ 1.
By [89, Exercise 6.4], it is enough to prove that (L m+1 p : Rp L p ) = L m p for every p ∈ Ass R (R/L m ). In view of Corollary 7.6.5, we may consider the following cases:

L p ) = L m p . Case 2. p ∈ Ass R (R/L m )\Min(L m ).
It follows readily from Corollary 7.6.5 that p = (x i 1 , . . . , x ir ) such that the induced subgraph on {i 1 , . . . , i r } is an odd cycle, say C r . Now, set x e := (x λ , x ρ ) for each edge e = {λ, ρ} ∈ E(θ n 1 ,...,n k ). We thus have the following equalities

L p = ( e∈E(θn 1 ,...,n k ) x e ) p = e∈E(θn 1 ,...,n k ) (x e ) p .
Here, one can consider the following subcases:

Subcase 2.1. {a, b} ∈ E(C r ). Pick an edge e = {α, β} ∈ E(θ n 1 ,...,n k ) \ E(C r ).
Want to demonstrate that (x e ) p =R p . To see this, it is sufficient for us to verify that p / ∈ Supp(R/x e ). Suppose, on the contrary, that p ∈ Supp(R/x e ). Because Supp(R/x e ) = V (Ann(R/x e )), this yields p ∈ V (Ann(R/x e )), and so p ∈ V (x e ).

Hence, x e ⊆ p. By virtue of e / ∈ E(C r ) and {a, b} ∈ E(C r ), this gives rise to a contradiction. We therefore get p / ∈ Supp(R/x e ), and thus (x e ) p =R p . Accordingly, one can conclude the following equalities

L p = e∈E(Cr) (x e ) p = ( e∈E(Cr) x e ) p .
Note that e∈E(Cr) x e is equal to the cover ideal of the induced graph on {i 1 , . . . , i r }, that is, e∈E(Cr) x e = J(C r ). It follows from Proposition 3.3.26(i) that J(C r ) has the strong persistence property, that is,

((J(C r )) m+1 : R J(C r )) = (J(C r )) m . This implies that ((J(C r )) m+1 p : Rp J(C r ) p ) = (J(C r )) m
p , and so (L m+1 p : Rp L p ) = L m p . We therefore deduce that L p has the strong persistence property, as required.

Subcase 2.2. {a, b} ∈ E(θ n 1 ,...,n k ) \ E(C r ).
A similar argument gives the following equality

L p = ((x a , x b ) ∩ e∈E(Cr) x e ) p .
Notice that {a, b} ∪ E(C r ) can be viewed as the edge set of the cycle graph C r with one chord {a, b}, say C. This yields that (x a , x b ) ∩ e∈E(Cr) x e equals the cover ideal of C, that is, J(C) = (x a , x b ) ∩ e∈E(Cr) x e . Furthermore, one can obtain from Proposition 3.3.26(ii) that J(C) has the strong persistence property, that is,

((J(C)) m+1 : R J(C)) = (J(C)) m . Thus, ((J(C)) m+1 p : Rp J(C) p ) = (J(C)) m
p , and so (L m+1 p : Rp L p ) = L m p . Therefore, L p has the strong persistence property, as desired. This completes our discussion.

On the normality of the cover ideals of theta graphs

In general, there are well-known relations among the persistence property, strong persistence property, and normality. In fact, every normal monomial ideal has the strong persistence property (see Theorem 3.2.34), and the strong persistence property implies the persistence property (see Proposition 3.1.12). However, there exists some square-free monomial ideal which does not satisfy the persistence property. To see such an example, consider the following square-free monomial ideal in the polynomial ring R = K[x 1 , . . . , x 12 ], I = (x 2 x 4 x 5 x 7 , x 1 x 3 x 6 x 8 , x 3 x 6 x 8 x 9 , x 1 x 4 x 7 x 10 , x 4 x 5 x 7 x 10 , x 1 x 3 x 8 x 10 , x 2 x 4 x 5 x 11 , x 1 x 4 x 6 x 11 , x 1 x 6 x 8 x 11 , x 2 x 8 x 9 x 11 , x 6 x 8 x 9 x 11 , x 2 x 5 x 7 x 12 , x 3 x 6 x 9 x 12 , x 2 x 7 x 9 x 12 , x 3 x 5 x 10 x 12 , x 5 x 7 x 10 x 12 ). [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], we obtain m = (x 1 , . . . , x 12 ) ∈ Ass R (R/I 3 ) \ Ass R (R/I 4 ), that is, I does not satisfy the persistence property. It should be noted that this counterexample is different from the counterexample which has been presented in Proposition 3.2. 5.

By using Macaulay2

In what follows, our target is to probe the normality of the cover ideals of the theta graphs. Indeed, we found that proving the normality of odd cycle graphs, theta graphs, and bipartite graphs follow the same scheme. This is illustrated in the following theorem which introduces a class of normal monomial ideals. Note that R = K[x 1 , . . . , x m ] is a polynomial ring over a field K and [m] = {1, . . . , m}. Theorem 7.6.10. Let I be a monomial ideal in R. Let A h ⊆ [m] × [m] for all h = 1, . . . , s with s ∈ N. For each t ≥ 1, let c t and d t be arbitrary positive integers. Assume that the following statement is true

x l 1 1 • • • x lm m ∈ I t ⇐⇒        (i) l p + l q ≥ c t for each {p, q} ∈ A h and each h ∈ [s]; (ii) {p,q}∈A h [(l p + l q ) -c t ] ≥ d t for some h ∈ Ω ⊆ [s],
with Ω possibly empty. Then I is normal.

Proof. Let α = x α 1 1 • • • x αm m ∈ I t .
Our goal is to prove that α ∈ I t . Assume that {ω j } r j=1 ⊂ N m 0 is the set of exponents of the minimal generators of I t , and let ω j,i be the i-th component of ω j . Corollary 1.4.3 of [START_REF] Herzog | Monomial Ideals[END_REF] implies that (α 1 , . . . , α m ) belongs to the Newton polyhedron of I. This means that there exist λ 1 , . . . , λ r ∈ Q ≥0 with r j=1 λ j = 1 such that α i ≥ r j=1 λ j ω j,i for each i ∈ [m]. If {p, q} ∈ A h and h ∈ [s], then we obtain

α p + α q ≥ r j=1 λ j (ω j,p + ω j,q ) ≥ r j=1 λ j c t = c t . (7.6.2)
In addition, if Ω is not empty, then for every h ∈ Ω, we have

{p,q}∈A h [(α p + α q ) -c t ] ≥ {p,q}∈A h r j=1 λ j (ω j,p + ω j,q ) -c t = r j=1 λ j   {p,q}∈A h [(ω j,p + ω j,q ) -c t ]   ≥ r j=1 λ j d t = d t . (7.6.3)
It follows from (7.6.2) and (7.6.3) that α ∈ I t , and hence I is normal.

As an immediate consequence of Theorem 7.6.10, we get the corollary below. Corollary 7.6.11. Let G = (V (G), E(G)) be a finite simple graph with the cover ideal J ⊂ R, and

V (G) = [m] = s h=1 V h with ϕ ⊆ V h ⊆ V (G).
For each t ≥ 1, let c t and d t be arbitrary positive integers. Assume that the following statement is true

x l 1 1 • • • x lm m ∈ J t ⇐⇒                  (i) l p + l q ≥ c t for each {p, q} ∈ E(G) with p, q ∈ V h and each h ∈ [s]; (ii) {p,q}∈E(G) with p,q∈V h [(l p + l q ) -c t ] ≥ d t for some h ∈ Ω ⊆ [s],
with Ω possibly empty. Then J is normal.

Proof. The claim can be proved by direct application of Theorem 7.6.10 with

A h := {{p, q} ∈ E(G) : p, q ∈ V h } for each h ∈ [s].
In Lemma 3.1.10 it has been proved that if J is the cover ideal of the odd cycle graph C of order 2n + 1, then

x l 1 1 • • • x l 2n+1
2n+1 ∈ J t if and only if l i + l i+1 ≥ t for every i ∈ [2n + 1] and 2n+1 i=1 [(l i + l i+1 ) -t] ≥ t. In Theorem 7.1.7, it has been shown that J is normal. The proof in [4] is essentially the same as the proof of the above corollary but with s

= 1, V 1 = V (C) = [2n + 1]
, and c t = d t = 1. As an application of Corollary 7.6.11 above, we give in the following theorem a generalization such that we prove that cover ideals of theta graphs are normal.

Theorem 7.6.12. Let J be the cover ideal of the theta graph θ n 1 ,...,n k . Then J is normal.

Proof. Assume that C 2e 1 +1 , . . . , C 2ez+1 are all induced odd cycle subgraphs of θ n 1 ,...,n k . Set c t = d t = t for all t ≥ 1, V h := V (C 2e h +1 ) for each h = 1, . . . , z, and

V z+1 := V (G) \ z h=1 V (C 2e h +1
). Now, the assertion can be deduced from Theorem 7.6.3 and Corollary 7.6.11.

To give another application of Corollary 7.6.11, we can re-prove a well-known result which is related to the cover ideals of bipartite graphs, see [START_REF] Villarreal | Monomial algebras[END_REF]Corollary 14.6.25]. Proposition 7.6.13. Let G = (V (G), E(G)) be a bipartite graph with the cover ideal J, and

V (G) = [m]. Then J is normal.
Proof. Since G is bipartite, it follows from [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6] that J is normally torsion-free, and [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.6] implies that J t = J (t) for all t, where J (t) denotes the t-th symbolic power of J. By virtue of J (t) = p∈Min(J) p t , one obtains J t = p∈Min(J) p t . This yields that x ℓ 1 1 • • • x ℓm m ∈ J t if and only if ℓ p + ℓ q ≥ t for each {p, q} ∈ E(G). In view of Corollary 7.6.11, we can derive that J is a normal square-free monomial ideal, as desired.

We terminate this argument by stating two theorems which introduce some classes of normal monomial ideals. Theorem 7.6.14. Let I be a monomial ideal in R = K[x 1 , . . . , x m ] with m + 1 = 1. For each t ≥ 1, let c t be an arbitrary positive integer. Assume there are some fixed subsets S ⊆ [m] and W ⊆ [m] such that the following statement is true

x l 1 1 • • • x lm m ∈ I t ⇐⇒          (i) i∈S l 2i ≥ c t ; (ii) for each z ∈ W there is T z ⊆ S with i∈Tz l 2i ≤ l 2z+1 .
Then I is normal.

Proof. Let α = x α 1 1 • • • x αm m ∈ I t .
Assume that {ω j } r j=1 ⊂ N m 0 is the set of exponents of the minimal generators of I t , and also assume that ω j,i is the i-th component of ω j . Due to α ∈ I t , one can deduce from [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.4.3] that (α 1 , . . . , α m ) belongs to the Newton polyhedron of I. This gives that there exist λ 1 , . . . , λ r ∈ Q ≥0 with r j=1 λ j = 1 such that α i = r j=1 λ j ω j,i for each i ∈ We can now combine (7.6.4) and (7.6.5) to conclude α ∈ I t , and hence I is normal. This finishes the proof.

To clarify Theorem 7.6.14, we present the example below.

Example 7.6.15. Let R = K[x 1 , . . . , x 2n ] be the polynomial ring over a field K, and I be the square-free monomial ideal in R with

G(I) = {x 2i-3 x 2i-1 x 2i x 2i+1 : i = 1, . . . ,
T z = {z, z + 1, z + 2} for each z ∈ W .
More hypotheses can be made on the powers of square-free monomial ideals so that normality is deduced. For instance, we have the following theorem whose proof mimics the proof of Theorem 7.6.14.

Theorem 7.6.16. Let I be a monomial ideal in R = K[x 1 , . . . , x m ] with x m+1 represents x 1 . For each t ≥ 1, let c t be an arbitrary positive integer. Assume there are some fixed subsets S ⊆ [m], W ⊆ [m], and T ⊆ S such that the following statement is true

x l 1 1 • • • x lm m ∈ I t ⇐⇒          (i) i∈S l 2i ≥ c t ; (ii) i∈T l 2i ≤ z∈W l 2z+1 .
Then I is normal.

Powers of closed neighborhood and dominating ideals of graphs

Sharifan and Moradi, in [START_REF] Sharifan | Closed neighborhood ideal of a graph[END_REF], introduced the notion of closed neighborhood ideals and dominating ideals of graphs, refer to Section 1.2 for recalling the definitions. The authors calculated some algebraic invariants of these ideals such as regularity and projective dimension in terms of the information from the underlying graph. Next, Honeycutt and Sather-Wagstaff, in [START_REF] Honeycutt | Closed neighborhood ideals of finite simple graphs[END_REF], investigated the Cohen-Macaulay, unmixed, and complete intersection properties of closed neighborhood ideals.

In this section, we concentrate on the normality, strong persistence property, persistence property, and symbolic strong persistence property of closed neighborhood ideals and dominating ideals of some classes of graphs. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Normality and associated primes of Closed neighborhood ideals and dominating ideals[END_REF]. with s i=1 p i = p, r j=1 q j = q, p + q = tk, ε ≥ 0, and β is some monomial in R such that x 1 ∤ β. As x 1 ∤ β, x 1 ∤ δ, and gcd(v, x 1 ) = 1 for all v ∈ G(I) ∪ G(H), we obtain bk = cq + ε, in particular, bk ≥ cq. In addition, suppose that p is minimal according to the membership δ k ∈ I p H q . Now, one can deduce from (7.7.1) that (7.7.2)

δ k = s i=1 u p i i r j=1 h q j j β ∈ (I + H) tk .
It

u l i i r j=1 h z j j γ,
with s i=1 l i = l, r j=1 z j = z, l + z = t, and γ is some monomial in R. Observe that x 1 ∤ γ due to x 1 ∤ δ.

We have shown above that p + q = tk, lk + zk = tk, and p ≤ lk. It follows from this that q ≥ zk. We have also shown above that bk ≥ cq. Thus, bk ≥ czk, and since k ≥ 1, we have b ≥ cz. Multiplying (7.7.3) by x b 1 yields

x b 1 δ = s i=1 u l i i x b 1 r j=1 h z j j γ.
The latter can be rewritten as

s i=1 u l i i r j=1 (x c 1 h j ) z j (x b-cz 1 γ) ∈ I l (x c 1 H) z ⊆ L t .
Therefore, α = x b 1 δ ∈ L t , which makes L normal, completing the proof. (ii)-(iv) can be established similar to the proof of Corollary 6.2.3.

To formulate Theorem 7.7.3, one requires the next auxiliary result. We first should recall the definition of linear relation graph, see Definition 6.3.10. As an immediate consequence of Theorem 6.5.3, we get the following corollary. The following theorem is crucial for us to prove Theorem 7.8.1 and Lemma 7.9.3. Proof.

(i) Let h = x c 1 1 • • • x cm m .
We use induction on m. It follows from Theorem 7.7.1 that the claim is true for the case in which m = 1. Suppose now that m > 1 and also the assertion is true for m -1, that is,

I + x c 1 1 • • • x c m-1 m-1 H is normal. Since I, I + x c 1 1 • • • x c m-1
m-1 H, and

I + (I + x c 1 1 • • • x c m-1
m-1 H) are normal, one can rapidly conclude from Theorem 7.7.

1 that I + x cm m (I + x c 1 1 • • • x c m-1 m-1 H) is normal as well. Due to I + x cm m (I + x c 1 1 • • • x c m-1 m-1 H) = I + x c 1 1 • • • x cm m H,
L := IS ∩ (x n , x ℓ n+1 ) ⊂ S = R[x n+1 ] with ℓ > 1, is normal?
As an application of Theorem 7.7.1, we provide an affirmative answer to above open question in the following proposition:

Proposition 7.7.6. Let I ⊂ R = K[x 1 , . . . , x n ] be a normal square-free monomial ideal with G(I) ⊂ R. Then L := IS ∩ (x n , x ℓ n+1 ) ⊂ S = R[x n+1 ] with ℓ ≥ 1, is normal.
Proof. It is routine to check that

L = I ∩ (x n ) + I ∩ (x ℓ n+1 ) = x n (I : S x n ) + x ℓ n+1 I.
Since I is normal, [START_REF] Villarreal | Monomial algebras[END_REF]Proposition 12.2.3] implies that (I : S x n ) is normal, and hence x n (I : S x n ) is normal as well. In addition, x n (I : S x n ) + I = I is normal. Now, one can rapidly deduce from Theorem 7.7.1 that L is normal, as claimed.

To see an application of Proposition 7.7.6, we give the subsequent corollary.

Corollary 7.7.7. Let I be a normal square-free monomial ideal in R = K[x 1 , . . . , x n ] with G(I) ⊂ R. Then the square-free monomial ideal

L := IS ∩ (x n , x n+1 , . . . , x n+m ) ⊂ S = R[x n+1 , . . . , x n+m ],
is normal.

Proof. We demonstrate the claim by using induction on m. In view of Proposition 7.7.6, we deduce that the claim is true for the case in which m = 1. Suppose that the claim holds for m -1. Put J := I ∩ (x n , x n+1 , . . . , x n+m-1 ). We thus have L = J + x n+m I. The inductive hypothesis implies that J is normal. In the light of Theorem 7.7.3, one can conclude that L is normal. This completes the inductive step, and so the claim has been shown by induction.

The next theorem will be used to establish Lemma 7.9.3. To accomplish this, it is enough to show that L t ⊆ L t for all t ≥ 1. To do this, pick an arbitrary monomial α in L t and write α = x b n+1 δ for some integer b and some monomial δ ∈ R with x n+1 ∤ δ. Since α ∈ L t , [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.2] gives that α k ∈ L tk = (x n F + x n+1 G) tk for some positive integer k. This implies that α k ∈ (x n F ) p (x n+1 G) q for some integers p and q with p + q = tk. Assume that q is maximal according to this membership. This means that p is minimal according to this membership. Observe that if p = 0, then α k ∈ (x n+1 G) tk , and hence α ∈ (x n+1 G) t = (x n+1 G) t ⊂ L t . Henceforth, let p > 0. Thanks to x n F is normal, a similar argument shows that one may assume q > 0 too. Write (7.7.4)

α k = x bk n+1 δ k = s i=1 f p i i x p n s j=1 g q j j x q n+1 β,
with s i=1 p i = p, s j=1 q j = q, and β some monomial in S. Let x n+1 | β. On account of either f j x n = u j or f j = u j , this contradicts the maximality of q. In addition, if there exists some f j with p j > 0 such that x n+1 | f j , then once again this leads to a contradiction to the maximality of q. Therefore, we can assume in (7.7.4) that x n+1 ∤ β, and also x n+1 ∤ f j with p j > 0. Consequently, one can derive that q = bk, in particular, δ k ∈ (x n F ) p G q . Now, we get the following equality

δ k = s i=1 f p i i x p n s j=1 g q j j β ∈ (x n F + G) tk .
Thus, δ ∈ (x n F + G) t , and so δ ∈ (x n F + G) t . Hence, one obtains δ ∈ (x n F ) l G h for some l and h with l + h = t, in particular, δ k ∈ (x n F ) lk G hk . It follows from the minimality of p that lk ≥ p. Because lk ≥ p and p + q = lk + hk = tk, we gain q ≥ hk. Since k ≥ 1, one can deduce from q = bk and q ≥ hk that b ≥ h.

It follows now from δ ∈ (x n F ) l G h and b ≥ h that α = x b n+1 δ ∈ (x n F + x n+1 G) t = L t ,
and the argument is over.

Closed neighborhood and dominating ideals of complete bipartite graphs

Here, we are going to provide some applications of the previous section. In fact, we study the closed neighborhood ideals and dominating ideals of complete bipartite graphs. We begin with the following theorem which says that closed neighborhood ideals of complete bipartite graphs are normal, and so satisfy the (strong) persistence property.

Theorem 7.8.1. Let K r,s be a complete bipartite graph. Then the following statements hold:

(i) N I(K r,s ) is normal.
(ii) N I(K r,s ) has the strong persistence property.

(iii) N I(K r,s ) has the persistence property.

Proof. (i) For convenience of notation, put

L := N I(K r,s ). Let V (K r,s ) = V 1 ∪ V 2 , where V 1 = {x 1 , . . . , x r } and V 2 = {x r+1 , . . . , x r+s }. If r = 1, then L = (x 1 s+1 i=2 x i , x 1 x 2 , . . . , x 1 x s+1 ) = x 1 (x 2 , . . . , x s+1 ).
It is well-known that any monomial prime ideal is normal, and by virtue of Remark 7.1.1, one has L is normal. Similarly, if s = 1, then L is normal too. Accordingly, we assume that r, s > 1. Set g := r i=2 x i and f := r+s i=r+1 x i . It is easy to check that G(L) = {x 1 f, . . . , x r f, x 1 x r+1 g, . . . , x 1 x r+s g}. Hence, one can write L = I + x 1 H, where H := (f, x r+1 g, . . . , x r+s g) and I := (x 2 , . . . , x r )f . A similar argument shows that I is normal. Since H = g(x r+1 , . . . , x r+s ) + f R, where R = K[x 1 , . . . , x r , x r+1 , . . . , x r+s ], it follows from Theorem 7.1.3 that H is normal. It is not hard to check that I + H = H, and so I + H is normal as well. Note that gcd(x 1 , v) = 1 for all v ∈ G(I) ∪ G(H). Now, the claim can be deduced immediately from Theorem 7.7.3.

(ii) and (iii) can be proved similar to the proof of Corollary 6.2.3. This finishes the proof.

In the following proposition, we investigate when the unique homogeneous maximal ideal appears in the associated primes set of powers of closed neighborhood ideals of complete bipartite graphs. Proposition 7.8.2. Let K r,s with r, s > 1 be a complete bipartite graph, and m = (x 1 , . . . , x r , x r+1 , . . . , x r+s ) ⊂ R = K[x 1 , . . . , x r , x r+1 , . . . , x r+s ], be the unique homogeneous maximal ideal. Then, for all s ≥ 3, m ∈ Ass(N I(K r,s ) s ) \ Ass(N I(K r,s )) ∪ Ass(N I(K r,s ) 2 ) .

In particular, depth(R/N I(K r,s ) s ) = 0 for all s ≥ 3, and

lim k→∞ depth(R/N I(K r,s ) k ) = 0. Proof. Set L := N I(K r,s ). Let V (K r,s ) = V 1 ∪ V 2 , where V 1 = {x 1 , . . . , x r } and
V 2 = {x r+1 , . . . , x r+s }. Put g := r i=1 x i and f := r+s i=r+1 x i . It is not hard to investigate that G(L) = {x 1 f, . . . , x r f, x r+1 g, . . . , x r+s g}. Due to L is a square-free monomial ideal, this implies that Ass(L) = Min(L), and hence m / ∈ Ass(L). We show that m / ∈ Ass(L 2 ). Suppose, on the contrary, that m ∈ Ass(L 2 ). This gives that m = (L 2 : v) for some monomial v ∈ R. By virtue of Proposition 10.2.16, one has deg x i (v) ≤ 1 for all i = 1, . . . , r + s. Let x d ∈ m for some 1 ≤ d ≤ r + s. Hence,

x d v ∈ L 2 ,
and so there exists some monomial w ∈ G(L 2 ) such that w | x d v. In addition, it follows from G(L) = {x 1 f, . . . , x r f, x r+1 g, . . . , x r+s g} that there exist i and j with 1

≤ i = j ≤ r + s such that deg x i (w) = deg x j (w) = 2. Since deg x i (v) ≤ 1 and deg x j (v) ≤ 1,
this leads to a contradiction. Consequently, m / ∈ Ass(L 2 ). Based on Theorem 7.8.1(iii), the ideal L satisfies the persistence property. Hence, it is enough for us to show that m ∈ Ass(L 3 ). To accomplish this, set h := r+s i=1 x 2 i . We claim that h / ∈ L 3 and m ⊆ (L 3 : h). We show that

x k h ∈ L 3 for all 1 ≤ k ≤ r + s. Let 1 ≤ k ≤ r. Set α := (x r+1 g)(x r+2 g)(x k f ). Because α ∈ L 3 and α | x k h, one can derive that x k h ∈ L 3 . Now, let r + 1 ≤ k ≤ r + s. Put β := (x 1 f )(x 2 f )(x k g).
Since β ∈ L 3 and β | x k h, this implies that x k h ∈ L 3 . Therefore, we deduce that m ⊆ (L 3 : h). On the contrary assume that h ∈ L 3 . Thus, there exist monomials

u 1 , u 2 , u 3 ∈ G(L) such that u 1 u 2 u 3 | h. If u 1 , u 2 , u 3 ∈ {x 1 f, . . . , x r f } (respectively, u 1 , u 2 ,
u 3 ∈ {x r+1 g, . . . , x r+s g}), then f 3 | h (respectively, g 3 | h), which contradicts the fact that deg x i (h) = 2 for all i. Let u 1 , u 2 ∈ {x r+1 g, . . . , x r+s g}, and u 3 ∈ {x 1 f, . . . , x r f }, say u 3 = x 1 f . Then, we get x 3 1 | u 1 u 2 u 3 , and so x 3 1 | h, which is a contradiction. Finally, let u 1 , u 2 ∈ {x 1 f, . . . , x r f }, and u 3 ∈ {x r+1 g, . . . , x r+s g}, say u 3 = x r+1 g. This gives rise to x 3 r+1 | u 1 u 2 u 3 , and so x 3 r+1 | h, which is a contradiction. Accordingly, h / ∈ L 3 , and thus m = (L 3 : h). This terminates the proof.

We are in a position to state another main result of this section in the subsequent theorem.

Theorem 7.8.3. The dominating ideals of complete bipartite graphs are nearly normally torsion-free.

Proof. Let K r,s be a complete bipartite graph, R = K[x 1 , . . . , x r , x r+1 , . . . , x r+s ], and m = (x 1 , . . . , x r , x r+1 , . . . , x r+s ) be the unique homogeneous maximal ideal. Let V (K r,s ) = V 1 ∪ V 2 , where V 1 = {x 1 , . . . , x r } and V 2 = {x r+1 , . . . , x r+s }. Put L := DI(K r,s ), I := (x i : i = 1, . . . , r), J := (x i : i = r + 1, . . . , r + s), g := r i=1 x i , and f := r+s i=r+1 x i . According to [START_REF] Sharifan | Closed neighborhood ideal of a graph[END_REF]Lemma 2.2] and [133, Exercise 6.1.23], we get the following equalities:

L = r i=1 (J + x i R) ∩ r+s i=r+1 (I + x i R) =(J + r i=1 x i R) ∩ (I + r+s i=r+1 x i R) =J ∩ I + r i=1 x i R + r+s i=r+1 x i R =JI + gR + f R.
In what follows, our aim is to use Corollary 6.1.3. To do this, without loss of generality, it is enough to show that L(m \ {x 1 }) is normally torsion-free. Since

I(m\{x 1 }) = R, J(m\{x 1 }) = J, f R(m\{x 1 }) = f R, and gR(m\{x 1 }) = r i=2 x i R, we obtain L(m \ {x 1 }) = J + r i=2 x i R + f R = J + r i=2 x i R.
It follows now from Theorem 5.3.4 that L(m \ {x 1 }) is normally torsion-free. Therefore, the claim can be deduced from Corollary 6.1.3, and the proof is done.

Dominating ideals of h-wheel graphs

The main point of this section is to explore dominating ideals of a class of graphs which are called h-wheel graphs. For this purpose, we start by stating the definition of h-wheel graphs. Definition 7.9.1. ( [22,Definition 1.6]) A graph G with the vertex set V (G) is called an h-wheel if V (G) can be written as the union of two disjoint sets, the set of rim vertices R G and the set of center vertices C G , such that the following conditions hold:

(1) The subgraph induced by C G is the complete graph on h vertices.

(2) The subgraph induced by R G is an odd cycle.

(3) There exist x i 1 , . . . ,

x i k ∈ R G with k ≥ 3 such that N R G (y) = {x i 1 , . . . , x i k } for all y ∈ C G .
(4) For every y ∈ C G , the vertex y belongs to at least two odd cycles in the subgraph induced by y and N R G (y).

The k is called the radial number for G. Also, for each j = 1, . . . , k -1, set ℓ i as the length of the path along the subgraph induced by R G from x i j to x i j+1 , and set ℓ k as the length from x i k to x i 1 . The positive integers ℓ 1 , . . . , ℓ k are called the radial lengths.

It should be noted that, in [78, Page 265], the authors studied the 1-wheel, which we call a wheel for simplicity. In fact, given an h-wheel G and a vetex y ∈ C G , the subgraph induced by y and R G is a wheel. Example 7.9.2. We give a 4-wheel graph G in the following figure. x 5

x 4

x 3

x 2

4-wheel graph

Note that C G = {y 1 , y 2 , y 3 , y 4 }, R G = {x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 }, and Then the following statements hold:

N R G (y 1 ) = N R G (y 2 ) = N R G (y 3 ) = N R G (y 4 ) = {x 1 , x
(i) H is normal.

(ii) H has the strong persistence property.

(iii) H has the persistence property.

Proof. (i) Clearly, n ≥ 3. Let {1, . . . , n} \ {ℓ 1 , . . . , ℓ k } = {i 1 , . . . , i n-k }. To simplify the notation, set u j := (x i j -1 , x i j , x i j +1 ) ⊂ R for each j = 1, . . . , n -k, where

R = K[x 1 , . . . , x n ].
We use induction on s := n -k. If s = 0, then there is nothing to prove. Let s = 1.

Then H = u 1 = (x i 1 -1 , x i 1 , x i 1 +1
), which is certainly normal. Now, let s > 1, and the claim has been shown for all integers less than s. Put J := s-1 j=1 u j . Note that the inductive hypothesis implies that J is normal. Moreover, H = J ∩ u s . Here, one may consider the following cases: Case 1. supp(J) ∩ supp(u s ) = ∅. Then Lemma 7.4.5 yields that H is normal.

(J) ∩ supp(u s ) = {x is-1 , x is }. This implies that H = J ∩ (x is-1 , x is ) + x is+1 J. It is routine to check that one can write J ∩ (x is-1 , x is ) = s-1 j=1, where x is-1 x is / ∈u j u j ∩ (x is-1 , x is ).
The inductive hypothesis yields that s-1 j=1, where x is-1 x is / ∈u j u j is normal. Notice that any two consecutive integers can be appeared at most in two u i 's. Also, recall that any integer can be appeared at most in three u i 's. It follows now from Corollary 7.7.7 that s-1 j=1, where x is-1 x is / ∈u j u j ∩(x is-1 , x is ) is normal too. On account of Theorem 7.7.3(i), we deduce that H is normal.

Subcase 3.2. |a -b| = 2, that is, supp(J) ∩ supp(u s ) = {x is-1 , x is+1 }. We thus have H = J ∩ (x is-1 , x is+1 ) + x is J. Want to show that J ∩ (x is-1 , x is+1 ) is normal.
Our strategy is to use Theorem 7.7.8. For this purpose, it is sufficient to prove that J ∩ (x is-1 ) + (J : x is+1 ) is normal. If (x is-3 , x is-2 , x is-1 ) or (x is+1 , x is+2 , x is+3 ) or both of them do not appear in J, then Corollary 7.7.7 yields that J ∩ (x is-1 , x is+1 ) is normal. Hence, assume that (x is-3 , x is-2 , x is-1 ) and (x is+1 , x is+2 , x is+3 ) appear in J. This means that we can write

J = J 1 ∩ (x is-3 , x is-2 , x is-1 ) ∩ (x is+1 , x is+2 , x is+3 ).
Note that the inductive hypothesis gives that J 1 is normal. Now, we obtain

J ∩ (x is-1 ) + (J : x is+1 ) =J 1 ∩ (x is+1 , x is+2 , x is+3 ) ∩ (x is-1 ) +J 1 ∩ (x is-3 , x is-2 , x is-1 ) =J 1 ∩ (x is-3 , x is-2 , x is-1 ), where supp(J 1 )∩{x is-1 , x is+1 } = ∅. Since x is-1 / ∈ J 1 , this implies that the normality of J 1 ∩ (x is-3 , x is-2 , x is-1
) can be concluded from Case 1 or Case 2 or Subcase 3.1. Therefore, J ∩ (x is-1 ) + (J : x is+1 ) is normal. It follows from Theorem 7.7.8 that J ∩ (x is-1 , x is+1 ) is normal. On account of Theorem 7.7.3(i), we deduce that H is normal.

It has already been introduced, in [17], the notion of partial t-cover ideals of finite simple graphs. We first recall the definition of partial t-cover ideals in the following definition. Definition 7.9.5. ( [17,Definition 1.1]) Suppose that G is a finite simple graph on the vertex set V (G) = {x 1 , x 2 , . . . , x n } with the edge set E(G). Also, for any

x ∈ V (G), let N (x) = {y : {x, y} ∈ E(G)} denote the set of neighbors of x. Fix an integer t ≥ 1. The partial t-cover ideal of G is the monomial ideal

J t (G) = x∈V (G)   {x i 1 ,...,x i t }⊆N (x) (x, x i 1 , . . . , x it )   .
When t = 1, above construction is simply the cover ideal of a finite simple graph G. It has already been shown in [17,Theorem 1.2] that if T is a tree, then, for any t ≥ 1, J t (T ) satisfies the persistence property. The next corollary states that the partial 2-cover ideal of any cycle graph satisfies the persistence property as well.

Corollary 7.9.6. Let C n be a cycle graph and J 2 (C n ) be its partial 2-cover ideal. Then the following statements hold:

(i) J 2 (C n ) is normal.
(ii) J 2 (C n ) has the strong persistence property.

(iii) J 2 (C n ) has the persistence property.

We are ready to express the main result of this section in the following theorem. Theorem 7.9.7. Let G be an h-wheel graph with rim R G and center C G such that V (C G ) = {y 1 , . . . , y h } and V (R G ) = {x 1 , . . . , x 2m+1 }, where m ≥ 2. Also, let x ℓ 1 , . . . , x ℓ k be the radial vertices such that there exist at least three consecutive numbers among them. Let DI(G) denote the dominating ideal of G. Then the following statements hold:

(i) DI(G) is normal.
(ii) DI(G) has the strong persistence property.

(iii) DI(G) has the persistence property.

Proof. (i) To simplify the notation, put F := (x ℓ 1 , . . . , x ℓ k )S and J := (y 1 , . . . , y h )S, where S = K[x 1 , . . . , x 2m+1 , y 1 , . . . , y h ]. Since there exist at least three consecutive numbers among ℓ 1 , . . . , ℓ k , this gives rise to k i=1 (x ℓ i -1 , x ℓ i , x ℓ i +1 ) ⊆ F . Hence, by remembering this fact that x 0 (respectively, x 2m+2 ) represents x 2m+1 (respectively,

x 1 ), one can deduce the following equalities

DI(G) =(F + J) ∩ k i=1 ((x ℓ i -1 , x ℓ i , x ℓ i +1 ) + J) ∩ j∈{1,...,2m+1}\{ℓ 1 ,...,ℓ k } (x j-1 , x j , x j+1 ) = k i=1 (x ℓ i -1 , x ℓ i , x ℓ i +1 ) + J ∩ j∈{1,...,2m+1}\{ℓ 1 ,...,ℓ k } (x j-1 , x j , x j+1 ) = 2m+1 j=1 (x j-1 , x j , x j+1 ) + J j∈{1,...,2m+1}\{ℓ 1 ,...,ℓ k } (x j-1 , x j , x j+1 ) =DI(R G ) + J j∈{1,...,

Chapter 8

On the set defining minimal vertex covers and dominating sets

Sets defining minimal vertex covers of uniform hypergraphs

One motivation for this argument comes from a nice result in [START_REF] Martí-Farré | Sets defining minimal vertex covers[END_REF] which is devoted to finite simple graphs. In fact, Theorem 2.1 in [START_REF] Martí-Farré | Sets defining minimal vertex covers[END_REF] says that if V is a finite nonempty set of n elements, r a positive integer, and A = {A 1 , . . . , A r } a non-empty family of r subsets of V , then under certain conditions, there exists a unique simple graph G with vertex set V whose minimal vertex cover sets are the elements of A.

In what follows, we generalize the above mentioned result to d-uniform hypergraphs (see Theorem 8.1.3). To review the definition of hypergraphs, refer to Subsection 1.3. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Sets defining minimal vertex covers of uniform hypergraphs[END_REF]. Before stating the following definition, remember that an m×n matrix M = (a ij ) is called binary if a ij ∈ {0, 1} for each i = 1, . . . , m and j = 1, . . . , n. Definition 8.1.1. Let M = (a ij ) be an m × n binary matrix, and d be a positive integer with 2 ≤ d ≤ n. We say that M satisfies the condition: When d = 2, the above definition coincides with conditions (P1), (P2) and (P3) in [START_REF] Martí-Farré | Sets defining minimal vertex covers[END_REF].

(P1 ) when, for each 1 ≤ i 1 , i 2 ≤ m with i 1 = i 2 ,
In the next definition, we correspond a matrix to a non-empty family of some subsets of a non-empty set. Definition 8.1.2. Suppose that V = {1, . . . , n} is a non-empty set of n elements, and let A = {A 1 , . . . , A m } be a non-empty family of m subsets of V . We define the incidence matrix associated to A, denoted by M (A), as the binary matrix M (A) = (a ij ) with m rows and n columns such that a ij = 0 if j / ∈ A i , and a ij = 1 if j ∈ A i for all i = 1, . . . , m and j = 1, . . . , n.

We are now ready to present the main theorem of this section. If there exists such a hypergraph H, then it is unique and its edge set is given by

E(H) = {{j 1 , . . . , j d } : a i,j 1 + • • • + a i,j d ≥ 1 for all i = 1, . . . , m,
with j 1 , . . . , j d are distinct positive integers}.

Proof. For necessity, assume that there exists a d-uniform hypergraph H with V (H) = V and A is the set whose elements are the minimal vertex cover sets of H. We first prove that M (A) satisfies the condition (P1 ). To do this, let 1 ≤ i 1 , i 2 ≤ m with i 1 = i 2 . Suppose, on the contrary, that (P1 ) does not hold. This implies that, for each j = 1, . . . , n, we have a i 1 ,j ≤ a i 2 ,j . Thus, one derives A i 1 ⊆ A i 2 , a contradiction to the minimality of A i 2 . Consequently, M (A) satisfies the condition (P1 ).

We now show that M (A) satisfies the condition (P2 )(d). To achieve this, suppose that 1 ≤ j 1 , . . . , j d-1 ≤ n are d-1 distinct positive integers. On the contrary, assume that for each 1 ≤ i ≤ m, one has a i,j 1 + • • • + a i,j d-1 = 0. Set p i := (x w : w ∈ A i ) for each i = 1, . . . , m. Thus, one can show that I(H) = m i=1 p i , where I(H) denotes the edge ideal of the hypergraph H. It follows from the assumption that {j 1 , . . . , j d-1 } ∩ A i = ∅ for each i = 1, . . . , m. This implies that {x j 1 , . . . , x j d-1 } ∩ p i = ∅ for each i = 1, . . . , m. Accordingly, x j 1 • • • x j d-1 ∈ p i for each i = 1, . . . , m, and hence

x j 1 • • • x j d-1 ∈ m i=1 p i .
This implies that there exists an element u ∈ G(I(H)) such that u|x j 1 • • • x j d-1 . Consequently, we deduce that degu ≤ d -1, which contradicts the fact that degu = d. Therefore, M (A) satisfies the condition (P2 )(d).

To complete the argument, one has to demonstrate that M (A) satisfies the condition (P3 )(d). Suppose that for each ℓ ≥ d distinct positive integers 1 ≤ j 1 , . . . , j ℓ ≤ n, we have a i,j 1 + • • • + a i,j ℓ ≥ 1 for all i = 1, . . . , m. Our aim is to verify that there are at least d distinct integers j α 1 , . . . , j α d ∈ {j 1 , . . . , j ℓ } such that

a i,jα 1 + • • • + a i,jα d ≥ 1 for all i = 1, . . . , m.
Set p i := (x w : w ∈ A i ) for each i = 1, . . . , m. It is routine to check that

a i,j 1 + • • • + a i,j ℓ ≥ 1 for all i = 1, . . . ,
with degu = d such that u|x j 1 • • • x j ℓ . Assume that u = x jα 1 • • • x jα d . Thus, j α 1 , . . . , j α d ∈ {j 1 , . . . , j ℓ }. Also, u ∈ G(I(H)) yields that x jα 1 • • • x jα d ∈ p i for all i = 1, . . . , m. Accordingly, a i,jα 1 + • • • + a i,jα d ≥ 1 for all i = 1, . . . , m, as required.
To conclude the proof, we need to verify sufficiency. To do this, suppose that the matrix M (A) satisfies conditions (P1 ), (P2 )(d), and (P3 )(d). We define a hypergraph H with the vertex set V (H) = V , and the following edge set

E(H) = {{j 1 , . . . , j d } : a i,j 1 + • • • + a i,j d ≥ 1 for all i = 1, . . . , m,
with j 1 , . . . , j d are distinct positive integers}.

Here, we need to demonstrate E(H) = ∅. Since A i = ∅ for all i = 1, . . . , m, it follows that for each row i, there exists an integer j such that a ij = 1. Hence, a i,1 + • • • + a i,n ≥ 1 for all i = 1, . . . , m. Now, the condition (P3 )(d) yields that there exist at least d distinct integers j 1 , . . . , j d ∈ {1, . . . , n} such that a i,j 1 + • • • + a i,j d ≥ 1 for all i = 1, . . . , m. Accordingly, {j 1 , . . . , j d } ∈ E(H), and so E(H) = ∅. Certainly,

H is a d-uniform hypergraph.
Hence, our goal is to show that every element of A is a minimal vertex cover set of H. To accomplish this, we first prove that A 1 , . . . , A m are vertex cover sets of H. To do this, consider {j 1 , . . . , j d } ∈ E(H). This implies that a i,j 1 + • • • + a i,j d ≥ 1 for all i = 1, . . . , m, and so {j 1 , . . . , j d } ∩ A i = ∅ for all i = 1, . . . , m. Therefore, A i is a vertex cover set of H for all i = 1, . . . , m.

In the following, we claim that if C is a vertex cover set of H, then A i ⊆ C for some 1 ≤ i ≤ m. Suppose, on the contrary, that A i ⊈ C for each i = 1, . . . , m. This implies that there exists an element f s ∈ A s \ C for all s = 1, . . . , m. Assume that {f 1 , . . . , f m } = {y 1 , . . . , y r } such that y 1 , . . . , y r are r distinct integers. Due to {f 1 , . . . , f m } = {y 1 , . . . , y r }, for each i = 1, . . . , m, we get there exists a positive integer k i with 1 ≤ k i ≤ r such that y k i ∈ A i , and hence a i,y 1 + • • • + a i,yr ≥ 1 for each i = 1, . . . , m. We claim that r ≥ d. On the contrary, let r ≤ d -1. It follows that one can pick d -1 -r distinct integers z 1 , . . . , z d-1-r ∈ {1, . . . , n} \ {y 1 , . . . , y r }. Since a i,y 1 + • • • + a i,yr ≥ 1 for each i = 1, . . . , m, we obtain

a i,y 1 + • • • + a i,yr + a i,z 1 + • • • + a i,z d-1-r ≥ 1,
for each i = 1, . . . , m. This contradicts the condition (P2 )(d), and so one derives r ≥ d. Now, the condition (P3 )(d) follows that there exist at least d distinct integers y α 1 , . . . , y α d ∈ {y 1 , . . . , y r } such that a i,yα 1 + • • • + a i,yα d ≥ 1 for all i = 1, . . . , m. We thus gain {y α 1 , . . . , y α d } ∈ E(H). This implies that {y α 1 , . . . , y α d } ∩ C = ∅, and so {y 1 , . . . , y r } ∩ C = ∅. Therefore, one has {f 1 , . . . , f m } ∩ C = ∅, which is a contradiction. Consequently, one derives that if C is a vertex cover set of H, then

A i ⊆ C for some 1 ≤ i ≤ m.
Moreover, the condition (P1 ) gives that A i ⊈ A j for each 1 ≤ i, j ≤ m with i = j. Accordingly, we deduce that A 1 , . . . , A m are the minimal vertex cover sets of H, and thus one obtains I(H) = m i=1 p i , where p i := (x w : w ∈ A i ) for all i = 1, . . . , m. By virtue of [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.3.6], there exists only one square-free monomial ideal such that the set of its associated prime ideals is {p 1 , . . . , p m }. This establishes that the hypergraph H is unique, and the proof is over.

M (A) =            0 0 1 0 1 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1            .
Since a 2,j ≥ a 4,j , for each 1 ≤ j ≤ 6, one derives the condition (P1 ) does not hold, and so there is no uniform hypergraph such as H with V (H) = V , and with A is the set whose elements are the minimal vertex cover sets of H.

It is routine to check that the condition (P1 ) holds. Moreover, one can see that for any d -1 = 2 distinct integers 1 ≤ j 1 , j 2 ≤ 7, there exists an integer 1 ≤ i ≤ 5 such that a i,j 1 + a i,j 2 = 0. So the matrix M (A) satisfies the condition (P2 )(d) for d = 3. Furthermore, we have a i,1 + a i,4 + a i,6 + a i,7 ≥ 1 for all i = 1, 2, 3, 4, 5, while there are no distinct integers j 1 , j 2 , j 3 ∈ {1, 4, 6, 7} such that a i,j 1 + a i,j 2 + a i,j 3 ≥ 1 for all i = 1, 2, 3, 4, 5. This means that the condition (P3 )(d) for d = 3, does not hold. Therefore, we are not able to build any 3-uniform hypergraph such as H with V (H) = V , and with A is the set whose elements are the minimal vertex cover sets of H.

M (A) =                  0 0 1 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 0 0 1 0 0                  .
One can easily see that the condition (P1 ) holds. In addition, for investigating the condition (P2 )(d) for d = 3, one has to check that for any two distinct integers 1 ≤ j 1 , j 2 ≤ 6, there exists a positive integer 1 ≤ i ≤ 7 such that a i,j 1 + a i,j 2 = 0. To do this, we only need to examine it 6 2 = 15 times. This shows that the matrix M (A) satisfies the condition (P2 )(d) for d = 3. Also, for checking the condition (P3 )(d) for d = 3, we must check that for each 6 ≥ ℓ ≥ 3 and distinct integers 1 ≤ j 1 , . . . , j ℓ ≤ 6 such that a i,j 1 + • • • + a i,j ℓ ≥ 1 for all i = 1, . . . , 7, then there exist at least three distinct integers j α 1 , j α 2 , j α 3 ∈ {j 1 , . . . , j ℓ } such that a i,jα 1 + a i,jα 2 + a i,jα 3 ≥ 1 for all i = 1, . . . , 7. When ℓ = k, where 3 ≤ k ≤ 6, one is required to investigate the condition (P3 )(d) for d = 3, exactly 6 ℓ times. This implies that we have to check the condition (P3 )(d) for d = 3, 6 3 + 6 4 + 6 5 + 6 6 = 42 times. After checking them, we deduce that the condition (P3 )(d) for d = 3, holds . Now, based on Theorem 8.1.3, it follows that the hypergraph H with V (H) = {1, 2, 3, 4, 5, 6}, and the following edge set

E(H) = {{1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 4, 5}, {3, 4, 6}},
is the unique 3-uniform hypergraph such that A = {A i | i = 1, . . . , 7} is the set whose elements are the minimal vertex cover sets of H.

Sets defining minimal dominating sets of regular graphs

Suppose that V = {1, . . . , n} is a non-empty set of n elements, S = {S 1 , . . . , S m } a non-empty set of m non-empty subsets of V . In this section, by using some algebraic notions in commutative algebra, we investigate the question arises whether there exists an undirected finite simple graph G with V (G) = V , where S is the set whose elements are the minimal dominating sets of G.

It should be noted that the results of this section can be found in [START_REF] Nasernejad | An algebraic approach to sets defining minimal dominating sets of regular graphs[END_REF].

Before asserting the main result of this section, one needs to review the following defnitions. Remember that an m×n matrix M = (a ij ) is called binary if a ij ∈ {0, 1} for each i = 1, . . . , m and j = 1, . . . , n. Definition 8.2.1. Let M = (a ij ) be an m × n binary matrix, and r be a positive integer with 2 ≤ r ≤ n. We say that M satisfies the condition: (i) when, for any 1 ≤ i 1 , i 2 ≤ m with i 1 = i 2 , there exists a positive integer

1 ≤ j ≤ n such that a i 1 ,j > a i 2 ,j .
(ii) when, for any r -1 distinct positive integers 1 ≤ j 1 , . . . , j r-1 ≤ n, there exists a positive integer

1 ≤ i ≤ m such that a i,j 1 + • • • + a i,j r-1 = 0.
(iii) when, for any ℓ ≥ r distinct positive integers 1 ≤ j 1 , . . . , j ℓ ≤ n such that

a i,j 1 + • • • + a i,j ℓ ≥
with degu = r such that u | x j 1 • • • x j ℓ . Let u = x jα 1 • • • x jα r .
Accordingly, we get j α 1 , . . . , j αr ∈ {j 1 , . . . , j ℓ }. One can derive from u ∈ G(N I(G)) that x jα 1 • • • x jα r ∈ p i for all i = 1, . . . , m. Therefore, a i,jα 1 +• • •+a i,jα r ≥ 1 for all i = 1, . . . , m, as claimed.

To finish the argument, we show that M (S) satisfies condition (iv). By virtue of G is an (r -1)-regular graph, this implies that there exist n subsets N G [1]

(G) = (x θ t,1 • • • x θt,r : t = 1, . . . , n). Since N I(G) = ∩ m i=1 p i
, where p i = (x w : w ∈ S i ) for each i = 1, . . . , m, this gives that x θ t,1 • • • x θt,r ∈ p i for each i = 1, . . . , m and t = 1, . . . , n. Fix 1 ≤ t ≤ n. Hence, we get {θ t,1 , . . . , θ t,r } ∩ S i = ∅ for each i = 1, . . . , m, and so a i,θ t,1 + • • • + a i,θt,r ≥ 1 for each i = 1, . . . , m. Now, assume that z 1 , . . . , z r are r distinct integers such that a i,z 1 + • • • + a i,zr ≥ 1 for i = 1, . . . , m. This implies that {z 1 , . . . , z r } ∩ S i = ∅ for each i = 1, . . . , m, and so {x z 1 , . . . , x zr } ∩ p i = ∅ for all i = 1, . . . , m. We can deduce that x z 1 • • • x zr ∈ p i for all i = 1, . . . , m, and hence Moreover, since G is an (r -1)-regular graph, we derive there exist r sets Conversely, assume that the matrix M (S) satisfies conditions (i)-(iv). Our aim is to verify the existence of an (r -1)-regular graph G with V (G) = V where S is the set whose elements are the minimal dominating sets of G. Condition (iv) gives that there exist n subsets Γ t = {θ t,1 , . . . , θ t,r } ⊆ {1, . . . , n} for any integer 1 ≤ t ≤ n such that a i,θ t,1 + • • • + a i,θt,r ≥ 1 for i = 1, . . . , m and t = 1, . . . , n, and also for any integer 1 ≤ t ≤ n there exist r sets Γ i 1 , . . . , Γ ir with t ∈ Γ i k for all k = 1, . . . , r, t ∈ Γ t , and θ t,i ∈ Γ t if and only if t ∈ Γ θ t,i for all i = 1, . . . , r. Hence, we define an (r -1)-regular graph G with N G [t] := Γ t = {θ t,1 , . . . , θ t,r } for any integer 1 ≤ t ≤ n. In other words, conditions (iv)(a)-(d) guarantee the existence of such a graph G. In particular, one can conclude from the definition of closed neighborhood ideal of G that N I(G) = (x θ t,1 Here, we claim that if D is an arbitrary dominating set of G, then S i ⊆ D for some 1 ≤ i ≤ m. On the contrary, assume that S i ⊈ D for each i = 1, . . . , m. Thus there exists an element f i ∈ S i \ D for all i = 1, . . . , m. Suppose that ∪ m i=1 {f i } = {y 1 , . . . , y λ } such that y 1 < • • • < y λ are λ distinct integers. Because ∪ m i=1 {f i } = {y 1 , . . . , y λ }, we get f i ∈ {y 1 , . . . , y λ } for all i = 1, . . . , m. Thus, we deduce that there exists a positive integer k i with 1 ≤ k i ≤ λ such that f i = y k i for all i = 1, . . . , m. Due to f i ∈ S i for all i = 1, . . . , m, this implies that y k i ∈ S i for all i = 1, . . . , m. Since, for all i = 1, . . . , m, the i-th row of the matrix M (S) is associated to S i ,

x z 1 • • • x zr ∈ ∩ m i=1 p i . As N I(G) = ∩ m i=1 p i and N I(G) = (x θ t,1 • • • x θt,r : t = 1, . . . , n), we get there exists an element x θ s,1 • • • x θs,r ∈ G(N I(G)) for some 1 ≤ s ≤ n such that x θ s,1 • • • x θs,r | x z 1 • • • x zr . Since both of x θ s,1 • • • x θs,
N G [i 1 ], . . . , N G [i r ] with t ∈ N G [i k ] for all k = 1, . . . ,
y k i ∈ S i , and 1 ≤ k i ≤ λ, this yields that a i,y 1 + • • • + a i,y λ ≥ 1 for each i = 1, . . . , m.
We prove that λ ≥ r. Suppose, on the contrary, that λ ≤ r -1.

If λ = r -1, then a i,y 1 + • • • + a i,y r-1 ≥ 1 for each i = 1, . . . , m
, which is a contradiction with condition (ii). Now, let λ < r -1, then this gives that one can choose r -1-λ distinct integers z 1 , . . . , z r-1-λ ∈ {1, . . . , n} \ {y 1 , . . . , y λ }. On account of a i,y 1 + • • • + a i,y λ ≥ 1 for each i = 1, . . . , m, one obtains a i,y 1 + • • • + a i,y λ + a i,z 1 + • • • + a i,z r-1-λ ≥ 1 for each i = 1, . . . , m. This contradicts condition (ii), and hence we derive λ ≥ r. It follows from condition (iii) that there exist at least r distinct integers y α 1 , . . . , y αr ∈ {y 1 , . . . , y λ } such that a i,yα 1 + • • • + a i,yα r ≥ 1 for all i = 1, . . . , m. In view of condition (iv)(a), this implies that {y α 1 , . . . , y αr } = Γ s = N G [s] for some 1 ≤ s ≤ n. Thus, {y α 1 , . . . , y αr } ∩ D = ∅, and so {y 1 , . . . , y λ } ∩ D = ∅. This yields that (∪ m i=1 {f i }) ∩ D = ∅, which is a contradiction. Hence, one can conclude that if D is an arbitrary dominating set of G, then S i ⊆ D for some 1 ≤ i ≤ m. On the other hand, condition (i) gives that S i ⊈ S j for each 1 ≤ i, j ≤ m with i = j. Consequently, we derive that S 1 , . . . , S m are the minimal dominating sets of G, and so the proof is over.

The next example illustrates how we can employ Theorem 8.2.3. Example 8.2.4. Assume that V = {x 1 , . . . , x 6 } and the S i 's are a non-empty set of V as follows:

S 1 = {x 1 , x 2 }, S 2 = {x 1 , x 4 }, S 3 = {x 1 , x 6 }, S 4 = {x 2 , x 3 }, S 5 = {x 2 , x 5 }, S 6 = {x 3 , x 4 }, S 7 = {x 3 , x 6 }, S 8 = {x 4 , x 5 }, S 9 = {x 5 , x 6 }, S 10 = {x 1 , x 3 , x 5 }, S 11 = {x 2 , x 4 , x 6 }.
It is easy to investigate that condition (i) holds. In order to show condition (ii), we should check that for each r -1 = 3 distinct integers 1 ≤ j 1 , j 2 , j 3 ≤ 6, there exists a positive integer 1 ≤ i ≤ 11 such that a i,j 1 + a i,j 2 + a i,j 3 = 0. To do this, we have to examine it 6 3 = 20 times. This proves that M (S) satisfies condition (ii). Moreover, for verifying condition (iii), one must check that for each 4 ≤ ℓ ≤ 6 and distinct integers 1 ≤ j 1 , . . . , j ℓ ≤ 6 such that a i,j 1 +• • •+a i,j ℓ ≥ 1 for all i = 1, . . . , 11, then there exist at least four distinct integers j α 1 , j α 2 , j α 3 , j α 4 ∈ {j 1 , . . . , j ℓ } such that a i,jα 1 + a i,jα 2 + a i,jα 3 + a i,jα 4 ≥ 1 for all i = 1, . . . , 11. Hence, we have to check condition (iii) exactly 6 4 + 6 5 + 6 6 = 22 times. After checking them, we derive that condition (iii) holds. Finally, direct computations show that after checking Since conditions (i)-(iv) hold, according to Theorem 8.2.3, one can deduce that the closed neighborhood ideal of G is given by N I(G) = (x 1 x 2 x 4 x 6 , x 1 x 2 x 3 x 5 , x 2 x 3 x 4 x 6 , x 1 x 3 x 4 x 5 , x 2 x 4 x 5 x 6 , x 1 x 3 x 5 x 6 ). We can easily see that G is isomorphic to G 2 .

Since G 1 has no induced odd cycle, this implies that G 1 is bipartite, while G 2 has an induced odd cycle, and so is non-bipartite. This yields that G 1 and G 2 are non-isomorphic. . . , T ℓ . Put W i := W ∩ V (T i ) for all i = 1, . . . , ℓ. It follows that there exists a unique vertex w i ∈ V (T i ) such that {v, w i } ∈ E(T ) for all i = 1, . . . , ℓ. Hence, for every i = 1, . . . , ℓ, we may consider one of the following four cases: Case 1. deg T w i ≥ 3. We show that W i is a minimal path cover set of T i . Let r 1 , . . . , r s be a maximal path in T i . It is easy to see that L T i = L T ∩ V (T i ). Hence, r 1 , . . . , r s is a maximal path in T . It follows that {r 1 , . . . , r s } ∩ W i = ∅, and so W i is a path cover set of T i . Now, let V ⊊ W i be a minimal path cover set of T i . Put

V ′ := {v} ∪ ( ℓ j=1,j̸ =i W j ) ∪ V.
It is easy to see that V ′ is a path cover set of T and V ′ ⊊ W , which is the required contradiction. This shows that W i is a minimal path cover set of T i .

Case 2. deg T w i = 2 and T i is a path graph. By minimality of W , it follows that W i = ∅. Case 3. deg T w i = 2 and T i is not a path graph. Then T i is the union of a path P i and a tree T ′ i such that if v ′ i is the cut-point of this union, then deg T v ′ i ≥ 3. Hence, we can assume that V (P i ) = {v ′ i , v 1 , . . . , v t , w i }. By minimality of W , we have W ∩ (V (P i ) \ {v ′ i }) = ∅, and so

W i = W ∩ V (T ′ i ). Since L T ′ i = L T ∩ V (T i ) and deg T v ′ i ≥ 3,
. Since W = ( ℓ i=1 W i ) ∪ {v}, we get |W | = ℓ i=1 |W i | + 1. This implies that |W | = t i=1 |W i | + k i=t+1 |W i | + m i=k+1 |W i | + ℓ i=m+1 |W i | + 1 ≤ t i=1 d i -t + 0 + m i=k+1 d i -m + k + 0 + 1.

262

On the other hand, we have the following equalities

d = ℓ i=1 d i = t i=1 d i + k -t + m i=k+1 d i + ℓ -m.
This implies that |W | ≤ d -ℓ + 1. Since v / ∈ L T , one has ℓ ≥ 2. Therefore, |W | ≤ d -1, and the proof is complete.

Next, we are going to find a better bound for the index of covering of a tree T which is not a path graph. For this purpose, we require the following lemma. Lemma 9.1.9. Let T be a tree, P be a maximal path, and P ′ be a path in T . If

V (P ′ ) ⊈ V (P ) and V (P ) ∩ V (P ′ ) = ∅, then there exists v ∈ V (P ) ∩ V (P ′ ) with deg T v ≥ 3.
Proof. The proof is straightforward. Proposition 9.1.10. Suppose that T is a tree which is not a path graph. Also, let S T be the set of all vertices with degree greater than two in T . Then S T is a path cover set of T . In particular,

I T ≤ |S T | ≤ |L T | -2.
Proof. Let T be an arbitrary tree. It is easy to see that |S T | ≤ |L T | -2. So, it is enough to show that S T is a path cover set of T when T is not a path graph. Suppose, on the contrary, that there exists a maximal path P : v 1 , . . . , v s in T such that V (P ) ∩ S T = ∅. Hence, deg T v j = 2 for all j = 2, . . . , s -1. Choose an arbitrary vertex v j in V (P ). Since T is not a path graph, it follows that there is a vertex such as w in V (T ) such that deg T w ≥ 3, and so w / ∈ V (P ). By connectivity of T , one can conclude that there exists unique path P ′ in T between w and v j . Clearly, V (P ′ ) ∩ V (P ) = ∅. Due to V (P ′ ) ⊈ V (P ), by virtue of Lemma 9.1.9, there exists v ∈ V (P ) ∩ V (P ′ ) with deg T v ≥ 3, which is a contradiction. Therefore, S T is a path cover set of T , as claimed.

Max-path and path cover ideals

In this subsection, we turn to the study of monomial ideals arising from maximal paths of a graph G.

= (x i 1 • • • x it : i 1 , . . . , i t is a maximal path in G) ⊂ R = K[x 1 , . . . , x n ],
Let p = (x j 1 , . . . , x j ℓ ) be a minimal prime ideal of P I(G). We prove that {j 1 , . . . , j ℓ } is a minimal path cover set of G. To do this, let i 1 , . . . , i t be an arbitrary maximal path in G. Then x i 1 • • • x it ∈ P I(G). Since P I(G) ⊆ p, it follows that there exists an integer 1 ≤ r ≤ ℓ such that x jr |x i 1 • • • x it . Thus, x jr ∈ {x i 1 , . . . , x it }, and so j r ∈ {i 1 , . . . , i t }. This implies that {j 1 , . . . , j ℓ } ∩ {i 1 , . . . , i t } = ∅. Therefore, {j 1 , . . . , j ℓ } is a path cover set of G. Now, suppose, on the contrary, that {j 1 , . . . , j ℓ } is not a minimal path cover set. Hence, there exists a minimal path cover set W such that W ⊊ {j 1 , . . . , j ℓ }. It is easy to see that

P I(G) ⊆ (x ℓ : ℓ ∈ W ) ⊊ p,
which is a contradiction with the minimality of p. (iii) It follows that from part (i).

Proposition 9.1.13. Suppose that T is a tree which has exactly d leaves. Then we have (i) ht(P C(T )) = min{d(i, j) + 1 : i, j ∈ L T }, where by d(i, j) we mean the distance between two vertices i and j.

(ii) If i 1 , . . . , i r is a maximal path in T , then

x i 1 • • • x ir ∈ G(P I(T )).
(iii) If T is not a path graph, then ht(P I(T )) ≤ d -2.

(iv) ν(P I(T )) = d 2 , where by ν(I) we mean the minimal number of homogeneous generators of a graded ideal I.

(v) htp ≤ d -1 for all p ∈ Ass(P I(T )).

(vi) If L T = {i 1 , . . . , i d }, then, for all j = 1, . . . , d, we have

p j = (x i 1 , . . . , x i j , . . . , x i d ) ∈ Ass(P I(T )).
Proof. It is enough to apply Propositions 9.1.12, 9.1.6, 9.1.7, 9.1.10, and Theorem 9.1.8.

In the following corollary, we characterize trees in which the max-path ideals are unmixed (or Cohen-Macaulay). Proof. Suppose that T has the vertex set [n] and also has exactly d leaves. Parts (i) =⇒ (ii) and (ii) =⇒ (iii) are clear. Hence, the only non-trivial part is (iii) =⇒ (i).

To do this, assume that P I(T ) is an unmixed ideal. If d = 2, then T is a path graph, and so there is nothing to prove. Hence, assume that d ≥ 3, and thus T is not the path graph. By Proposition 9.1.13(iii), we have ht(P I(T )) ≤ d -2 and by Proposition 9.1.13(vi), P I(T ) has some associated primes of height d -1, which is a contradiction.

Next corollary is another application of Propositions 9.1.12 and 9.1.13. Corollary 9.1.15. Let T be a tree on the vertex set [n] and v ∈ V (T ). Then there exists a minimal path cover set of T which contains v.

Proof. If v ∈ V (T ), then it is clear that we can find at least one maximal path in T such as i 1 , . . . , i r that passes through v. So x v |x i 1 • • • x ir . Due to Proposition 9.1.13, it follows that x i 1 • • • x ir ∈ G(P I(T )). On the other hand, according to Proposition 9.1.12, one can conclude that P I(T ) = {j 1 ,...,j ℓ } is a minimal path cover set of T (x j 1 , . . . , x j ℓ ), is a minimal primary decomposition. Thus, there exists a minimal path cover set of T which contains v, as claimed. Let G be a graph on the vertex set [n] and t ≤ n is a positive integer. We recall that P t (G), the path ideal of length t, is the monomial ideal generated by all monomials

x i 1 • • • x i t+1 , where i 1 , . . . , i t+1 is a path of length t in G.
In a similar way, we define the ideals P I t (G) and P C t (G) in the following

P I t (G) = (x i 1 • • • x i t+1 : i 1 , . . . , i t+1 is a path of length t in G),
and

P C t (G) = (x j 1 • • • x j ℓ : {j 1 , . . . , j ℓ } is a (t + 1)-path vertex cover of G).
It is necessary to note that P I t (G) = P t (G) for all t ∈ N.

P I t (G) = {j 1 ,...,j ℓ } is a minimal (t+1)-path vertex cover of G (x j 1 , . . . , x j ℓ )
is the minimal primary decomposition of P I t (G).

(ii

) If i 1 , . . . , i t+1 is a path of length t, then x i 1 • • • x i t+1 ∈ G(P I t (G)).
(iii) P C t (G) = P I t (G) ∨ , where I ∨ means the Alexander dual ideal of I.

(iv) ht(P

I t (G)) = ψ t+1 (G). (v) ht(P C t (G)) = t + 1.
(vi) If G is a tree, then ht(P I t (G)) ≤ n/(t + 1).

Proof. (i) Let p = (x j 1 , . . . , x j ℓ ) be a minimal prime ideal of P I t (G). We prove that {j 1 , . . . , j ℓ } is a minimal (t + 1)-path vertex cover of G. For this purpose, assume that i 1 , . . . , i t+1 is an arbitrary path of order t + 1 in G. Then

x i 1 • • • x i t+1 ∈ P I t (G).
Since P I t (G) ⊆ p, this implies that there exists an integer 1 ≤ r ≤ ℓ such that The reverse inclusion is clear by the definition of P I t (G).

x jr |x i 1 • • • x i t+1 .
(ii) Suppose, on the contrary, that

x i 1 • • • x i t+1 ∈ P I t (G) but x i 1 • • • x i t+1 / ∈ G(P I t (G)).
Hence, there exists a path of length t, say j 1 , . . . , j t+1 , in G such that we have (iv) It follows from part (i).

x j 1 • • • x j t+1 |x i 1 • • • x i t+1 . Then {x j 1 , . . . , x j t+1 } ⊆ {x i 1 , . . . , x i t+1 }, and so {x j 1 , . . . , x j t+1 } = {x i 1 , . . . , x i t+1 }, this is a contradiction. ( iii
(v) Since P C t (G) = P I t (G) ∨ , it follows that

P C t (G) = i 1 ,...,i t+1 is a path of length t in G (x i 1 , . . . , x i t+1 ).
This implies that ht(P C t (G)) = t + 1.

(vi) According to [20,Theorem 2], it follows that ψ t+1 (G) ≤ |V (G)| t+1 . Now, by part (iv), one can conclude that ht(P I t (G)) ≤ n/(t + 1).

On account of the following equalities

d(j 1 , j t+1 ) = d(j 1 , i 1 ) = d(j 1 , i t+1 ) = d(j t+1 , i 1 ) = d(j t+1 , i t+1 ) = d(i 1 , i t+1 ),
one can conclude that d(u, v) = 0, which is a contradiction. This implies that T has at most one vertex of degree greater than 2 and because T is not a path graph, it has exactly one vertex of degree greater than 2. This gives that T is starlike.

(iv) =⇒ (i). The proof is based on the definition of path ideals.

Next theorem is the basic step to characterize trees with the maximal path ideal of height 1. Proof. (i) =⇒ (ii). Let v ∈ V (T ) be a vertex of degree d and N T (v) = {v 1 , . . . , v d }. Then any vertex v i is connected to a tree such as T i for all i = 1, . . . , d. We show that T i is a path graph for all i = 1, . . . , d. Suppose, on the contrary, that there exists a positive integer 1 ≤ i ≤ d such that T i is not a path graph. Hence, the number of leaves of T exceed d, which is a contradiction. This implies that {v} is a minimal path cover set, and so I T = 1. By Proposition 9.1.12(iii), one can conclude that ht(P I(T )) = 1, as claimed.

(ii) =⇒ (iii). Assume that ht(P I(T )) = 1. Then by Proposition 9.1.12(iii), there exists a minimal path cover set such as W = {v} with v ∈ V (T ). We first prove that deg T v ≥ 3. To do this, let i, j, k ∈ L T , P : i = i 1 , i 2 , . . . , i r = j be the maximal path between i and j, and P ′ : i = j 1 , j 2 , . . . , j s = k be the maximal path between i and k. Then by Lemma 9.1.9, put ℓ = max{m : i m ∈ V (P )∩V (P ′ ) and deg

T i m ≥ 3}. It is clear that V (P ) ∩ V (P ′ ) = {i 1 , . . . , i ℓ }. So, v ∈ {i 1 , . . . , i ℓ }.
Suppose that P ′′ is the maximal path between j and k. It is easy to check that V (P ′′ ) ∩ {i 1 , . . . , i ℓ } = {i ℓ }. This implies that v = i ℓ , and hence deg T v ≥ 3. Now, assume that u ∈ V (T ) with u = v and deg T u ≥ 3. Let P ′′′ be the path between u and v, u ′ , u ′′ ∈ N T (u) \ V (P ′′′ ), v ′ , v ′′ ∈ N T (v) \ V (P ′′′ ). Then, similar to the proof of Corollary 9.1.18 (iii) =⇒ (iv), if we extend the path u ′ , u, u ′′ to a maximal path Q, it follows that V (Q) ∩ {v} = ∅, which is a contradiction. This implies that v is the only vertex of degree greater than 2, and thus T is starlike.

(iii) =⇒ (iv). Suppose that T is starlike. Then there exists a vertex v ∈ V (T ) such that deg T v = d. If d = 3, then htp = 2 and the proof is complete. We therefore suppose that d ≥ 4. Let P 1 , . . . , P d be the path graphs which are connected to the vertex v. Assume that p ∈ Ass(P I(T )) and htp > 1. By Proposition 9.1.13(v), we have htp ≤ d -1. Suppose that htp ≤ d -2, and seek a contradiction. Then, by Proposition 9.1.12(i), it follows that p = (x j 1 , . . . , x js ) such that W = {j 1 , . . . , j s } is a minimal path cover set with 2 ≤ s ≤ d -2. By minimality of W , it follows that

|W ∩ V (P i )| ≤ 1 for all i = 1, . . . , d. Since htp > 1, we obtain v / ∈ W . Without loss of generality, suppose that j k ∈ V (P k ) for k = 1, . . . , s. This implies that W ∩ ({v} ∪ V (P s+1 ) ∪ V (P s+2 )) = ∅, a contradiction.
(iv) =⇒ (i). If there exists a minimal path cover set such as W such that |W | = 1, then ht(P I(T )) = 1, and so T is starlike. Since |L T | = d, we get T has a vertex of degree d and the proof is complete. Suppose that, for each minimal path cover set W , we have |W | > 1. Then, according to our hypothesis, we conclude that |W | = d -1 for all minimal path cover set of T . So, ht(P I(T )) = d -1 which is a contradiction with Proposition 9.1.13(iii).

In the following corollary, we characterize trees in which ht(P I(T )) = 1. Here, we characterize trees in which the max-path ideals have height 2. (i) E is a path graph between the centers of S 1 and S 2 .

(ii) E is the union of two path graphs P and P ′ such that

|V (P ) ∩ V (S i )| = 1 for i = 1, 2, and V (P ) ∩ V (P ′ ) = {v} with deg T v = 3.
Proof. First, suppose that T is as described in the theorem and z i is the center of the starlike tree S i for i = 1, 2. For sufficiency, the set W = {z 1 , z 2 } is a minimal path cover set, and so ht(P I(T )) = 2.

For necessity, assume that ht(P I(T )) = 2. Then by Proposition 9.1.13(iii), we have d ≥ 4 and also, there exists a minimal path cover set such as W = {z 1 , z 2 }. Let ℓ 1 := deg T z 1 and ℓ 2 := deg T z 2 . If ℓ 1 ≤ 2 and ℓ 2 ≤ 2, then by considering all possible shapes of maximal paths we get a contradiction. Hence, without loss of generality, assume that ℓ 1 ≥ 3.

Then T \ {z 1 } has ℓ 1 connected components such as T 1 , . . . , T ℓ 1 and for each ) and E satisfies one of the conditions (i) or (ii).

1 ≤ i ≤ ℓ 1 ,
In the subsequent corollary, we find ht(P I(T )) for a class of trees which can be considered as a generalization of the class of trees introduced in Theorem 9.1.21. Corollary 9.1.22. Let T be a tree on the vertex set [n]. Suppose also that

T = S 1 ∪ E 1 ∪ S 2 ∪ E 2 ∪ • • • ∪ E m-1 ∪ S m ,
such that S 1 , . . . , S m are starlike trees and E i , for all i = 1, . . . , m -1, satisfies one of the following conditions. (i) E i is a path graph between the centers of S i and S i+1 .

(ii) E i is the union of two path graphs P i and

P ′ i such that |V (P i ) ∩ V (S k )| = 1 for k = i, i + 1, and V (P i ) ∩ V (P ′ i ) = {v i } with deg T v i = 3.
Then ht(P I(T )) = m.

Proof. Suppose that z i is the center of the starlike tree S i for all i = 1, . . . , m. Then the set W = {z 1 , . . . , z m } is a minimal path cover set of T , and so ht(P I(T )) ≤ m. Also, note that we can find some maximal paths in

T like Q 1 , . . . , Q m such that V (Q 1 ) ⊆ V (S 1 ), V (Q 2 ) ⊆ V (E 1 ) ∪ V (S 2 ), . . . , V (Q m ) ⊆ V (E m-1 ) ∪ V (S m ).
Since each of the Q i s is a maximal path, the V (Q i )s (1 ≤ i ≤ m) are pairwise disjoint. Let W ′ be an arbitrary path cover set of T . Then for each

1 ≤ i ≤ m, W ′ ∩ V (Q i ) = ∅. This shows that |W ′ | ≥ m. Therefore, ht(P I(T )) = m.
It is necessary to note that the converse of Corollary 9.1.22 is not valid. To observe this, consider the following tree T on a vertex set with thirteen elements.

It is routine to check that ht(P I(T )) = 4, while we cannot write T based on the format which has been stated in Corollary 9.1.22.

Path cover sets among some graphs

In this section, we are going to discuss the path cover sets among graphs. We begin with the following theorem. In fact, the next theorem links an arbitrary minimal path cover set of the union of a cycle and a connected graph with the minimal path cover set of the connected graph. It should be noted that the result of this section can be found in [START_REF] Khashyarmanesh | On the minimal path cover sets of union of certain graphs[END_REF].

Recall that a separating set in a connected graph G is a set of vertices whose deletion disconnects G. If a separating set contains only one vertex v, we call v a cut-point.

(i) W G is a minimal path cover set of H. (ii) W G \ {v ′ } is a minimal path cover set of H. (iii) (W G ∪ {v}) \ {v ′ } is a minimal path cover set of H.
Proof. If W G is a minimal path cover set of H, then (i) holds. So, suppose that W G is not a minimal path cover set of H. Thus, one can consider the following two cases: Case 1. W G is a path cover set of H, but it is not minimal. It follows that there exists an element such as w in W G such that W G \ {w} is a path cover set of H. We show that w = v ′ . Suppose, on the contrary, that w = v ′ . We prove that W G \ {w} is a path cover set of G. Suppose that there exists a maximal path such as P in G such that V (P ) ∩ (W G \ {w}) = ∅, and seek a contradiction. Hence,

V (P ) ∩ W G ⊆ {w}. Since V (P ) ∩ W G = ∅, we have V (P ) ∩ W G = {w}. If w ∈ V (C), then w ∈ V (C) ∩ W G ,
and so w = v ′ , which is a contradiction. This implies that w / ∈ V (C), and hence w = v. If v / ∈ V (P ), then P is a maximal path in H, and so

V (P ) ∩ (W G \ {w}) = ∅, which is a contradiction. Therefore, v ∈ V (P ). If v ∈ W G , then v ∈ V (P ) ∩ W G , and hence v = w, which is a contradiction. This implies that v / ∈ W G , and so v = v ′ . If v ′ ∈ V (P ), then v ′ ∈ V (P ) ∩ W G ,
and hence v ′ = w, which is a contradiction. So v ′ / ∈ V (P ), and one has V (P ) ∩ V (C) = {v}. Thus, P is a maximal path in H. It follows that V (P ) ∩ (W G \ {w}) = ∅, which is a contradiction. This implies that w = v ′ , and so W G \ {v ′ } is a minimal path cover set of H. Therefore, (ii) holds.

Case 2. W G is not a path cover set of H. Set W ′ := W G ∪ {v}. We first show that W ′ is a path cover set of H. Suppose, on the contrary, that there exists a maximal path such as P in H such that V (P ) ∩ W ′ = ∅. Thus, v / ∈ V (P ) and

V (P ) ∩ W G = ∅. Since v / ∈ V (P ), one has V (P ) ∩ V (C) = ∅, and so P is a maximal path in G. Accordingly, V (P ) ∩ W G = ∅, which is a contradiction. It follows that W ′ is a path cover set of H. If v ∈ W G , then W ′ = W G is a path cover set of H,
which is a contradiction. This implies that v / ∈ W G , and thus v = v ′ . We next prove that W ′ \ {v ′ } is a path cover set of H. Suppose, on the contrary, that there exists a maximal path such as P in H such that

V (P )∩(W ′ \{v ′ }) = ∅. Therefore, v / ∈ V (P ) and V (P ) ∩ W G ⊆ {v ′ }. Due to v / ∈ V (P ), one has V (P ) ∩ V (C) = ∅, and so P is a maximal path in G. It follows that V (P ) ∩ W G = ∅, and hence V (P ) ∩ W G = {v ′ }.
This implies that v ′ ∈ V (P ), and so V (P ) ∩ V (C) = ∅, which is a contradiction. Therefore, W ′ \ {v ′ } is a path cover set of H. Since W G is not a path cover set of H, it follows that there exists a maximal path such as

Q in H such that V (Q)∩W G = ∅. Due to W ′ \ {v ′ } is a path cover set of H, one has V (Q) ∩ (W ′ \ {v ′ }) = ∅. Hence, v ∈ V (Q) and v ′ / ∈ V (Q). If deg H v ≥ 2, then Q is a maximal path in G, and so V (Q) ∩ W G = ∅, which is a contradiction. This implies that deg H v = 1. If W ′ \ {v ′ }
is a minimal path cover set of H, then the proof is complete. So, we assume that W ′ \ {v ′ } is not minimal. It follows that there exists an element such as

w in W ′ \ {v ′ } such that W ′ \ {v ′ , w} is a path cover set of H with w = v ′ . If w = v, then W ′ \ {v ′ , w} = W G \ {v ′ }
is a path cover set of H, which is a contradiction. This implies that w = v, and so w ∈ W G . In the following, we verify that W G \ {w} is a path cover set of G. Suppose, on the contrary, that there exists a maximal path such as P in G such that

V (P ) ∩ (W G \ {w}) = ∅. Hence, V (P ) ∩ W G = {w}. If v /
∈ V (P ), then P is a maximal path in H, and thus V (P )∩(W ′ \{v ′ , w}) = ∅. Since V (P ) ∩ W ′ = {w}, we get V (P ) ∩ (W ′ \ {v ′ , w}) = ∅, which is a contradiction. This implies that v ∈ V (P ). By virtue of v ′ / ∈ V (P ) and deg H v = 1, one can conclude that P is not a maximal path in G, which is a contradiction. So W G \ {w} is a path cover set of G, which is a contradiction. Accordingly, W ′ \ {v ′ } is a minimal path cover set of H. Therefore, (iii) holds, and the proof is complete.

(ii) W G \ {t} is a minimal path cover set of H. (iii) (W G \ {t}) ∪ {v} is a minimal path cover set of H. Proof. If v = w, then G = H. So, W G
is a minimal path cover set of H, and hence (i) holds. Thus, we may assume that v = w. If W G is a minimal path cover set of H, then (i) holds. Therefore, assume that W G is not a minimal path cover set of H. Hence, we can consider the following two cases: Case 1. W G is a path cover set of H, but it is not minimal. This implies that there exists an element z in W G such that W G \ {z} is a path cover set of H. In the following, our goal is to prove that z = t. Suppose, on the contrary, that z = t. We show that W G \ {z} is a path cover set of G. Suppose, on the contrary, that there exists a maximal path such as P in G such that V (P ) ∩ (W G \ {z}) = ∅. Thus,

V (P ) ∩ W G ⊆ {z}. By V (P ) ∩ W G = ∅, one has V (P ) ∩ W G = {z}. If v /
∈ V (P ), then P is a maximal path in H, and we obtain

V (P ) ∩ (W G \ {z}) = ∅, which is a contradiction. It follows that v ∈ V (P ). Let deg H v = 1.
Then one can assume that

P : i 1 , . . . , i k , v, r 1 , . . . , r s , w with {i 1 , . . . , i k } ⊆ V (H). Since deg H v = 1, it follows that Q : i 1 , . . . , i k , v is a maximal path in H. Due to V (P ) ∩ (W G \ {z}) = ∅ and V (Q) ⊆ V (P ), we get V (Q) ∩ (W G \ {z}) = ∅, which is a contradiction. Therefore, deg H v ≥ 2. Since z = t and z ∈ W G , one has z / ∈ V (L). If w / ∈ V (P ), by virtue of deg H v ≥ 2, we have P is a maximal path in H. Thus, V (P ) ∩ (W G \ {z}) = ∅,
which is a contradiction. Therefore, w ∈ V (P ). According to z / ∈ V (L), one can conclude that z = v and z = w. This yields that w / ∈ W G and v / ∈ W G . Since t ∈ V (L), v ∈ V (P ), and w ∈ V (P ), we have t ∈ V (P ). According to t ∈ W G , one has t ∈ V (P ) ∩ W G . This implies that t = z. This leads to a contradiction. Therefore, W G \ {z} is a path cover set of G, which is a contradiction. Accordingly, one has z = t, and hence W G \ {t} is a minimal path cover set of H. Therefore, (ii) holds.

Case 2. W G is not a path cover set of H. Put W ′ := W G ∪ {v}. We first show that W ′ is a path cover set of H. Suppose, on the contrary, that there exists a maximal path such as P in H such that V

(P ) ∩ W ′ = ∅. Hence, v / ∈ V (P ) and V (P ) ∩ W G = ∅. Since v / ∈ V (P ), it follows that P is a maximal path in G. Accordingly, V (P )∩W G = ∅, which is a contradiction. Therefore, W ′ is a path cover set of H. If v ∈ W G , then W ′ = W G is a path cover set of H, which is a contradiction. This implies that v / ∈ W G .
In the following, our aim is to prove that W ′ \ {t} is a path cover set of H. Suppose, on the contrary, that there exists a maximal path such as P in H such that V

(P ) ∩ (W ′ \ {t}) = ∅. Since v / ∈ W G , we have v = t. It follows that v / ∈ V (P ) and V (P ) ∩ W G ⊆ {t}. Due to v / ∈ V (P )
, one has P is a maximal path in G, and thus V (P ) ∩ W G = ∅. This implies that V (P ) ∩ W G = {t}, and so t ∈ V (P ). Since P is a maximal path in G and t ∈ V (L), we get v ∈ V (P ), which is a contradiction. Therefore, W ′ \ {t} is a path cover set of H. If W ′ \ {t} is a minimal path cover set of H, then the proof is complete. So, suppose that W ′ \ {t} is not minimal. This implies that there exists an element like z in W ′ \ {t} such that

W ′ \ {z, t} is a path cover set of H. If z = v, then W ′ \ {z, t} = W G \ {t} is a path cover set of H, which is a contradiction. It follows that z = v. Since z ∈ W ′ \ {t} and z = v, one has z ∈ W G . According to W G is not a path cover set of H, one can conclude that there exists a maximal path like Q in H such that V (Q) ∩ W G = ∅. By W ′ \ {t} is a path cover set of H, it follows that V (Q) ∩ (W ′ \ {t}) = ∅. Hence, v ∈ V (Q). If deg H v ≥ 2, then Q is a maximal path in G, and so V (Q) ∩ W G = ∅,
which is a contradiction. This implies that deg H v = 1. We now verify that W G \ {z} is a path cover set of G. Suppose, on the contrary, that there exists a maximal path such as

Γ in G such that V (Γ) ∩ (W G \ {z}) = ∅. Due to V (Γ) ∩ W G = ∅, one has V (Γ) ∩ W G = {z}. If v / ∈ V (Γ), then Γ is a maximal path in H, and so V (Γ) ∩ (W ′ \ {z, t}) = ∅. Since V (Γ) ∩ W ′ = {z}, we have V (Γ) ∩ (W ′ \ {z, t}) = ∅, which is a contradiction. This yields that v ∈ V (Γ). Since deg H v = 1, one can conclude that V (L) ⊆ V (Γ). According to t ∈ V (L), one has t ∈ V (Γ), and hence t ∈ V (Γ) ∩ W G . Since z = t, this leads to a contradiction. Therefore, W G \ {z} is a path cover set of G, which is a contradiction. Thus, W ′ \ {t} is a minimal path cover set of H. According to v = t, one can conclude that W ′ \ {t} = (W G \ {t}) ∪ {v}.
Therefore, (iii) holds, and the proof is complete.

It is necessary to observe that Theorem 9.2.4 may be false if we omit the assumption that V (L) ∩ W G must be non-empty. As a counterexample, consider the graph below, where V (G) := [8], V (H) := [5], V (L) := {3, 6, 7, 8}, and v := 3 is the cut-point. It is easy to see that W G := {2, 4} is a minimal path cover set of G, but neither W G nor W G ∪ {3} is a minimal path cover set of H.

(i) W G is a minimal path cover set of H. (ii) W G \ S is a minimal path cover set of H. Since V (T ) = V (L) ∪ V (T ′ ), one can easily check that S ′ ∪ S 1 = S, S ′ ∪ S 2 = S, and S ′ ∪ S 3 = S ∪ {z}. If (1) holds, then we get (i). Let (2) hold. It follows that W G \ S 1 = (W G \ S) ∪ ((W G \ S 1 ) ∩ (S ′ \ S 1 )
), and so one has

W G \ S 1 = (W G \ S) ∪ (S ′ \ S 1 ). Since W G \ S 1 is a minimal path cover set of H, we can deduce that W G \ S 1 ⊆ V (H), and hence S ′ \ S 1 ⊆ V (H). By S ′ \ S 1 ⊆ V (L) and V (H) ∩ V (L) ⊆ {v}, one can conclude that S ′ \ S 1 ⊆ {v}. It is easy to see that S ′ \ S 1 = ∅. This implies that W G \ S 1 = W G \ S,
and so we obtain (ii).

In the statement (3), by using a similar argument which has been stated in statement (2), we have W G \ S 1 = (W G \ S) ∪ (S ′ \ S 1 ), and thus

(W G \ S 1 ) ∪ {v} = (W G \ S) ∪ (S ′ \ S 1 ) ∪ {v}.
By virtue of (W G \ S 1 ) ∪ {v} is a minimal path cover set of H, one can conclude that (W G \ S 1 ) ∪ {v} ⊆ V (H), and so

S ′ \ S 1 ⊆ V (H). Since S ′ \ S 1 ⊆ V (L) and V (H) ∩ V (L) ⊆ {v}, it follows that S ′ \ S 1 ⊆ {v}. If S ′ \ S 1 = {v}, then v ∈ S ′ and v / ∈ S 1 . Since v ∈ S ′ , it follows that v ∈ W G , and so v ∈ S 1 , which is a contradiction. Accordingly, S ′ \ S 1 = ∅. Hence, (W G \ S 1 ) ∪ {v} = (W G \ S)
∪ {v}, and so we have (iii).

Let (4) hold. One can easily check that

W G \ S ′ = (W G \ S) ∪ (S 1 \ S ′ ). Since W G \ S ′ is a minimal path cover set of H, one can conclude that W G \ S ′ ⊆ V (H), and hence S 1 \ S ′ ⊆ V (H). According to S 1 \ S ′ ⊆ V (T ′ ) and V (H) ∩ V (T ′ ) = {v}, it follows that S 1 \ S ′ ⊆ {v}. If S 1 \ S ′ = ∅, then one has (ii), but if S 1 \ S ′ = {v},
then we get (iii). Statement (5) implies that (W G \ S ′ ) \ S 2 = W G \ S, and so we have (ii). It follows from ( 6) that ((W G \ S ′ ) \ S 2 ) ∪ {v} = (W G \ S) ∪ {v}, and hence we obtain (iii).

In the statement (7), according to the argument has been stated in (4), one has

W G \S ′ = (W G \S)∪(S 1 \S ′ ), and so (W G \S ′ )∪{z} = (W G \S)∪(S 1 \S ′ )∪{z}. Since (W G \S ′ )∪{z} is a minimal path cover set of H, we have (W G \S ′ )∪{z} ⊆ V (H), and hence z ∈ V (H). This implies that z = v. Due to S 1 \ S ′ ⊆ V (H), S 1 \ S ′ ⊆ V (T ′ ), and
V (H) ∩ V (T ′ ) = {v}, one has S 1 \ S ′ ⊆ {v}. Thus, we can conclude that (W G \ S ′ ) ∪ {z} = (W G \ S) ∪ {v}.
Therefore, we get (iii). Let the statement (8) hold. Since S ′ ∪ S 3 = S ∪ {z}, one can conclude that

((W G \ S ′ ) ∪ {z}) \ S 3 = (W G \ (S ∪ {z})) ∪ ({z} \ S 3 ).
Since z ∈ S 3 , it follows that {z}\S 3 = ∅, and so

((W G \S ′ )∪{z})\S 3 = W G \(S∪{z}). If z ∈ S, then W G \ (S ∪ {z}) = W G \ S. If z / ∈ S, then z / ∈ W G , and hence W G \ (S ∪ {z}) = W G \ S. This implies that ((W G \ S ′ ) ∪ {z}) \ S 3 = W G \ S.
Accordingly, we have (ii).

It follows from ( 5) that (W G \ S ′ ) \ S 2 = W G \ S, and so we have (ii). Statement (6) implies that ((W G \ S ′ ) \ S 2 ) ∪ {v} = (W G \ S) ∪ {v}, and thus we obtain (iii).

In the statement (7), by virtue of the argument has been stated in (4), we have

W G \ S ′ = (W G \ S) ∪ (S 1 \ S ′ ), and hence (W G \ S ′ ) ∪ {v} = (W G \ S) ∪ (S 1 \ S ′ ) ∪ {v}. Since (W G \S ′ )∪{v} is a minimal path cover set of H, it follows that (W G \S ′ )∪{v} ⊆ V (H), and so S 1 \ S ′ ⊆ V (H). Due to S 1 \ S ′ ⊆ V (T ) and V (H) ∩ V (T ) = {v}, one has S 1 \ S ′ ⊆ {v}. If S 1 \ S ′ = {v}, then v ∈ S 1 and v / ∈ S ′ . Since v ∈ S 1 , it follows that v ∈ W G ,
and so v ∈ S ′ , which is a contradiction. Thus, S 1 \ S ′ = ∅, and hence

(W G \ S ′ ) ∪ {v} = (W G \ S) ∪ {v}. Therefore, we get (iii).
Let statement (8) hold. Since S ′ ∪ S 3 = S ∪ {v}, one can conclude that

((W G \ S ′ ) ∪ {v}) \ S 3 = (W G \ (S ∪ {v})) ∪ ({v} \ S 3 ).
Since v ∈ S 3 , it follows that {v}\S 3 = ∅, and so

((W G \S ′ )∪{v})\S 3 = W G \(S∪{v}). If v ∈ S, then W G \ (S ∪ {v}) = W G \ S. If v / ∈ S, then v / ∈ W G , and hence W G \ (S ∪ {v}) = W G \ S. This implies that ((W G \ S ′ ) ∪ {v}) \ S 3 = W G \ S.
Therefore, one has (ii).

In the statement (9), based on {v} \ S 3 = ∅, one can conclude that

(((W G \ S ′ ) ∪ {v}) \ S 3 ) ∪ {v} = (W G \ (S ∪ {v})) ∪ {v}. If v / ∈ S, then v / ∈ W G , and thus (((W G \ S ′ ) ∪ {v}) \ S 3 ) ∪ {v} = (W G \ S) ∪ {v}. If v ∈ S, then v ∈ W G . Accordingly, (((W G \ S ′ ) ∪ {v}) \ S 3 ) ∪ {v} = (W G \ S) ∪ {v}.
So, we have (iii). This is enough to complete the inductive step.

The following proposition says that the size of an arbitrary minimal path cover set of the union of two cycles is at most two. Proposition 9.2.7. Suppose that G is the union of two cycles C and C ′ such that

V (C) ∩ V (C ′ ) is a path and W G is an arbitrary minimal path cover set of G. Then either |W G | = 1 or |W G | = 2. Proof. Let Q be the path such that V (Q) = V (C) ∩ V (C ′ )
with starting point v and ending point v ′ . Now, pick an arbitrary maximal path P in G. By maximality of P , one can consider the following four cases:

Case 1. V (P ) = V (C). Case 2. V (P ) = V (C ′ ). Case 3. V (P ) = V (G). Case 4. V (P ) = (V (G) \ V (Q)) ∪ {v, v ′ }.
It is straightforward to check that W G may be one of the following four types:

(A) W G = {w, w ′ }, where w ∈ V (C) \ V (Q) and w ′ ∈ V (C ′ ) \ V (Q). (B) W G = {w, w ′ }, where w ∈ V (C) \ V (Q) and w ′ ∈ V (Q) \ {v, v ′ }. (C) W G = {w, w ′ }, where w ∈ V (C ′ ) \ V (Q) and w ′ ∈ V (Q) \ {v, v ′ }. (D) W G = {w}, where w ∈ {v, v ′ }.

It follows that either |W

G | = 1 or |W G | = 2, as claimed.

Max-path ideals with certain heights

One of the problems arising in this context is to explore the graphs which have a minimal path cover set with size one. As mentioned in Example 9.1.2, if G is a cycle or a complete graph or a path graph, then for each v ∈ V (G), the set {v} is a minimal path cover set of G. Nevertheless, there exist some graphs such that none of their vertices is a minimal path cover set, even for Hamiltonian graphs. For instance, consider the following Hamiltonian graph. Since P 1 : 1, 2, 7, 8 and P 2 : 4, 3, 6, 5 are two maximal paths, we cannot find a singleton minimal path cover. In the following, our aim is to introduce a large class of finite simple graphs which have a minimal path cover set with size one. To achieve this, we need the following two propositions.

i=1 V (G i ) be a path. If v ∈ m i=1 V (G i ), then {v} is a minimal path cover set of the union of G 1 , . . . , G m .
Proof. Let v be an arbitrary element of m i=1 V (G i ). Suppose, on the contrary, that there exists a maximal path such as

P : v 1 , . . . , v k in the union of G 1 , . . . , G m such that v / ∈ V (P ). Since v k ∈ m i=1 V (G i ), there exists a positive integer j with 1 ≤ j ≤ m such that v k ∈ G j . Due to G j is a complete graph and v ∈ G j , one can conclude that {v, v k } ∈ E(G j ).
This implies that P is not a maximal path in the union of G 1 , . . . , G m , which is a contradiction. Proposition 9.2.9. Let G 1 , . . . , G m be some subgraphs of a finite simple graph G, such that m i=1 V (G i ) = {v} and the set {v} is a minimal path cover set of the graphs G 1 , . . . , G m . Then {v} is a minimal path cover set of the union of G 1 , . . . , G m .

Proof. Let P be an arbitrary maximal path in the union of G 1 , . . . , G m . If there exists a positive integer j with 1 ≤ j ≤ m such that V (P ) ⊆ V (G j ), then P is a maximal path in the graph G j , and so v ∈ V (P ). So, assume that there are two positive integers r and s with 1 ≤ r, s ≤ m and r = s such that V

(P ) ∩ V (G r ) = ∅ and V (P ) ∩ V (G s ) = ∅. Suppose that w ∈ V (P ) ∩ V (G r ) and w ′ ∈ V (P ) ∩ V (G s ).
Let Q : w, j 1 , . . . , j k , w ′ be the path between w and w ′ with V (Q) ⊆ V (P ). By virtue of m i=1 V (G i ) = {v}, one has v ∈ V (Q), and so v ∈ V (P ). This completes the proof.

The conclusion of the above proposition does not hold true if m i=1 V (G i ) is a path and v ∈ m i=1 V (G i ) such that the set {v} is a minimal path cover set of graphs G 1 , . . . , G m . As a counterexample, consider graphs below. One can see that

V (G 1 ) ∩ V (G 2 ) is a path. 1 2 3 4 5 6 7 8 3 4 5 7 8 1 2 3 6 7 G 1 G 2 G 1 ∪ G 2
It is straightforward to check that W := {3} is a minimal path cover set of both graphs G 1 and G 2 . Now, consider the maximal path P : 5, 4, 8, 7, 6, 2, 1. Due to W ∩ V (P ) = ∅, one can conclude that W is not a minimal path cover set of the union of G 1 and G 2 .

We next state the definition of tadpole graphs. (3,4)-tadpole graph

The following proposition can be easily proved based on the definition of minimal path cover sets.

Proposition 9.2.11. Let G be a cycle graph and let

T = C 3 ∪ P n be a (3, n)-tadpole graph for some natural number n such that E(G) ∩ E(C 3 ) = {u, v} with deg T u = 2 and deg T v = 2. Then W is a minimal path cover set of G if and only if W is a minimal path cover set of G ∪ T .
It is important to note that Proposition 9.2.11 is not valid if we drop the assumption that the graph G is cycle. To see this, one just has to check that W = {5} is a minimal path cover set of G, as shown below, but it is not a minimal path cover set of the union of G and the (3, 3)-tadpole graph T , by virtue of the maximal path 9,8,7,4,3,2,1. G i is the union of all the G i and there is a vertex v ∈ s i=1 V (G i ) such that the vertex v is the cut-point of this union and the set {v} is a minimal path cover set of G i for all i = 1, . . . , s. Then ht(P I(G)) = 1.

Proof. We can combine together Propositions 9.2.8, 9.2.9, and 9.2.11 to obtain the set {v} is a minimal path cover set of G, and so ht(P I(G)) = 1. This finishes the proof.

Along the argument has been stated in the proof of Theorem 9.1.19, we provide the proposition below. Proposition 9.2.13. Suppose that T is a tree which is not a path graph. Also, suppose that ht(P I(T )) = k. Then T has at least k vertices with degree greater than or equal to 3.

Proof. Let S T be the set of all vertices with degree greater than 2 in T . Due to Proposition 9.1.10, one has I T ≤ |S T | ≤ |L T | -2, where I T is the index of covering of T . On the other hand, it follows from Proposition 9.1.12 that ht(P I(T )) = I T . This implies that |S T | ≥ k, and the proof is complete.

Assume that T is a tree which is not a path graph, V (T ) = [n], L T = {r 1 , . . . , r d }, and ht(P I(T )) = k. We thus can conclude that there exists a minimal path cover set of T with size k, say {j 1 , . . . , j k } ⊆ [n]. Put p = (x j 1 , . . . , x j k ). Therefore, p is a minimal prime ideal of P I(T ) such that htp = k. Now, consider W ⊆ L T with |W | = d -1, and set p = (x j : j ∈ W ). Due to Proposition 9.1.7, one has p is a minimal prime ideal of P I(T ) such that htp = d -1.

Another question that can be asked in this context is that does there exist a monomial prime ideal p ∈ Ass(P I(T )) with htp = r for all r = k + 1, . . . , d -2? We provide an example that refutes this assertion. To do this, we consider a tree T on the vertex set V (T ) := [8] with the following edge set

E(T ) := {{1, 2}, {2, 3}, {2, 4}, {4, 5}, {4, 6}, {4, 7}, {4, 8}}.
It is easy to see that P I(T ) = (x 1 x 2 x 4 x 5 , x 1 x 2 x 4 x 6 , x 1 x 2 x 4 x 7 , x 1 x 2 x 4 x 8 , x 3 x 2 x 4 x 5 , x 3 x 2 x 4 x 6 , x 4 x 7 x 8 , x 3 x 2 x 4 x 7 , x 3 x 2 x 4 x 8 , x 1 x 2 x 3 , x 4 x 5 x 6 , x 4 x 5 x 7 , x 4 x 5 x 8 , x 4 x 6 x 7 , x 4 x 6 x 8 ).

Hence, one can deduce that Ass(P I(T )) is given by {(x 4 , x 2 ), (x 4 , x 1 ), (x 8 , x 6 , x 5 , x 2 ), (x 8 , x 7 , x 5 , x 2 ), (x 8 , x 6 , x 5 , x 3 , x 1 ), (x 8 , x 7 , x 6 , x 2 ), (x 7 , x 6 , x 5 , x 2 ), (x 7 , x 6 , x 5 , x 3 , x 1 ), (x 8 , x 7 , x 5 , x 3 , x 1 ), (x 8 , x 7 , x 6 , x 5 , x 3 ), (x 8 , x 7 , x 6 , x 3 , x 1 ), (x 8 , x 7 , x 6 , x 5 , x 1 ), (x 4 , x 3 )}.

This implies that there is no prime monomial ideal p in Ass(P I(T )) with htp = 3.

In the following, our goal is to find the whole of the minimal prime ideals of P I(T ) such as p 1 , . . . , p k such that I T = htp i for all i = 1, . . . , k, where I T is the index of covering of tree T .

Proposition 9.2.14. Let T be a tree which is not a path graph. Then T has a vertex u ∈ V (T ) with deg T u ≥ 3, and w 1 , w 2 ∈ L T with w 1 = w 2 , and

v 1 , . . . , v r , v r+1 , . . . , v s ∈ V (T ),
(possibly s = 0) with deg T v i = 2 for each i = 1, . . . , s such that

Q : w 1 , v 1 , . . . , v r , u, v r+1 , . . . , v s , w 2 ,
is a maximal path in T .

Proof. Since the tree T is not a path graph, one can conclude that |V (T )| ≥ 4.

We argue by induction on n := |V (T )|. If n = 4, then T is the claw graph. The claim is true in this case. Now, suppose, inductively, that n > 4 and that the result has been proved for any tree which is not a path graph and the size of its set of vertices is less than n. Let T be a tree which is not a path graph with

n = |V (T )| > 4.
If T has no vertex with degree two, then T is the union of some star graphs such as S 1 , . . . , S t such that, for each

i = 1, . . . , t -1, E(S i ) ∩ E(S i+1
) is an edge. Therefore, one can easily find a maximal path Q : w 1 , u, w 2 in T such that w 1 , w 2 ∈ L T with w 1 = w 2 and deg T u ≥ 3. Thus, we assume that there exists a vertex in T with degree two. Here, we claim that there exists a leaf such as w 1 in T such that N T (w 1 ) = {z} with deg T z = 2, where N T (v) denotes the neighbor set of v in T . On the contrary, assume that for any leaf w in T , one has

N T (w) = {z w } with deg T z w ≥ 3. Let L T = {w 1 , .
. . , w m }, and N T (w i ) = {z i } for each i = 1, . . . , m. If there exist positive integers i and j with 1 ≤ i = j ≤ m such that N T (w i ) = N T (w j ) = p, then the maximal path Q : w i , p, w j is the required path, and the proof is over. Thus, one may assume that z i = z j for all positive integers i and j with 1

≤ i = j ≤ m,
(T ) deg T v = 2n -2.
Accordingly, one has 2n -2 ≥ 2n, which is a contradiction. This shows that there exists a leaf such as

w 1 in T such that N T (w 1 ) = {z} with deg T z = 2. Now, set T 1 := T \ {w 1 }. Note that E(T 1 ) = E(T ) \ {w 1 , z}.
Due to |V (T 1 )| < n, it follows from the inductive hypothesis that the tree T 1 has a vertex u ∈ V (T 1 ) with deg T u ≥ 3, and w ′ , w 2 ∈ L T 1 with w ′ = w 2 , and

v 1 , . . . , v r , v r+1 , . . . , v s ∈ V (T 1 ) (possibly s = 0) with deg T 1 v i = 2 for each i = 1, . . . , s such that Q : w ′ , v 1 , . . . , v r , u, v r+1 , . . . , v s , w 2 is a maximal path in T 1 . Note L T 1 = (L T \ {w 1 }) ∪ {z} since deg T 1 z = 1. If z / ∈ {w ′ , w 2 }, then Q is the required path in T . Let z ∈ {w ′ , w 2 }. If z = w ′ , then Q : w 1 , w ′ , v 1 , . . . , v r , u, v r+1 , . . . , v s , w 2 is a maximal path in T . If z = w 2 ,
then since T has no cycle, we have z = w ′ , and so deg T w ′ = 1; this yields that Q : w ′ , v 1 , . . . , v r , u, v r+1 , . . . , v s , w 2 , w 1 is a maximal path in T . This completes the inductive step, and so the claim has been proved by induction.

Theorem 9.2.15. Suppose that T is a tree which is not a path graph. Then there is exactly one minimal path cover set of T such that the degree of its vertices is greater than or equal to 3.

Proof. According to Proposition 9.1.10, the set of all vertices in T with degree greater than 2, denoted by S T , is a path cover set. We proceed by induction on n := |S T |. If n = 1, then T is a starlike graph with center z, and so the set {z} is the unique minimal path cover set of T which the degree of its vertex is greater than or equal to 3. Hence, the result in the case where n = 1 holds. Now, suppose, inductively, that n > 1 and that the result has been proved for all trees which are not path graphs and the size of the set of all vertices with degree greater than 2 is less than n. Assume that T is a tree such that is not a path graph and n = |S T | > 1.

It therefore becomes of interest to seek a contradiction. Suppose that H 1 and H 2 are two distinct minimal path cover sets of tree T such that the degree of their vertices are greater than or equal to 3. In the following, our aim is to prove that H 1 ∩ H 2 = ∅. Since T is a tree which is not a path graph, Proposition 9.2.14 yields that T has a vertex u ∈ V (T ) with deg T u ≥ 3, and w 1 , w 2 ∈ L T with w 1 = w 2 , and v 1 , . . . , v r , v r+1 , . . . , v s ∈ V (T ) (possibly s = 0) with deg T v i = 2 for each i = 1, . . . , s such that Q : w 1 , v 1 , . . . , v r , u, v r+1 , . . . , v s , w 2 is a maximal path in T . Due to the degree of each vertex of H i is greater than or equal to 3 and H i ∩ V (Q) = ∅ for all i = 1, 2, it follows that u ∈ H i for all i = 1, 2, and thus H 1 ∩ H 2 = ∅. Put α := deg T u. Hence, T \ {u} has exactly α connected components like T 1 , . . . , T α . Put B i,k := H i ∩ V (T k ) for each k = 1, . . . , α, and i = 1, 2. Furthermore, for each k = 1, . . . , α, there is unique vertex

f k ∈ V (T k ) such that {u, f k } ∈ E(T ).
Accordingly, for each k = 1, . . . , α, we may consider one of the following four cases:

Case

= B 2,k = ∅. Case 4. deg T f k = 1. Then B 1,k = B 2,k = ∅. Based on V (T ) = α k=1 V (T k ) ∪ {u}, u ∈ H 1 ∩ H 2 , and B 1,k = B 2,k for each k = 1, . . . , α, one can conclude that H 1 = H 2 ,
which is a contradiction. This completes the inductive step, and so the theorem has been proved by induction. Lemma 9.2.16. Suppose that T is a tree which is not a path graph and L is the set of all vertices of T with degree smaller than or equal to 2. Then (P I(T ) : v∈L x v ) is a minimal prime ideal of P I(T ).

Proof. In order to simplify the notation, we set J := (P I(T ) : v∈L x v ). Let L = {j 1 , . . . , j d }. We first prove that v∈L x v / ∈ P I(T ). Suppose, on the contrary, that v∈L x v ∈ P I(T ). Hence, there exists an element u in the unique minimal set of monomial generators of P I(T ) such that u divides v∈L x v . By the definition of the max-path ideal, it follows that u

= x i 1 • • • x i k and Q : i 1 , . . . , i k is a maximal path in T . Since u divides v∈L x v , one has deg T i s ≤ 2 for all s = 1, . . . , k.
Choose an arbitrary vertex i s in V (Q). Since tree T is not a path graph, we obtain there exists a vertex v in V (T ) such that deg T v ≥ 3, and thus v / ∈ V (Q). It follows from connectivity of the tree T that there is unique path Q ′ in T which connects v and

i s . According to i s ∈ V (Q) ∩ V (Q ′ ) and v / ∈ V (Q), we have V (Q) ∩ V (Q ′ ) = ∅ and V (Q ′ ) ⊈ V (Q). In view of Lemma 9.1.9, there exists v ′ ∈ V (Q) ∩ V (Q ′ ) with deg T v ′ ≥ 3,
which is a contradiction. This implies that v∈L x v / ∈ P I(T ). By virtue of Proposition 9.1.12, one can conclude that

P I(T ) = W is a minimal path cover set of T (x k : k ∈ W ),
is the minimal primary decomposition of P I(T ). Let W be an arbitrary minimal path cover set of T . It is easy to see that if W ∩ L = ∅, then

(x k : k ∈ W ) : v∈L x v = (1).
This implies that J has the minimal primary decomposition J = ℓ i=1 p i such that, for all i = 1, . . . , ℓ, we have p i = (x t i,1 , . . . , x t i,s ) which the set {t i,1 , . . . , t i,s } is a minimal path cover set of T and deg T t i,k ≥ 3 for each k = 1, . . . , s. On the other hand, Theorem 9.2.15 yields that ℓ = 1. Therefore, we get (P I(T ) : v∈L x v ) is a prime monomial ideal, as desired. Theorem 9.2.17. Suppose that T is a tree which has a vertex u with deg T u ≥ 4 and w 1 , . . . , w k ∈ L T , where k = deg T u -1, are such that, for all i = 1, . . . , k, if u, r 1 , . . . , r s , w i is the path between u and w i , then deg T r j = 2 for all j = 1, . . . , s. Also, suppose that L is the set of all vertices of T with degree smaller than or equal to 2. Then I T = htp, where p = (P I(T ) : v∈L x v ).

I(T ) = (x v • • • x u • • • x v ′ : v, v ′ ∈ L T ).
It follows from Theorem 2.1.27 that (x u ) ∈ Ass R (R/P I(T )), and so ht(P I(T )) = 1. Also, since L = {i k : k = 1, . . . , s}, we get (P I(T ) : v∈L x v ) = (x u ). Therefore, the assertion holds for n = 4. Now, suppose, inductively, that n > 4 and that the result has been proved for any tree which satisfies the hypothesis of theorem and that the number of leaves is less than n. Let T be a tree that |L T | = n and it has a vertex u with deg T u ≥ 4 and w 1 , . . . , w k ∈ L T , where k = deg T u -1, are such that, for all i = 1, . . . , k, if u, r 1 , . . . , r s , w i is the path between u and w i , then deg T r j = 2 for all j = 1, . . . , s. Also, suppose that L is the set of all vertices of T with degree smaller than or equal to two. Pick an arbitrary vertex in L T , say w, such that if u, v 1 , . . . , v s , w is the path between u and w, then deg

T v i = 2 for all i = 1, . . . , s. Set T 1 := T \ {w} and L 1 := {x v : deg T 1 v ≤ 2}. Let P I(T 1 ) = q 1 ∩ q 2 ∩ • • • ∩ q ℓ be
the minimal primary decomposition of P I(T 1 ), and P I(T ) = p 1 ∩ p 2 ∩ • • • ∩ p t be the minimal primary decomposition of P I(T ) such that p 1 = (x j 1 , . . . , x j k ) with deg T j i ≥ 3 for all i = 1, . . . , k. Let N T (w) = {w 1 } with deg T w 1 = 2. In the following, our aim is to prove that ht(P I(T )) = ht(P I(T 1 )). According to deg T w 1 = 2, one can conclude that (P I(T ) : R x w ) = P I(T 1 ), and so t j=1 (p j : R x w ) = ℓ i=1 q i . Without loss of generality, we may assume that x w / ∈ p j only for j = 1, . . . , α with 1 ≤ α ≤ t. Hence, 288 α j=1 p j = ℓ i=1 q i . It follows immediately from the Second Uniqueness Theorem for Primary Decomposition that α = ℓ and, after possibly reordering, p i = q i for each i = 1, . . . , ℓ. Also, it is clear that P I(T ) = ℓ i=1 q i ∩ t j=α+1 p j is the minimal primary decomposition of P I(T ). Now, let p j = (x i 1 , . . . , x ir ) with α + 1 ≤ j ≤ t. Therefore, Γ = {i 1 , . . . , i r } is a minimal path cover set of T . On the other hand, we can consider T as the union of T 1 and a path graph L ′ with V (L ′ ) = {w 1 , w} such that w 1 is the cut-point of this union. According to V (L ′ ) ∩ Γ = {w} is a non-empty set, applying Theorem 9.2.4 yields the following three cases: Case 1. Γ is a minimal path cover set of T 1 . Since w / ∈ V (T 1 ), this case is impossible.

Case 2. Γ \ {w} is a minimal path cover set of T 1 . Without loss of generality, assume that i 1 = w, and put Γ 1 := Γ \ {w}. Then Γ 1 = {i 2 , i 3 , . . . , i r }, and so (x i 2 , x i 3 , . . . , x ir ) ∈ Ass(P I(T 1 )). By Ass(P I(T 1 )) = {q 1 , q 2 , . . . , q ℓ }, it follows that there exists a positive integer k with 1 ≤ k ≤ ℓ such that q k = (x i 2 , x i 3 , . . . , x ir ). Since Ass(P I(T 1 )) ⊆ Ass(P I(T )), we get q k ∈ Ass(P I(T )), which contradicts the minimality of (x i 1 , x i 2 , x i 3 , . . . , x ir ). Therefore, this case is impossible. Case 3. (Γ \ {w}) ∪ {w 1 } is a minimal path cover set of T 1 . Without loss of generality, we may assume that i 1 = w. Set Γ 2 := (Γ \ {w}) ∪ {w 1 }. This implies that Γ 2 = {w 1 , i 2 , i 3 , . . . , i r }, and so (x w 1 , x i 2 , x i 3 , . . . , x ir ) ∈ Ass(P I(T 1 )). Due to Ass(P I(T 1 )) = {q 1 , q 2 , . . . , q ℓ }, one has there exists a positive integer k with 1 ≤ k ≤ ℓ such that q k = (x w 1 , x i 2 , x i 3 , . . . , x ir ). Therefore, htp j = htq k . Since ht(P I(T )) = min{htq i , htp j : i = 1, . . . , ℓ, j = α + 1, . . . , t}, one can conclude that ht(P I(T )) = ht(P I(T 1 )). According to L 1 = L \ {w} and (P I(T ) : R x w ) = P I(T 1 ), it follows that (P I(T ) : R v∈L x v ) = (P I(T 1 ) : R v∈L 1 x v ).

We therefore shall reduce to the situation where N

T (u) = {v 1 , v 2 , . . . , v m+1 } with m + 1 = deg T u ≥ 4 and deg T v i = 1 for any i = 2, . . . , m + 1. Let w ∈ N T (u) with deg T w = 1. Once again, set T 1 := T \ {w}, and thus |L T 1 | = |L T | -1. Let P I(T 1 ) = q 1 ∩ q 2 ∩ • • • ∩ q ℓ be
the minimal primary decomposition of P I(T 1 ), and P I(T ) = p 1 ∩ p 2 ∩ • • • ∩ p t be the minimal primary decomposition of P I(T ) such that p 1 = (x j 1 , . . . , x j k ) with deg T j i ≥ 3 for all i = 1, . . . , k. By virtue of the inductive hypothesis, we get ht(P I(T 1 )) = htq 1 , where q 1 = (P I(T 1 ) : R v∈L

1 x v ) with L 1 = {v ∈ V (T 1 ) : deg T 1 v ≤ 2}.
Without loss of generality, we may assume that x w / ∈ p j only for j = 1, . . . , α with 1 ≤ α ≤ t. Furthermore, it is routine to check that (P I(T ), x w ) = (P I(T 1 ), x w ). This implies that

α j=1 (p j , x w ) ∩ t j=α+1 p j = ℓ i=1 (q i , x w ).
Pick a positive integer j with 1 ≤ j ≤ α. Then ℓ i=1 (q i , x w ) ⊆ (p j , x w ), and so there exists a positive integer i 1 with 1 ≤ i 1 ≤ ℓ such that (q i 1 , x w ) ⊆ (p j , x w ).

Hence, q i 1 ⊆ p j . It follows from the inductive hypothesis that htq 1 ≤ htq i 1 , and thus htq 1 ≤ htp j . Now, pick a positive integer j with α + 1 ≤ j ≤ t. Then ℓ i=1 (q i , x w ) ⊆ p j , and hence there exists a positive integer i 2 with 1 ≤ i 2 ≤ ℓ such that (q i 2 , x w ) ⊆ p j . So q i 2 ⊆ p j . The inductive hypothesis yields that htq 1 ≤ htq i 2 , and one has htq 1 ≤ htp j . Therefore, we have htq 1 ≤ htp j for any j = 1, . . . , t. In the following, our aim is to show that htq 1 = htp 1 . According to p 1 ∈ Ass(P I(T )), one can conclude that Γ := {j 1 , . . . , j k } is a minimal path cover set of T . Also, we can consider T as the union of T 1 and a path graph L ′ with V (L ′ ) = {u, w} such that u is the cut-point of this union. Since deg T j i ≥ 3 for all i = 1, . . . , k, we have the following two cases:

(A) u ∈ Γ. Due to V (L ′ ) ∩ Γ = {u} is a non-empty set, it follows from Theorem 9.2.4 that one has the following three subcases:

(i) Γ is a minimal path cover set of T 1 . Hence, p 1 ∈ Ass(P I(T 1 )). Since deg T j i ≥ 3 for all i = 1, . . . , k, we obtain p 1 = q 1 , and so htp 1 = htq 1 .

(ii) Γ \ {u} is a minimal path cover set of T 1 . We now verify that this case is impossible. Without loss of generality, we may assume j

1 = u. Set Γ 1 := Γ \ {u}. Then Γ 1 = {j 2 , j 3 , . . . , j k }. Since deg T u = m + 1 ≥ 4, it follows that deg T 1 u = deg T u -1 ≥ 3,
and so there exist at least two distinct vertices such as

v, v ′ ∈ N T (u) with deg T v = deg T v ′ = 1.
Therefore, Γ 1 does not cover the maximal path v, u, v ′ in T 1 , this yields a contradiction.

(iii) (Γ \ {u}) ∪ {u} is a minimal path cover set of T 1 . Due to the argument has been stated in (i), one can conclude that p 1 = q 1 , and hence htp 1 = htq 1 .

(

B) u / ∈ Γ. Because of deg T u = m + 1 ≥ 4, one has deg T 1 u = deg T u -1 ≥ 3, and
so there exist at least two distinct vertices such as v, v ′ ∈ N T (u) with deg T v = deg T v ′ = 1, and hence Γ has to cover the maximal path v, u, v ′ in T . This implies that Γ must contain a vertex with degree one, which is a contradiction.

In order to complete the proof, we note that p 1 = (P I(T ) : v∈L x v ), by virtue of Lemma 9.2.16.

We conclude this section with an example which illuminates Theorem 9.2.17.

Example 9.2.18. Let T be a tree, as shown in the following figure, on the vertex set [12]. 1 Direct computation shows that (x 8 , x 5 , x 2 ) ∈ Ass(P I(T )). On the other hand, Theorem 9.2.17 yields that ht(P I(T )) = ht(x 8 , x 5 , x 2 ) = 3, where (x 8 , x 5 , x 2 ) = P I(T ) : R x 1 x 3 x 6 x 7 x 9 x 10 x 11 x 12 .

On a class of polymatroidal ideals

Matroid theory is one of the most attractive areas in combinatorics and is deeply rooted in graph theory and linear algebra. The well-known examples of matroids include uniform matroids, graphical matroids, and linear matroids. We refer to [START_REF] Welsh | Matroid Theory[END_REF], [START_REF] Oxley | Matroid Theory[END_REF], and [START_REF] Herzog | Discrete polymatroids[END_REF] for basic definitions and notions in matroid theory. The polymatroids originated in [START_REF] Edmonds | Submodular functions, matroids, and certain polyhedra[END_REF] and the discrete polymatroids appeared in [START_REF] Herzog | Discrete polymatroids[END_REF] as a multiset analogue of matroids. In commutative algebra, the matroidal ideals as the square-free version of polymatroidal ideals, hold a very special place due to their nice algebraic and homological properties. A monomial ideal I ⊂ K[x 1 , . . . , x n ] is called polymatroidal, if the set of exponent vectors of the minimal generating set of I corresponds to the set of bases of a discrete polymatroid. In particular, a square-free polymatroidal ideal is simply referred to as a matroidal ideal, because in this case, the set of exponent vectors of the minimal generating set of I corresponds to the set of bases of a matroid. The algebraic and homological properties of polymatroidal ideals have been studied by many authors, for example, see [START_REF] Herzog | Resolutions by mapping cones, The Roos Festschrift[END_REF], [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF], [START_REF] Herzog | Cohen-Macaulay polymatroidal ideals[END_REF], [12], [START_REF] Herzog | Persistence and stability properties of powers of ideals[END_REF]. It is known from [31] and [START_REF] Herzog | Resolutions by mapping cones, The Roos Festschrift[END_REF] that the product of polymatroidal ideals is again polymatroidal. Moreover, polymatroidal ideals have linear quotients, and hence linear resolutions, see [31]. In particular, all powers of a polymatroidal ideal have linear resolutions. If I is a polymatroidal ideal, then the Rees algebra R(I) is normal, see [START_REF] Villarreal | Rees cones and monomial rings of matroids[END_REF]Proposition 3.11], [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF]Theorem 3.4]. The stability indices related to associated primes and depth of powers of polymatroidal ideals are of particular interest and have been discussed in several papers, for example, see [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF] and [START_REF] Herzog | Persistence and stability properties of powers of ideals[END_REF]. In fact, polymatroidal ideals are known to have the persistence property, see [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF]Proposition 3.3], and they even have the strong persistence property, see [START_REF] Herzog | Persistence and stability properties of powers of ideals[END_REF]Proposition 2.4].

In this section, we introduce another class of matroidal ideals, namely, the class of the t-path ideals of complete multipartite graphs. Let G be a finite simple graph with n vertices. We refer to a path of length t -1 in G as a t-path. Let S be a polynomial ring over a field K in n variables. To simplify the notation, throughout this text, we identify the vertices of G with the variables in S. The t-path ideal

I t (G) ⊂ S = K[x 1 , . . . , x n ] is generated by those monomials x i 1 • • • x it such that x i 1 , . . . , x it is a t-path in G.
The t-path ideals of graphs were introduced by Conca and De Negri in [30] and later on discussed by several authors for different classes of graphs, in particular for directed trees, trees and, cycles, see [START_REF] He | Algebraic properties of the path ideal of a tree[END_REF], [19], [25]. It should be noted that the results of this section can be found in [START_REF] Khashyarmanesh | On the matroidal path ideals[END_REF].

In what follows, we first recall some basis definition and notion related to matroids. Let Given any subset C of P ([n]), there exists a matroid on [n] with C as its sets of bases if and only if C satisfies the exchange property.

Let S = K[x 1 , . . . , x n ] be a polynomial ring over a field K and I be a monomial ideal in S. For a given a = (a 1 , . . . , a n ) with nonnegative entries, x a denotes the monomial x a 1 1 • • • x an n in S. A monomial ideal I ⊂ S generated in a single degree is called polymatroidal if for all monomials x a , x b ∈ G(I) with a i > b i , there exists j with a j < b j such that x j (x a /x i ) ∈ G(I). A square-free polymatroidal ideal is simply referred to as a matroidal ideal. In other words, a monomial ideal I in S is matroidal, if G(I) can be identified as a set of bases of a matroid.

Here, we recall some basic definitions and notions from graph theory. Throughout this section, all considered graphs will be simple, undirected, and finite. Let G be a graph with the vertex set V (G) and the edge set E(G). A subset A ⊆ V (G) is called independent if no vertices in A are adjacent in G. A graph G is called complete r-partite, if V (G) can be partitioned into r independent sets V 1 , . . . , V r such that a and b are adjacent for all a ∈ V i and all b ∈ V j with 1 ≤ i = j ≤ r. Such a partition is called an r-partition of G. Moreover, if |V i | = n i for all i = 1, . . . , r, then the complete r-partite graph is denoted by K n 1 ,...,nr . In this section, we focus on I(K n 1 ,...,nr ).

Let G be a graph with |V (G)| = n. From now on, we identify the vertices of the graph G with the variables in the polynomial ring S = K[x 1 , . . . , x n ]. Definition 9.3.1. The t-path ideal of G, denoted by I t (G), is given by

I t (G) := (x i 1 • • • x it : x i 1 , . . . , x it is a t-path in G) ⊂ S.
When t = 2, then I 2 (G) is simply the edge ideal I(G) of G. If there is no t-path in G, then we set I t (G) = 0.

The following example illustrates the definitions that have been stated above. Then I 4 (K 1,2,3 ) has the following minimal monomial generators:

x 1 x 2 x 3 x 4 , x 1 x 2 x 3 x 5 , x 1 x 2 x 3 x 6 , x 1 x 2 x 4 x 5 , x 1 x 2 x 4 x 6 , x 1 x 2 x 5 x 6 , x 1 x 3 x 4 x 5 , x 1 x 3 x 4 x 6 , x 1 x 3 x 5 x 6 , x 2 x 3 x 4 x 5 , x 2 x 3 x 4 x 6 , x 2 x 3 x 5 x 6 .

Remember that for a monomial ideal I ⊂ S, if supp(I) = {x 1 , . . . , x n }, then I is said to be fully supported. Furthermore, if G(I) = {u 1 , . . . , u m }, then gcd(I) = gcd(u 1 , . . . , u m ).

Below we give a list of remarks that will be used frequently throughout this text. For any integer a, the notations a and a denote the ceiling and floor functions of a, respectively. Let G = K n 1 ,...,nr and V 1 , . . . , V r be the r-partition of G. Then we have the following remarks. Remark 9.3.3. Let P : x i 1 , x i 2 , . . . , x it be a t-path in G, and u = x i 1 x i 2 • • • x it ∈ G(I t (G)). If x i j ∈ V s for some 1 ≤ s ≤ r and 2 ≤ j ≤ t -1, then x i j-1 , x i j+1 / ∈ V s . This leads to the following conclusions: (i) A t-path in G can have at most t/2 vertices in V s . (ii) Let x i j ∈ V s for some 1 ≤ j ≤ t. If there exists x k ∈ V s such that x k / ∈ V (P ), then x i j can be replaced by x k in P to obtain a new t-path P ′ in G. In particular, (u/x i j )x k ∈ G(I t (G)).

(iii) If there exists some V s such that V (P ) ∩ V s = ∅, then any x i j ∈ V (P ) can be replaced in P by any x k ∈ V s , to obtain a new path P ′ in G. In particular, (u/x i j )x k ∈ G(I t (G)), for all x k ∈ V s and for each j = 1, . . . , t.

(iv) If I t (G) = 0, then I t (G) is fully supported. This can be easily seen due to statements (ii) and (iii).

(v) Let I t (G) = 0. Then I t (G) is a principal ideal if and only if t = n 1 + • • • + n r . This also follows from statements (ii) and (iii).

Matroidal path ideals of graphs

It is proved in [12, Theorem 1.1] that a monomial ideal I ⊂ S = K[x 1 , . . . , x n ] is polymatroidal if and only if I : u is polymatroidal for all monomials u in S. In particular, (I : x i ) is polymatroidal if I is polymatroidal. The following theorem gives further information about (I : x i ). . Thus, J 1 = I 1 + x i J 2 . Once again, from the proof of [12, Theorem 1.1] we obtain I 1 ⊆ J 2 , and thus I 1 ⊆ I 2 . Therefore,

I 0 ⊆ I 1 ⊆ I 2 . Now, set J k = d j=k I j x j-k i
, where 1 ≤ k ≤ d. By continuing this procedure, we can conclude that I 0 ⊆ I 1 ⊆ • • • ⊆ I d . This finishes the proof.

If I ⊂ K[x 1 , . . . , x n ] is a matroidal ideal, then max{deg x i u : u ∈ G(I)} is either 0 or 1. For each variable x i ∈ {x 1 , . . . , x n }, we set I 0,i := (u : x i / ∈ supp(u)) and I 1,i := (u/x i : x i ∈ supp(u)). Then, we get I = I 0,i + x i I 1,i , for all i = 1, . . . , n. With this notation, we obtain the following specialization of Theorem 9.3.4. (i) for each i = 1, . . . , n, we have G(I 1,i ) ⊆ n t=1, t̸ =i G(I 1,t ).

(ii) if I 1,i ⊆ I 1,j for some 1 ≤ i, j ≤ n with i = j, then I 1,i = I 1,j .

Proof. (i) The assumption that I is fully supported implies that I 1,i = 0 for each i = 1, . . . , n. Moreover, gcd(I) = 1 gives that I 0,i = 0 for each i = 1, . . . , n. Let u ′ ∈ G(I 1,i ) and v ∈ G(I 0,i ). Then it follows from the definition of I 0,i and I 1,i that v ∈ G(I) and u := u ′ x i ∈ G(I). Since deg x i u > deg x i v, one can conclude from the exchange property that there exists some s with deg xs u < deg xs v such that u ′ x s =

x s (u/x i ) ∈ G(I). Thus, u ′ ∈ G(I 1,s ). We therefore get G(I 1,i ) ⊆ n t=1, t̸ =i G(I 1,t ), as claimed.

(ii) Suppose, on the contrary, that there exists some monomial u ∈ G(I 1,j ) \ G(I 1,i ). Thus, x j u ∈ G(I). If x i ∈ supp(u), then (x j u)/x i ∈ G(I 1,i ). Using the assumption I 1,i ⊆ I 1,j , we obtain (x j u)/x i ∈ I 1,j which gives (x 2 j u)/x i ∈ I. Since I is matroidal, (x j u)/x i ∈ I. This yields a contradiction to x j u ∈ G(I). Hence, x i / ∈ supp(u). Since x j u ∈ G(I), we conclude that x j u ∈ G(I 0,i ). From Theorem 9.3.4, we know that I 0,i ⊆ I 1,i . Then there exists some monomial w ∈ G(I 1,i ) such that w divides x j u. Again, by using the assumption, I 1,i ⊆ I 1,j , we have x j / ∈ supp(w), and therefore w|u. Since both w and u are monomials of degree d -1, we conclude that u = w, as required.

It follows from [28,Theorem 3.2] that if I(G) is a matroidal ideal, then G is a complete multipartite graph. The converse follows as a corollary of [START_REF] Asghar | On the weakly polymatroidal property of the edge ideals of hypergraphs[END_REF]Theorem 2.2]. We give another straightforward proof of this fact in the language of edge ideals of complete multipartite graphs. Theorem 9.3.6. Let G be a graph without isolated vertices. The edge ideal I(G) is a matroidal ideal if and only if G is a complete r-partite graph for some r ≥ 2.

Proof. We first assume that G is a complete r-partite graph, that is, G = K n 1 ,..,nr with a vertex partition V 1 , . . . , V r and |V i | = n i for all i = 1, . . . , r. Let u := x k x l ∈ I(G) and v := x i x j ∈ I(G) such that x k does not divide v. We need to show that either x i (u/x k ) = x i x l ∈ I(G) or x j (u/x k ) = x j x l ∈ I(G). Let x l ∈ V t for some 1 ≤ t ≤ r. The graph G is complete r-partite and {x i , x j } ∈ E(G), hence it follows that at least one of the vertices between x i and x j does not belong to V t . If x i / ∈ V t , then x i x l ∈ I(G) and similarly if x j / ∈ V t , then x j x l ∈ I(G), as required. Conversely, let I(G) be a matroidal ideal and |V (G)| = n. We can assume that G does not contain any isolated vertices, and hence I(G) is fully supported. If I(G) is a principal ideal, then there is nothing to prove and the assertion holds trivially. and every x j k ∈ V (P 2 ) \ V (P 1 ) is such that x j k ∈ V 1 . If there exists some 2 ≤ p ≤ t -2 such that x ip , x i p+1 / ∈ V 1 , then for any x j k ∈ V (P 2 ) \ V (P 1 ), we have {x ip , x j k }, {x j k , x i p+1 } ∈ E(G), and P : x i 2 , . . . , x ip , x j k , x i p+1 , . . . , x it is a t-path in G as given in (3). Finally, suppose that each of (1), (2), and (3) do not hold. Then we are in the following situation:

(i) x i 1 , x i t-1 ∈ V 1 ;
(ii) every x j k ∈ V (P 2 ) \ V (P 1 ) is such that x j k ∈ V 1 . This also gives that for every

x j k ∈ V (P 2 ) \ V 1 , we have x j k ∈ V (P 1 ); (iii) for every 2 ≤ p ≤ t -2 either x ip ∈ V 1 or x i p+1 ∈ V 1 . Since x i 1 ∈ V 1 and {x i 1 , x i 2 } ∈ E(G), this implies that x i 2 / ∈ V 1 .
It follows now from (iii) that x i 3 ∈ V 1 . By continuing this process, we obtain from (iii) that

x i t-2 / ∈ V 1 since x i t-1 ∈ V 1 and {x i t-2 , x i t-1 } ∈ E(G)
. This shows that P 1 starts and ends at vertices in V 1 , and all vertices in P 1 with odd indices are also in V 1 . Furthermore, all vertices in V (P 1 ) with even indices do not belong to V 1 . Hence, t-1 must be an odd integer, and in V (P 1 ), there are t/2-1 vertices that do not belong to V 1 , that is, |V (P 1 ) \ V 1 | = t/2 -1. Moreover, it follows from (ii) that every vertex in V (P 2 ) which is outside of V 1 must belong to V (P 1 ), that is,

V (P 2 ) \ V 1 ⊆ V (P 1 ) \ V 1 .
Since P 2 is a t-path with even number of vertices in a complete r-partite graph G, this gives that at least half of its vertices are not in V 1 , that is,

|V (P 2 ) \ V 1 | ≥ t/2. This contradicts |V (P 2 ) \ V 1 | ≤ |V (P 1 ) \ V 1 | = t/2 -1.
This yields that at least one of the statements in (1), (2), or (3) must hold.

We are now ready to state the main result of this section. Theorem 9.3.8. Let t ≥ 2. If I t (K n 1 ,...,nr ) = 0, then it is a matroidal ideal.

Proof. We set G := K n 1 ,...,nr and let V 1 , . . . , V r be the r-partition of G. The assertion is true for I 2 (G), as shown in Theorem 9.3.6. Let t ≥ 3 and I t (G) = 0. If I t (G) is a principal ideal, then there is nothing to prove and the assertion holds trivially. Let |G(I t (G))| ≥ 2. To prove that I t (G) is matroidal, we must show that if P 1 :

x i 1 , . . . , x it and P 2 : x j 1 , . . . , x jt are two t-paths with x i l / ∈ V (P 2 ) for some 1 ≤ l ≤ t, then there exists some x j k ∈ V (P 2 ) \ V (P 1 ), such that one obtains a new path by removing x i l from P 1 and inserting x j k in an appropriate position in P 1 .

We may assume that x i l ∈ V 1 . If x i 1 , . . . , x i l-1 , x i l+1 . . . , x it is a (t -1)-path in G, then we obtain the desired conclusion by using Lemma 9.3.7. In particular, if l = 1 or l = t, then removing x i l from P 1 gives a (t -1)-path in G. Now, assume that x i 1 , . . . , x i l-1 , x i l+1 . . . , x it is not a (t -1)-path in G. This is possible only if {x i l-1 , x i l+1 } / ∈ E(G), that is, x i l-1 , x i l+1 belong to V a for some 2 ≤ a ≤ r. If there exists some x j k ∈ V (P 2 ) \ V (P 1 ) such that x j k / ∈ V a , then

x i 1 , . . . , x i l-1 , x j k , x i l+1 , . . . , x it , is a t-path in G, and the proof is complete. Otherwise, every x j k ∈ V (P 2 ) \ V (P 1 ) is such that x j k ∈ V a . If x i 1 / ∈ V a , then {x i 1 , x i l+1 } ∈ E(G), and we get x i l-1 , x i l-2 , . . . , x i 1 , x i l+1 , . . . , x it as a (t -1)-path in G. We again use Lemma 9.3.7 to obtain the desired conclusion. A similar argument gives us the desired conclusion when x it / ∈ V a . Otherwise, we are in the following situation:

(1) x i l-1 , x i l+1 , x i 1 , x it ∈ V a ;

(2) every x j k ∈ V (P 2 ) \ V (P 1 ) is such that x j k ∈ V a . This implies that if x j k ∈ V (P 2 ) \ V a , then x j k ∈ V (P 1 ).

We claim that there must exist some p with either 2 ≤ p ≤ l -3 or l + 2 ≤ p ≤ t -2 such that x ip , x i p+1 / ∈ V a . Indeed, if our claim is true and we can find such p with 2 ≤ p ≤ l -3, then x i 1 , . . . , x ip , x i l-1 , x i l-2 , . . . , x i p+1 , x i l+1 , . . . , x it is a (t -1)-path in G, and the desired result can be deduced from Lemma 9.3.7. Similarly, if we can find such p with l + 2 ≤ p ≤ t -2, then x i 1 , . . . x i l-1 , x ip , x i p-1 , . . . , x i l+1 , x i p+1 , . . . , x it is a (t -1)-path in G, and one can conclude the desired result from Lemma 9.3.7.

To show our claim, suppose, on the contrary, that for every p with 2 ≤ p ≤ l -3 and l + 2 ≤ p ≤ t -2, we have either x ip ∈ V a or x i p+1 ∈ V a . Since x i 1 ∈ V a and {x i 1 , x i 2 } ∈ E(G), this yields that x i 2 / ∈ V a . It follows now from the assumption that x i 3 ∈ V a . Moreover, one can conclude from (1) that x i l-2 / ∈ V a since x i l-1 ∈ V a and {x i l-2 , x i l-1 } ∈ E(G). This forces l -1 to be an odd integer. Similarly, since x i l+1 , x it ∈ V a , we derive that the number of vertices in the path x i l+1 , . . . , x it is odd as well. Collectively, we obtain t itself must be odd. This shows that V (P 1 ) contains (t + 1)/2 vertices from V a and the other (t -1)/2 vertices are not from V a . Note that these (t -1)/2 vertices that are not from V a include x i l as well. Considering that, by (2), every vertex in V (P 2 ) \ V (P 1 ) belongs to V a and x i l / ∈ V (P 2 ), this yields that in V (P 2 ) there are at most (t -1)/2 -1 vertices that are not from V a , which is impossible by Remark 9.3.3(i). Hence, the claim holds, and the proof is over.

Cohen-Macaulay property of I t (K n 1 ,...,n r )

We now investigate the Cohen-Macaulay property for I t (K n 1 ,...,nr ). Let I be a monomial ideal in the polynomial ring S = K[x 1 , . . . , x n ]. The ideal I is called the Veronese ideal of degree d in S if I is generated by all monomials of degree d in S.

1 ≤ i 1 < • • • < i t ≤ n.
Moreover, from Remark 9.3.3(i), it follows that if there exists some V i with n i > t/2 , then for any subset T of V i with |T | = t/2 + 1, there does not exist any t-path in G that contains all the elements of T . Hence, if there exists some V i of G with n i > t/2 , then I(G) is not square-free Veronese, and hence not Cohen-Macaulay. Now, assume that n i ≤ t/2 for each i = 1, . . . , r. Let u = x i 1 • • • x it be a square-free monomial in S. We have to show that the elements in A := supp(u) can be interpreted as a t-path in G. Let A i = A ∩ V i and a i = |A i |. Since n i ≤ t/2 , we get a i ≤ t/2 . Without loss of generality, we may assume that the elements in {x i 1 , . . . , x it } are arranged such that the first a 1 elements lie in V 1 , the next a 2 elements lie in V 2 and so on. If t is odd, then Let (R, m) be a commutative Noetherian local ring or a standard graded algebra over a field K with the graded maximal ideal m. Let I ⊂ R be an ideal, which is graded if R is a standard graded K-algebra. It is known from Brodmann [21] that depth(R/I k ) stabilizes for large k, that is, depth(R/I k ) is constant for k 0. The smallest t > 0 such that depth(R/I t ) = depth(R/I k ) for all k ≥ t is called the index of depth stability of I and is denoted by dstab(I), as defined in [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF]. Moreover, depth(I dstab(I) ) is called the limit depth of I and is denoted by lim k→∞ depth(R/I k ), see [START_REF] Herzog | The depth of powers of an ideal[END_REF]. In this section, we study the dstab(S/I t (K n 1 ,...,nr )) and lim k→∞ depth(S/I t (K n 1 ,...,nr ) k ). In addition, the linear relation graphs of monomial ideals were introduced to help in understanding their analytic spreads, see Definition 6. Therefore, to compute the limit depth of t-path ideals of complete r-partite graphs, it is enough to compute the number of vertices and the number of connected components in their linear relation graphs.

x 1 ,
from [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF]Corollary 5.7] that depth(S 2 /L k ) = max{0, k( t/2 -q) + q -1} for all k and lim k→∞ depth(S 2 /L k ) = 0 with dstab(L) = (q -1)/(q -t/2 ) .

The fact that the ideals J and L are generated in different sets of variables facilitates to conclude that (I t (G)) k = J k L k , and depth(S/I t (G) k ) = depth(S 1 /J k )+ depth(S 2 /L k ) + 1, for all k, for example, see [START_REF] Hoa | On some invariants of a mixed product of ideals[END_REF]Lemma 2.2]. Therefore, depth(S/I t (G) k ) = p + max{0, k( t/2 -q) + q -1}.

Since lim k→∞ depth(S/I t (G) k ) = p and dstab(L) = (q -1)/(q -t/2 ) , it follows that dstab(I t (G)) = (q -1)/(q -t/2 ) .

(iii) Let p > t/2 and q > t/2 . Then it follows from Remark 9.3.3(ii) that {x i , x j } ∈ E(Γ) for all 1 ≤ i = j ≤ p, and {y i , y j } ∈ E(Γ) for all 1 ≤ i = j ≤ q. If t is even, then every path in G contains half variables from V 1 and the other half from V 2 . Consequently, given any t-path in G, we cannot replace any vertex from V 1 with any vertex from V 2 . This shows that {x i , y j } / ∈ E(Γ). Therefore, Γ has exactly two connected components, and ℓ(I) = p + q -2 + 1 = n -1. By using the equality in (9.3.2), we obtain lim k→∞ depth(S/I t (G) k ) = n -(n -1) = 1 and we conclude from (9.3.1) that dstab(I t (G)) < n -1. Remark 9.3.13 gives that depth(I t (G)) = t -1. Hence, dstab(I t (G)) ≥ 1 and the equality holds only if t = 2 due to Remark 9.3.12.

If t is odd, then set m = t/2 . The sequence x 1 , y 1 , x 2 , y 2 , . . . , x m , y m is a 1 • • • x bn 1 of degree d with b i ≤ a i , for all i = 1, . . . , n is called the ideal of Veronese type, and is denoted by I d;a 1 ,...,an . In [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF], the index of depth stability of Veronese type ideals is discussed in detail. Note that if a 1 = • • • = a n = 1, then I d;a 1 ,...,an is square-free Veronese ideal in n variables. According to Theorem 9.3.11 and results obtained in [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF], we conclude the following proposition. In view of Proposition 9.3.15, it is enough to consider those cases in which at least one of the n i 's is greater than t/2 . From Theorem 9.3.14, it is clear that limit depth of I t (G) depends on t and how vertices are distributed among the partition sets. We see this behavior in subsequent results too. Now, we will describe limit depth and dstab for I 3 (K n 1 ,...,nr ) when r ≥ 3. Note that since r ≥ 3, we must have In the rest of this section, m denotes the unique graded maximal ideal in the polynomial ring S whose variables correspond to the vertices of the graph G. Proof. From Remark 9.3.13, we know that depth(I(G)) = 1. Then by following Remark 9.3.12, it is enough to show that depth(I(G) 2 ) = 0. Given a graph H, in [57, Theorem 3.1], the equivalent condition for depth(I(H) 2 ) = 0 is described as follows: depth(I(H) 2 ) = 0 if and only if H contains a cycle of length 3 and every other vertex of G has a neighbor in this cycle. It can be easily seen that G satisfies this condition. Let V 1 , . . . , V r be the r-partition of G and since r ≥ 3, we can choose x 1 ∈ V 1 , x 2 ∈ V 2 , and x 3 ∈ V 3 . Then the subgraph induced by x 1 , x 2 , and x 3 is the cycle of length 3 in G. Moreover, it follows from the definition of G that any x j ∈ V (G) is adjacent to at least one of the vertices in {x 1 , x 2 , x 3 }. Hence, depth(I(G) 2 ) = 0, as required. Proof. From Remark 9.3.13, we know that depth(I 3 (G)) = 2, and then by following Remark 9.3.12, it is enough to show that depth(I 3 (G) 2 ) = 0. Let V 1 , . . . , V r be the r-partition of G.

x 1 ∈ V 1 , x 2 ∈ V 2 ,
and x 3 , x 4 , x 5 ∈ V r , and set u := x 1 x 2 x 3 x 4 x 5 . Then for any x i ∈ V r , we have ux i = (x i x 2 x 3 )(x 4 x 1 x 5 ) ∈ I 3 (G) 2 because x i , x 2 , x 3 and x 4 , x 1 , x 5 are 3-paths in G. Furthermore, for any x i ∈ V (G)\V r , we have ux i = (x 1 x 3 x 2 )(x 4 x i x 5 ) ∈ I 3 (G) 2 because x 1 , x 3 , x 2 and x 4 , x i , x 5 are 3-paths in G. Therefore, I 3 (G) 2 : u = m, and hence depth(I 3 (G) 2 ) = 0, as required. Now, assume that |V i |, |V j | ≥ 2, for some 1 ≤ i = j ≤ r. Take x 1 , x 2 ∈ V i ,

x 3 , x 4 ∈ V j , and x 5 ∈ V k . Set u := x 1 x 2 x 3 x 4 x 5 ; then u / ∈ I 3 (G) 2 . We claim that I 3 (G) 2 : u = m. Indeed, for any x m ∈ V i , we have ux m ∈ I 3 (G) 2 because x 1 , x 5 , x 2 and x 3 , x m , x 4 are 3-paths in G. Similarly, for any x m ∈ V j , we have ux m ∈ I 3 (G) 2 . Finally, if x m ∈ V (G) \ (V i ∪ V j ), then again ux m ∈ I 3 (G) 2 because x 1 , x m , x 2 and x 3 , x 5 , x 4 are 3-paths in G. This shows that depth(I 3 (G) 2 ) = 0, as claimed.

We are ready to give the final result of this subsection which shows that in the case of r ≥ 3 and t ≥ 3, if the number of vertices in each partition set is big enough, then limit depth of I t (G) is 0. To prove this, we will show that Γ is a complete graph with V (Γ) = V (G). Let V 1 , . . . , V r be the r-partition of G. Let x i , x j ∈ V (G), for some i = j. We can choose A ⊂ V (G) such that |A| = t, x i ∈ A, x j / ∈ A, and |A ∩ V k | < t/2 for all k = 1, . . . , r. Such a choice of A is possible because n i ≥ t/2 and r ≥ 3. Take B = (A \ {x i }) ∪ {x j }. Then |B| = t and |B ∩ V k | ≤ t/2 , for all k = 1, . . . , r. Then by following the similar construction of the path at the end of the proof of Theorem 9.3.11, we obtain a t-path P in G with vertices in A and a t-path P ′ in G with vertices in B. Let u be the monomial in G(I t (G)) that corresponds to the t-path P and v be the monomial in G(I t (G)) that corresponds to the t-path P ′ . Then x i v = x j u and {x i , x j } ∈ E(Γ), as required. This completes the proof.

Chapter 10

On the maximal associated prime ideal of monomial ideals 10.1 A criterion for existence the maximal ideal Suppose that I is a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K, m = (x 1 , . . . , x n ) is the graded maximal ideal of R, and x 1 , . . . , x n are indeterminates. One of the most important questions is the existence of the graded maximal ideal in the set of associated primes, see [27], [START_REF] Hà | Embedded associated primes of powers of square-free monomial ideals[END_REF], [START_REF] Nasernejad | On the Alexander dual of the path ideals of rooted and unrooted trees[END_REF] for more details. Unfortunately, little is known on this subject in literature. In this section, our aim is to detect that whether m ∈ Ass R (R/I) solely by using the elements of the unique minimal set of monomial generators of I, and without computing any minimal primary decomposition of I. It should be noted that the results of this section can be found in [START_REF] Nasernejad | Detecting the maximal-associated prime ideal of monomial ideals[END_REF].

The following lemma investigates the relation between the existence of the graded maximal ideal and the number of minimal generators. Proof. Suppose, on the contrary, that m < n. It follows from m ∈ Ass R (R/I) that there exists a monomial h in R such that m = (I : R h). Hence, one has x i h ∈ I for each i = 1, . . . , n. Note that n ≥ 2 since m ≥ 1. Because m < n, the Pigeonhole Principle implies that there exist positive integers 1 ≤ i, j ≤ n with i = j and 1 ≤ k ≤ m such that u k |x i h and u k |x j h. Accordingly, x i h = u k w and x j h = u k v for some monomials v, w ∈ R. This yields that x i u k v = x i x j h = x j u k w, and so x i v = x j w. Therefore, we get x i |w, and thus w = x i g for some monomial g in R. Consequently, one derives x i v = x j x i g, and hence v = x j g. This gives that 305

x j h = u k x j g, and so h = u k g. That is, h ∈ I, which is a contradiction since m = (I : R h). Therefore, we obtain m ≥ n, as claimed.

The proposition below says that any monomial ideal in a polymonial ring in two variables, which is generated by at least two minimal generators has the graded maximal ideal as an associated prime. Proof. Suppose that u i = x α i y β i with nonnegative integers α i and β i for each i = 1, . . . , m. Lexicographically order the u i 's such that u i < u j if either α i < α j or α i = α j and β i < β j . Then one may assume that u 1 < u 2 < • • • < u m . On account of the u i 's are minimal generators, one can conclude that α 1 < α 2 < • • • < α n and

β 1 > β 2 > • • • > β n .
Note that α 2 ≥ 1 and β 1 ≥ 1. Set h := x α 2 -1 y β 1 -1 . Want to show that h / ∈ I. On the contrary, assume that h ∈ I. This implies that there exists a positive integer 1 ≤ i ≤ m such that u i |h. Hence, α i ≤ α 2 -1 and β i ≤ β 1 -1, and so α i < α 2 and β i < β 1 . Accordingly, one must have i = 1 and i ≥ 2, which is a contradiction. Therefore, we get h / ∈ I. We now demonstrate that xh ∈ I and yh ∈ I. By virtue of xh = x α 2 y β 1 -1 = x α 2 y β 2 y β 1 -β 2 -1 and yh = x α 2 -1 y β 1 = x α 2 -α 1 -1 x α 1 y β 1 , we obtain xh ∈ I and yh ∈ I. This yields that m = (I : R h), and thus m ∈ Ass R (R/I), as desired.

The next theorem gives us a subtle tool for checking whether the set of associated primes contains the graded maximal ideal. (i) r it,j < r i j ,j for each t = j, where j = 1, . . . , n;

(ii) for each i ∈ {1, . . . , k} \ {i 1 , . . . , i n }, there exists a positive integer 1 ≤ t ≤ n such that r i,t ≥ r it,t .

Proof. We first assume that m ∈ Ass R (R/I). This implies that there exists a monomial h in R such that m = (I : R h). Let h = x s

• • • x r i t ,n n ∤ x s 1 1 • • • x sn n .
Hence, one has to derive r it,t > s t , and so r it,t = s t + 1. Accordingly, we get s t = r it,t -1 for each t = 1, . . . , n. Furthermore, by virtue of r it,j ≤ s j for each j ∈ {1, . . . , t -1, t + 1, . . . , n}, and also s j = r i j ,j -1, one has r it,j < r i j ,j . Consequently, the condition (i) is proved. This gives rise to that i 1 , . . . , i n are distinct. To establish the condition (ii), consider i ∈ {1, . . . , k} \ {i 1 , . . . , i n }. Due to h / ∈ I, we deduce that x r i,1

1 • • • x r i,n n ∤ x s 1 1 • • • x sn n .
This yields that there exists a positive integer 1 ≤ t ≤ n such that r i,t > s t . As s t = r it,t -1, we obtain r i,t > r it,t -1, and thus r i,t ≥ r it,t . Therefore, the condition (ii) holds.

Conversely, assume that there are distinct integers i 1 , . . . , i n ∈ {1, . . . , k} such that the conditions (i) and (ii) hold. Set h := x s 1 1 • • • x sn n , where s t = r it,t -1 for each t = 1, . . . , n. Our aim is to show that m = (I : R h). To do this, we first prove that h / ∈ I. Clearly, for each t = 1, . . . , n, one has s t = r it,t -1 < r it,t . Moreover, the condition (ii) gives that, for each i ∈ {1, . . . , k} \ {i 1 , . . . , i n }, there exists a positive integer 1 ≤ α ≤ n such that r i,α ≥ r iα,α , and hence r i,α > s α . Therefore, for each i = 1, . . . , k, we get x

r i,1 1 • • • x r i,n n
∤ h, and so h / ∈ I. In order to complete the proof, one needs only prove that hx t ∈ I for each t = 1, . . . , n. To accomplish this, fix 1 ≤ t ≤ n. Remembering the assumption that s j = r i j ,j -1 for each j = 1, . . . , n, one has the following equalities

hx t =x s 1 1 • • • x s t-1 t-1 x st+1 t x s t+1 t+1 • • • x sn n =x r i 1 ,1 -1 1 • • • x r i t-1 ,t-1 -1 t-1
x r i t ,t t

x r i t+1 ,t+1 -1 t+1

• • • x r in,n -1 n .
Moreover, based on the condition (i), for each j = 1, . . . , n, we have r it,j < r i j ,j for each t = j, and so r it,j ≤ r i j ,j -1. This yields that x

r i t ,1 1 • • • x r i t ,n n
|hx t , and thus hx t ∈ I. Therefore, one can conclude that m ∈ Ass R (R/I), as required.

We now give the following example which clarifies the main aim of Theorem 10.1.3.

In the subsequent lemma, we assume that the number of minimal generators and the number of variables are the same, and provide a criterion for the graded maximal ideal to be an associate prime. • • • x rn,n n }. Then m ∈ Ass R (R/I) if and only if there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , n} such that r it,j < r i j ,j for each t = j, where j = 1, . . . , n.

Proof. If m ∈ Ass R (R/I), in view of Theorem 10.1.3, one derives that there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , n} such that r it,j < r i j ,j for each t = j, where j = 1, . . . , n. And the argument is done.

Conversely, assume that there are distinct integers i 1 , . . . , i n ∈ {1, . . . , n} such that r it,j < r i j ,j for each t = j, where j = 1, . . . , n. Since i 1 , . . . , i n are distinct, one has {i 1 , . . . , i n } = {1, . . . , n}, and hence the condition (ii) in Theorem 10.1.3 is superfluous. Therefore, m ∈ Ass R (R/I), and the proof is over.

In order to verify the main result of this section, one relies on the following corollary which is an immediate consequence of Lemma 10.1.5. (ii) C i ∩ C j = ∅ for all i = j.

Proof. We first assume that m ∈ Ass R (R/I). Inspired by Lemma 10.1.5, we get there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , n} such that r it,j < r i j ,j for each t = j, where j = 1, . . . , n. Because i 1 , . . . , i n are distinct, this gives that i j is the unique number among {i 1 , . . . , i n } such that r i j ,j = max{r i 1 ,j , . . . , r in,j } for each j = 1, . . . , n. Also, by virtue of {i 1 , . . . , i n } = {1, . . . , n}, one derives that {r i 1 ,j , . . . , r in,j } = {r 1,j , . . . , r n,j }. Consequently, the conditions (i) and (ii) are proved.

To prove the converse, suppose that the conditions (i) and (ii) hold. It follows that there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , n} with i j ∈ C j for all j = 1, . . . , n, such that r it,j < r i j ,j for each t = j, where j = 1, . . . , n. Now, Lemma 10.1.5 implies that m ∈ Ass R (R/I), as required.

We are now in a position to prove the main result of this paper in the next theorem. In fact, it yields a necessary and sufficient condition whether the graded maximal ideal appears in the set of associated prime ideals.

to the height of I, that is, β 1 (I) = ht(I). Note that edge ideals of bipartite graphs form a large class of König ideals.

Results on the embedded associated prime ideals

In what follows, we focus on the embedded associated primes of monomial ideals. The main results of this section are Corollary 10.2.4 and Theorem 10.2.6. To achieve this goal, we start with the following proposition, which will be used in proving Theorem 10.2.2. This completes our argument.

The next theorem enables us to argue on the associated prime ideals of a monomial ideal which can be divided into two monomial ideals generated in disjoint sets of variables.

Q i + I 2 = s i=1 (Q i + I 2 ).
In addition, it is routine to check that there exists the following R-monomorphism θ : R/I →

s i=1 R/(Q i + I 2 ),
Proof. Since f and g are two I t -corner-elements for some positive integer t, this implies that f / ∈ I t (respectively, g / ∈ I t ) and x i f ∈ I t (respectively, x i g ∈ I t ) for each i = 1, . . . , n. We thus have x i ∈ (I t : f ) (respectively, x i ∈ (I t : g)) for each i = 1, . . . , n. Accordingly, m ⊆ (I t : f ) (respectively, m ⊆ (I t : g)). Because f / ∈ I t (respectively, g / ∈ I t ), this yields that m = (I t : f ) (respectively, m = (I t : g)). On the contrary, assume that f | g. This gives that g = f h for some monomial h in R. We thus get m = ((I t : f ) : h), and so m = R, a contradiction. A similar discussion shows that g ∤ f , and the proof is complete.

The following lemma will be useful in the proof of Corollary 10.2.14. Lemma 10.2.12. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. Let p = (I t : h) for some positive integer t and some monomial h in R such that x i ∤ h for some 1 ≤ i ≤ n. Then p \ x i = ((I \ x i ) t : h), and so p \ x i ∈ Ass(I \ x i ) t .

Proof. We show that p \ x i = ((I \ x i ) t : h). To do this, we first prove that p \ x i ⊆ ((I \ x i ) t : h). To see this, take x j ∈ p \ x i . Since p \ x i ⊆ p, this implies that x j ∈ p, and so x j h ∈ I t . Because x j = x i and x i ∤ h, one can conclude that x i ∤ x j h. Hence, x j h ∈ I t \x i . Since I t \x i = (I \x i ) t , we get x j h ∈ (I \x i ) t , and thus x j ∈ ((I \x i ) t : h). We therefore have p \ x i ⊆ ((I \ x i ) t : h). Conversely, let v / ∈ p \ x i . It suffices to show that v / ∈ ((I \ x i ) t : h). Suppose, on the contrary, that v ∈ ((I \ x i ) t : h). Then vh ∈ (I \ x i ) t . Since v ∈ p and v / ∈ p \ x i , one must have v = x ℓ i a for some positive integer ℓ and some monomial a / ∈ p. Hence, x ℓ i ah ∈ (I \ x i ) t , and so ah ∈ ((I \ x i ) t : x ℓ i ). It follows readily from Lemma 2.1.12 that ah ∈ (I \ x i ) t . Since I \ x i ⊆ I, we obtain ah ∈ I t , and so a ∈ (I t : h). Accordingly, one has a ∈ p, which is a contradiction. We thus have v / ∈ ((I \ x i ) t : h), and so ((I \ x i ) t : h) ⊆ p \ x i . Consequently, p \ x i = ((I \ x i ) t : h), and hence p \ x i ∈ Ass(I \ x i ) t .

We now express two main results of this subsection in the subsequent corollaries.

Corollary 10.2.13. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. If p = (I t : h) for some positive integer t and some monomial h in R such that p \ x i / ∈ Ass(I \ x i ) t for some 1 ≤ i ≤ n, then x i | h.

Corollary 10.2.14. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K and z be an I t -corner-element for some positive integer t such that m \ x i / ∈ Ass(I \ x i ) t for some 1 ≤ i ≤ n. Then x i | z.

The following question, which posed in [START_REF] Rajaee | Superficial ideals for monomial ideals[END_REF]Question 4.13], asks if Lemma 10.2.12 can be strengthened further.

Normally torsion-freeness and normality under polarization

We continue our argument with two open questions which are devoted to the normally torsion-freeness and normality of monomial ideals under the polarization operator. We should recall the definition of polarization operator as follows.

Definition 11.2.1. (see Definition 3.2.37) The process of polarization replaces a power x t i by a product of t new variables x (i,1) • • • x (i,t) . We call x (i,j) a shadow of x i . We will use I t to denote the polarization of I t , will use S t for the new polynomial ring in this polarization, and will use w to denote the polarization in S t of a monomial w in R = K[x 1 , . . . , x n ]. The depolarization of an ideal in S t is formed by setting x (i,j) = x i for all i, j.

In general, the polarization operator is more complicated in comparison with the other monomial operators. Let I be a non-square-free monomial ideal in the polynomial ring K[x 1 , . . . , x n ]. A natural question is about whether there is some relation between the normality (respectively, normally torsion-freeness) of a monomial ideal I and the normality (respectively, normally torsion-freeness) of its polarization I?

A partial answer to this question is that the normality of I does not necessarily imply that I is normal. This assertion is supported in the following example which we give a monomial ideal I which is not normal while its polarization I is normal.

Example 11.2.2. Let I = (x 1 x 2 2 x 3 , x 2 x 2 3 x 4 , x 3 x 2 4 x 5 , x 4 x 2 5 x 1 , x 5 x 2 1 x 2 ) be a monomial ideal in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 ]. In view of Example 3.2.41, one can conclude that I is normal. On the other hand, we show that I is not integrally closed, and a fortiori is not normal. To see this, let α := x 1 x 2 x 3 x 4 x 5 ; one derives α 2 ∈ I 2 but α / ∈ I; thus, α ∈ I \ I.

The above example shows that the normality of I and the normality of I are not equivalent. As an application of Theorem 7.2.5, we give in the following example a monomial ideal I such that both I and I are not normal.

Example 11.2.3. Consider the following monomial ideal I = (x 2 1 x 2 x 3 , x 4 1 x 2 x 4 , x 2 1 x 3 x 5 , x 2 1 x 4 x 6 , x 2 1 x 5 x 6 , x 2 x 3 x 6 , x 2 x 4 x 5 , x 2 x 5 x 6 , x 3 x 4 x 5 , x 3 x 4 x 6 ), in the polynomial ring R = K[x 1 , . . . , x 6 ]. To illustrate the non-normality of I, our strategy is to use Theorem 7.2.5. To do this, take the monomial ideal L ⊂ R as follows L = (x 1 x 2 x 3 , x 2 1 x 2 x 4 , x 1 x 3 x 5 , x 1 x 4 x 6 , x 1 x 5 x 6 , x 2 x 3 x 6 , x 2 x 4 x 5 , x 2 x 5 x 6 , x 3 x 4 x 5 , x 3 x 4 x 6 ). 323

Also, suppose that W : {x 1 , . . . , x 6 } → N is the weight over R with W (x 1 ) = 2, and W (x i ) = 1 for all i = 2, . . . , 6. This implies that I = L W . By Theorem 7.2.5, it suffices to verify that L is not normal. To see this, set α := x 1 x 2 x 3 x 4 x 5 x 6 . Due to α 2 = (x 1 x 2 x 3 )(x 1 x 5 x 6 )(x 2 x 4 x 5 )(x 3 x 4 x 6 ) ∈ (L 2 ) 2 , we get α ∈ L 2 . Moreover, it is routine to check that α / ∈ L 2 , and hence α ∈ L 2 \ L 2 . This yields that L is not normal, and so I is not normal. To complete our argument, we must demonstrate that I is not normal. Without loss of generality, one can write I = (x 1 x 2 x 3 x 7 , x 1 x 2 x 4 x 7 x 8 x 9 , x 1 x 3 x 5 x 7 , x 1 x 4 x 6 x 7 , x 1 x 5 x 6 x 7 , x 2 x 3 x 6 , x 2 x 4 x 5 , x 2 x 5 x 6 , x 3 x 4 x 5 , x 3 x 4 x 6 ), in the polynomial ring S = K[x 1 , . . . , x 9 ]. Let α := x 1 x 2 x 3 x 4 x 5 x 6 x 7 . It is routine to see that α / ∈ ( I) 2 , while α 2 ∈ ( I) 4 ; hence, α ∈ ( I) 2 \( I) 2 . Thus, ( I) 2 is not integrally closed, and so I is not normal.

We now turn our attention to the relation between the persistence property (respectively, the strong persistence property) of a non-square-free monomial ideal I and its polarization, that is, I Pol . This counterexample can be found in [START_REF] Nasernejad | Some counterexamples for (strong) persistence property and (nearly) normally torsion-freeness[END_REF]. Question 11.2.4. Let I be a non-square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K, and I Pol denote its polarization. Let I satisfy the persistence property (respectively, strong persistence property). Then can we deduce that I Pol satisfies the persistence property (respectively, strong persistence property)? Solution. Suppose the following monomial ideal I =(x 2 4 x 5 x 1 , x 2 x 3 x 6 , x 2 x 3 x 1 , x 2 x 4 x 5 , x 2 x 4 x 1 , x 2 x 5 x 6 , x 3 x 4 x 5 , x 3 x 4 x 6 , x 3 x 5 x 1 , x 4 x 6 x 1 , x 2 5 x 6 x 1 ), in the polynomial ring R = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ] over a field K. Then the polarization of I is the following square-free monomial ideal, I Pol =(x 7 x 4 x 5 x 1 , x 2 x 3 x 6 , x 2 x 3 x 1 , x 2 x 4 x 5 , x 2 x 4 x 1 , x 2 x 5 x 6 , x 3 x 4 x 5 , x 3 x 4 x 6 , x 3 x 5 x 1 , x 4 x 6 x 1 , x 8 x 5 x 6 x 1 ), in the polynomial ring S = K[x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 , x 8 ] over a field K. By using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], we detect that (x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 ) ∈ Ass(S/(I Pol ) 2 ) \ Ass(S/(I Pol ) 3 ), and also ((I Pol ) 3 : S I Pol ) = (I Pol ) 2 . This shows that I Pol does not satisfy both of the persistence property and strong persistence property. To finish the argument,

Closed neighborhood ideals and dominating ideals

In this section, we present three open questions, which are related to closed neighborhood ideals and dominating ideals of graphs.

Let G be a simple finite graph. We are first going to give an argument on the relation between monomial prime ideals associated to Ass(DI(G) s ) for some s ≥ 1, and monomial prime ideals associated to Ass(DI(G p ) s ), where G p stands for the induced graph on p, i.e., the graph with vertex set p, and the following edge set

E(G p ) = {e ∈ E(G) | e ⊆ p}.
Francisco, Hà, and Van Tuyl, in [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11], showed that if H is a finite simple hypergraph on V = {x 1 , . . . , x n } with cover ideal J(H) ⊆ R = k[x 1 , . . . , x n ], then for all d ≥ 1, P = (x i 1 , . . . , x ir ) ∈ Ass(R/J(H) d ) ⇔ P = (x i 1 , . . . , x ir ) ∈ Ass(k[P ]/J(H P ) d ),

where k[P ] = k[x i 1 , . . . , x ir ], and H P is the induced hypergraph of H on the vertex set P = {x i 1 , . . . , x ir } ⊆ V . Here, we provide a counterexample which shows that one cannot mimic this result for dominating ideals of simple finite graphs. To do this, let G = (V (G), E(G)) be a simple finite graph with the vertex set V (G) = {x 1 , x 2 , x 3 , x 4 , x 5 , x 6 } and the following edge set E(G) = {{x 1 , x 2 }, {x 1 , x 3 }, {x 1 , x 6 }, {x 2 , x 4 }, {x 2 , x 5 }, {x 3 , x 4 }, {x 3 , x 5 }, {x 4 , x 6 }}.

One can easily see that the dominating ideal of G is given by DI(G) =(x 1 , x 2 , x 3 , x 6 ) ∩ (x 1 , x 2 , x 4 , x 5 ) ∩ (x 1 , x 3 , x 4 , x 5 ) ∩ (x 2 , x 3 , x 4 , x 6 ) ∩(x 2 , x 3 , x 5 ) ∩ (x 1 , x 4 , x 6 ).

Take p := (x 1 , x 2 , x 3 , x 6 ) ∈ Ass(DI(G) 2 ). It follows from the definition that V (G p ) = {x 1 , x 2 , x 3 , x 6 } and E(G p ) = {{x 1 , x 2 }, {x 1 , x 3 }, {x 1 , x 6 }}. In particular, the dominating ideal of G p is given by DI(G p ) = (x 1 , x 2 ) ∩ (x 1 , x 3 ) ∩ (x 1 , x 6 ) = (x 1 , x 2 x 3 x 6 ).

We can conclude from Theorem 5.3.4 that DI(G p ) is normally torsion-free, and so p / ∈ Ass(DI(G p ) 2 ). On the other hand, q := (x 1 , x 6 ) ∈ Ass(DI(G p ) 2 ), while q / ∈ Ass(DI(G) 2 ).

We now discuss the nature of G p . To accomplish this, we first recall the definition of partial t-cover ideals which has already been introduced in [17].

. , x n ]

be a monomial of degree ℓ ≤ d, with 1 ≤ j 1 ≤ j 2 ≤ • • • ≤ j ℓ ≤ n. We say that u is a vector-spread monomial of type t, or simply a t-spread monomial, if j i+1 -j i ≥ t i for all i = 1, . . . , ℓ -1.

Note that any variable x j is t-spread, and we also assume that u = 1 is t-spread. In addition, we say that a monomial ideal I ⊂ R = K[x 1 , . . . , x n ] is a vector-spread monomial ideal of type t, or simply a t-spread monomial ideal, if all monomials u ∈ G(I) are t-spread. Definition 11.4.2. ( [START_REF] Ficarra | Vector-spread monomial ideals and Eliahou-Kervaire type resolutions[END_REF]Definition 2.2]) With the notation above, if t i = t ≥ 0 for all i = 1, . . . , d -1, then we say that a t-spread monomial (ideal) u ∈ Mon(R), (I ⊆ R), is called an ordinary or uniform t-spread monomial (ideal), which has been introduced in [START_REF] Ene | t-spread strongly stable monomial ideals[END_REF]. Definition 11.4.3. A t-spread monomial ideal I of R is called a vector-spread Borel ideal of type t or simply a t-spread strongly stable ideal if for any t-spread monomial u ∈ I, and all j < i such that x i divides u and x j (u/x i ) is t-spread, then x j (u/x i ) ∈ I. For t = 0 = (0, 0, . . . , 0) (t = 1 = (1, 1, . . . , 1)), one obtains the classical notion of strongly stable (square-free strongly stable) ideal, see [START_REF] Herzog | Monomial Ideals[END_REF]Definition 4.2.2] (Problem 6.9 on page 112 in [START_REF] Herzog | Monomial Ideals[END_REF]). Definition 11.4.4. A t-spread monomial ideal I of R is called a t-spread principal Borel ideal if there exists a t-spread monomial u ∈ G(I) such that I is the smallest t-spread strongly stable ideal which contains u. We denote it as I = B t (u).

It should be noted that for a t-spread monomial u = x i 1

A comparison between clutters and complement clutters

To formulate the following question, we need to recall the notion of a clutter. A clutter (or simple hypergraph) C with vertex set X = {x 1 , . . . , x n } is a family of subsets of X, called edges, none of which is included in another, see [START_REF] Villarreal | Monomial algebras[END_REF]Definition 6.3.33] for more details. The edge ideal of a clutter is defined in [START_REF] Villarreal | Monomial algebras[END_REF]Definition 6.3.35], and the cover ideal of a clutter can be defined as the ideal of all monomials M such that given any edge e of C there is some variable x i such that x i ∈ e and x i |M . Note that the vertices of these clutters become the variables of the ring in which the edge ideal and cover ideal are alocated. Given a clutter C on To illuminate the purpose of the above question, we give an example of a clutter C where both I(C) and I(C c ) do not have the strong persistence property.

To do this, consider the graph which has been presented in Proposition 3.2.5. Set F := J(H 4 ). By using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], F is given by F = (x 1 x 3 x 6 x 8 x 9 x 10 x 11 x 12 , x 2 x 4 x 5 x 7 x 9 x 10 x 11 x 12 , x 1 x 2 x 4 x 5 x 7 x 10 x 11 x 12 , x 2 x 3 x 5 x 6 x 8 x 9 x 11 x 12 , x 1 x 2 x 3 x 6 x 8 x 9 x 11 x 12 , x 2 x 4 x 5 x 6 x 7 x 9 x 11 x 12 , x 1 x 3 x 6 x 7 x 8 x 9 x 10 x 12 , x 2 x 3 x 5 x 7 x 8 x 9 x 10 x 12 , x 2 x 3 x 4 x 5 x 7 x 9 x 10 x 12 , x 1 x 3 x 4 x 5 x 6 x 7 x 10 x 12 , x 1 x 2 x 3 x 4 x 5 x 7 x 10 x 12 , x 1 x 3 x 4 x 6 x 8 x 9 x 10 x 11 , x 1 x 2 x 4 x 5 x 7 x 8 x 10 x 11 , x 1 x 3 x 4 x 5 x 6 x 8 x 10 x 11 , x 1 x 2 x 4 x 6 x 7 x 8 x 9 x 11 , x 1 x 2 x 3 x 4 x 6 x 8 x 9 x 11 ).

It has already been shown in [START_REF] Kaiser | Replication in critical graphs and the persistence of monomial ideals[END_REF] that F does not satisfy the persistence property (and hence does not satisfy the strong persistence property either). One can show, by using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], that in the polynomial ring R = K[x 1 , . . . , x 12 ] over a field K, m = (x 1 , . . . , x 12 ) ∈ Ass R (R/F 3 ) \ Ass R (R/F 4 ) and (F 4 : R F ) = F 3 . Now, we construct the clutter C on {x 1 , . . . , x 12 } whose edge ideal is F , that is, F = I(C), as follows C := {{x 1 , x 3 , x 6 , x 8 , x 9 , x 10 , x 11 , x 12 }, {x 2 , x 4 , x 5 , x 7 , x 9 , x 10 , x 11 , x 12 }, {x 1 , x 2 , x 4 , x 5 , x 7 , x 10 , x 11 , x 12 }, {x 2 , x 3 , x 5 , x 6 , x 8 , x 9 , x 11 , x 12 }, {x 1 , x 2 , x 3 , x 6 , x 8 , x 9 , x 11 , x 12 }, {x 2 , x 4 , x 5 , x 6 , x 7 , x 9 , x 11 , x 12 }, {x 1 , x 3 , x 6 , x 7 , x 8 , x 9 , x 10 , x 12 }, {x 2 , x 3 , x 5 , x 7 , x 8 , x 9 , x 10 , x 12 }, {x 2 , x 3 , x 4 , x 5 , x 7 , x 9 , x 10 , x 12 }, {x 1 , x 3 , x 4 , x 5 , x 6 , x 7 , x 10 , x 12 }, {x 1 , x 2 , x 3 , x 4 , x 5 , x 7 , x 10 , x 12 }, {x 1 , x 3 , x 4 , x 6 , x 8 , x 9 , x 10 , x 11 }, {x 1 , x 2 , x 4 , x 5 , x 7 , x 8 , x 10 , x 11 }, {x 1 , x 3 , x 4 , x 5 , x 6 , x 8 , x 10 , x 11 }, {x 1 , x 2 , x 4 , x 6 , x 7 , x 8 , x 9 , x 11 }, {x 1 , x 2 , x 3 , x 4 , x 6 , x 8 , x 9 , x 11 }}.

Then, F = I(C) does not satisfy the strong persistence property. On the other hand, one can deduce from the definition that the complement clutter of C, that is C c , is as follows C c := {{x 2 , x 4 , x 5 , x 7 }, {x 1 , x 3 , x 6 , x 8 }, {x 3 , x 6 , x 8 , x 9 }, {x 1 , x 4 , x 7 , x 10 }, {x 4 , x 5 , x 7 , x 10 }, {x 1 , x 3 , x 8 , x 10 }, {x 2 , x 4 , x 5 , x 11 }, {x 1 , x 4 , x 6 , x 11 }, {x 1 , x 6 , x 8 , x 11 }, {x 2 , x 8 , x 9 , x 11 }, {x 6 , x 8 , x 9 , x 11 }, {x 2 , x 5 , x 7 , x 12 }, {x 3 , x 6 , x 9 , x 12 }, {x 2 , x 7 , x 9 , x 12 }, {x 3 , x 5 , x 10 , x 12 }, {x 5 , x 7 , x 10 , x 12 }}. This implies that the edge ideal of I(C c ) is given by I(C c ) = (x 2 x 4 x 5 x 7 , x 1 x 3 x 6 x 8 , x 3 x 6 x 8 x 9 , x 1 x 4 x 7 x 10 , x 4 x 5 x 7 x 10 , x 1 x 3 x 8 x 10 , x 2 x 4 x 5 x 11 , x 1 x 4 x 6 x 11 , x 1 x 6 x 8 x 11 , x 2 x 8 x 9 x 11 , x 6 x 8 x 9 x 11 , x 2 x 5 x 7 x 12 , x 3 x 6 x 9 x 12 , x 2 x 7 x 9 x 12 , x 3 x 5 x 10 x 12 , x 5 x 7 x 10 x 12 ). By using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], one can check that (I(C c ) 4 : R I(C c )) = I(C c ) 3 and also m = (x 1 , . . . , x 12 ) ∈ Ass R (R/I(C c ) 3 ) \ Ass R (R/I(C c ) 4 ), that is, I(C c ) does not satisfy the strong persistence property and the persistence property.

Matroidal path ideals of graphs

It is known from [START_REF] Villarreal | Rees cones and monomial rings of matroids[END_REF]Proposition 3.11] that polymatroidal ideals are normal. In [START_REF] Restuccia | On the normality of monomial ideals of mixed products[END_REF]Corollary 2.11], it is shown that the t-path ideals of complete bipartite graphs are normal. In general, for a given t ≥ 3, one can find graphs that are not complete r-partite, but their t-path ideal I t (G) is matroidal. As a very simple example, let G itself be a path on t vertices. Then I t (G) is matroidal because it is a principal ideal, while G is not complete r-partite.

Theorems 9.3.6 and 9.3.8 together show that if I 2 (G) is matroidal, then I t (G) is also matroidal for all t ≥ 2. Therefore, it is natural to ask the following question: If I t (G) is matroidal for some t, then is it true that I k (G) is also matroidal for all k ≥ t? The following example shows that it is not true in general. Then direct computation gives that I 3 (G) = x 4 J = (x 1 x 2 x 4 , x 1 x 3 x 4 , x 1 x 4 x 5 , x 1 x 4 x 6 , x 2 x 3 x 4 , x 2 x 4 x 5 , x 2 x 4 x 6 , x 3 x 4 x 5 , x 3 x 4 x 6 , x 4 x 5 x 6 ), where J = (x 1 x 2 , x 1 x 3 , x 1 x 5 , x 1 x 6 , x 2 x 3 , x 2 x 5 , x 2 x 6 , x 3 x 5 , x 3 x 6 , x 5 x 6 ). It is straightforward to check that the ideal J is indeed the edge ideal of the complete graph on the vertex set {x 1 , x 2 , x 3 , x 5 , x 6 }, and by Theorem 9.3.6, J is matroidal. This shows that I 3 (G) is also matroidal because it is the product of two matroidal ideals, namely (x 4 ) and J. Now, set u := x 1 x 2 x 3 x 4 and v := x 3 x 4 x 5 x 6 . Then u, v ∈ I 4 (G), but both x 5 (u/x 1 ) and x 6 (u/x 1 ) do not belong to G. Hence, I 4 (G) is not matroidal.

In view of Theorem 9.3.8 and Example 11.6.1, we close this argument with the following questions. Question 11.6.2. For a given t > 2, characterize the graphs for which I t (G) is matroidal.

Question 11.6.3. For which graphs it is true that if I t (G) is matroidal for some t, then I k (G) is also matroidal for all k ≥ t?

A minimal counterexample to the Conforti-

Cornuéjols conjecture

We terminate this chapter with the following questions and conjecture which has been stated in [START_REF] Francisco | Powers of square-free monomial deals and combinatorics[END_REF]. To accomplish this, we recall the notion of minor of a square-free

Index

Lemma 2 . 1 . 1 .

 211 Let p 1 , . . . , p m be nonzero monomial prime ideals of R. Then there exists a monomial ideal I of R such that Ass R (R/I) = {p 1 , . . . , p m }.Proof. Without loss of generality, we may assume that |G(pi )| ≤ |G(p j )| for all 1 ≤ i < j ≤ m. Fix d ∈ N and put I := p d 1 ∩ p 2d 2 ∩ p 4d 3 ∩ • • • ∩ p 2 m-1 d m

Case 1 .p d 1 .with 2

 112 Let p 2d 2 ∩ p 4d 3 ∩ • • • ∩ p 2 m-1 d m ⊆ Then p 2 ∩ p 3 ∩ • • • ∩ p m ⊆ p 1 ,and thus there exists i ∈ N with 2 ≤ i ≤ m such that p i ⊆ p 1 , and so G(p i ) ⊆ G(p 1 ). Also, we have|G(p 1 )| ≤ |G(p i )|. Hence, G(p 1 ) = G(p i ), and therefore p 1 = p i , which is a contradiction. ≤ i ≤ m -1. For j = i + 1, . . . , m, since p i = p j and |G(p i )| ≤ |G(p j )|, there exists x ℓ j ∈ G(p j )\G(p i ). Also, for j = 1, . . . , i -1, we choose an arbitrary element x ℓ j in G(p j ). Now, set v := m j=1,j̸ =i x 2 j-1 d ℓ j . Then v ∈ m j=1,j̸ =i p 2 j-1 d j , and so v ∈ p 2 i-1 d i . Therefore, there exists u ∈ G(p 2 i-1 d i ) which divides v. Let u = 2 i-1 d t=1

⊆ p 2 m-1 d m with m ≥ 2 .

 22 If 1 ≤ j ≤ m-1, then since p j = p m and |G(p j )| ≤ |G(p m )|, there exists x ℓ j ∈ G(p m )\G(p j ). Now, put v := v ∈ p 2 m-1 d m . Therefore, there exists u ∈ G(p 2 m-1 d m ) such that u divides v. Then deg(u) ≤ deg(v), and so 2 m-1 d ≤ d + 2d + • • • + 2 m-2 d. Hence, 2 m-1 ≤ 1 + 2 + • • • + 2 m-2, which is a contradiction.

Theorem 2 . 1 . 4 .

 214 Let A = {p 1 , . . . , p m } and B = {p ′ 1 , . . . , p ′ t } be two arbitrary sets of monomial prime ideals of R. Then there exist monomial ideals I and J of R with the following properties:(i) Ass R (R/I) = A ∪ B, Ass R (R/J) = B;and, (ii) I ⊆ J, Ass R (J/I) = A\B. Proof. Set D := A ∪ B. Then, by Proposition 2.1.3, there exist monomial ideals I and J of R such that I ⊆ J, Ass R (R/I) = D = A ∪ B, Ass R (R/J) = B, and Ass R (J/I) = D\B. Since D\B = A\B, we obtain Ass R (J/I) = A\B.

( i )

 i Ass R (R/I) = A, Ass R (R/J) = B; and,(ii) I ⊆ J, Ass R (J/I) = A\B. Proof. Set I := p 1 ∩ • • • ∩ p m . Since I = √ I, by [60, Corollary 1.2.5], we obtain I is a square-free monomial ideal and alsoMin(I) = Ass R (R/I) = {p 1 , . . . , p m } = A.Now, assume that is the set of all R-submodules R/I of the form K/I, where

Figure 2

 2 Figure 2

  primary decomposition of I * . In particular, Ass(S * /I * ) = {p * : p ∈ Ass(S/I)}. Lemma 2.2.11. ([13, Lemma 1.1]) Let I and J be monomial ideals in a polynomial ring S. Then (i) f ∈ I * if and only π(f ) ∈ I, for all f ∈ S * ; (ii) (I + J) * = I * + J * ; (iii) (IJ) * = I * J * ; (iv) (I ∩ J) * = I * ∩ J * ; (v) (I : J) * = (I * : J * );

Proposition 2 . 3 . 8 .

 238 [START_REF] Herzog | Monomial Ideals[END_REF] Proposition 1.2.2]) Let I and J be monomial ideals. Then (I : J) is a monomial ideal, and

Definition 2 . 3 . 12 .

 2312 [START_REF] Sharifan | Minimal path cover sets and monomial ideals[END_REF] Definition 3.1]) Let G be a graph on the vertex set [n] := {1, . . . , n}. Then the ideal

Theorem 2 . 3 . 13 .

 2313 Let T be a starlike tree on the vertex set {z} ∪ [n] with the center z. Then P I(T ) ∨ has the persistence property, where I ∨ means the Alexander dual ideal of I.

( 3 )

 3 If S is a multiplicatively closed subset of R disjoint from I, then (I S ) * = (I * ) S . (4) I ⊆ I * ⊆ (I) a .

Definition 2 . 3 . 28 .Corollary 2 . 3 . 29 .

 23282329 The ideal I * is called the Ratliff-Rush ideal associated with I or the Ratliff-Rush closure of I. A regular ideal I for which I * = I is called Ratliff-Rush closed. Suppose that I satisfies condition (♯). Then I is a Ratliff-Rush closed ideal.

Definition 2 . 4 . 1 .

 241 Let Φ be a family of ideals of R. Then we say that Φ has the persistence property if there exists a relation ≤ on Φ such that (Φ, ≤) is a partially ordered set with the following properties:(i) For all b ∈ Φ, the set ( a∈Φ Ass R (R/a)) ∩ V (b) is finite, where for an ideal c of R, V (c) is the set of prime ideals contain c.(ii) For all ideals a, b ∈ Φ with a ≤ b, we have Ass R (R/a) ⊆ Ass R (R/b).

::

  and so (I ∩ Q) a = I a . By virtue of I has the strong persistence property, we get (I k+1 : R I) = I k , and hence (I k+1 a Ra I a ) = I k a . Therefore, (L k+1 a Ra L a ) = L k a , and the proof is complete.

Theorem 3 . 1 . 11 .

 3111 Suppose that C 2n+1 is a cycle graph with vertex set [2n+1]. Then the cover ideal associated to C 2n+1 satisfies the strong persistence property. Proof. Label the vertices of C 2n+1 in counterclockwise order with 1, . . . , 2n + 1, as shown in the figure below.

Proposition 3 . 1 . 15 .

 3115 Suppose that C 2n+1 is a cycle graph on the vertex set [2n + 1],

Figure 3 .

 3 Figure 3.2.1: H 4

  (i) (I + J)(p) = I(p) + J(p); (ii) (IJ)(p) = I(p)J(p); (iii) (I ∩ J)(p) = I(p) ∩ J(p);(iv) (I : R J)(p) = (I(p) : R(p) J(p));

  , where Ass ∞ (I) denotes the stable set of associated prime ideals of I; (x) If p ∈ V * (I), then astab(I) ≥ astab(I(p)), where astab(I) denotes the index of stability for the associated prime ideals of I. Proof. (i) It is easy to see that if u|v (respectively, v|u), then u(p)|v(p) (respectively, v(p)|u(p)) for all u ∈ G(I) and v ∈ G(J). Because I(p) (respectively, J(p)) is generated by {u(p) : u ∈ G(I)} (respectively, {v(p) : v ∈ G(J)}), and by virtue of G((I + J)(p)) ⊆ G(I(p)) ∪ G(J(p)), one can conclude that (I + J)(p) = I(p) + J(p). (ii) It is straightforward to show that (uv)(p) = u(p)v(p) for all u ∈ G(I) and v ∈ G(J). Now, the desired conclusion follows immediately from the fact that G((IJ)(p)) ⊆ G(I(p))G(J(p)). (iii) On account of I(p) (respectively, J(p)) is generated by {u(p) : u ∈ G(I)} (respectively, {v(p) : v ∈ G(J)}), this implies that I(p)∩J(p) is generated by the set {lcm(u(p), v(p)) : for all u ∈ G(I) and v ∈ G(J)}. By virtue of I ∩ J is generated by {lcm(u, v) : for all u ∈ G(I) and v ∈ G(J)}, one has (I ∩ J)(p) is generated by {lcm(u, v)(p) : for all u ∈ G(I) and v ∈ G(J)}. Now, the claim follows readily from the fact that lcm(u, v)(p) = lcm(u(p), v(p)) for all u ∈ G(I) and v ∈ G(J).

(

  iv) Since (I : R J) = v∈G(J) (I : R v), and by virtue of statement (iii), one has (I : R J)(p) = v∈G(J) (I : R v)(p). We prove that (I : R v)(p) = (I(p) : R(p) v(p)) for all v ∈ G(J). Due to I(p) = u∈G(I) u(p)R(p), one derives the following equalities (I(p) : R(p) v(p)) = u∈G(I) u(p)R(p) : R(p) v(p) = u∈G(I) u(p) : R(p) v(p) .

R x 1

 1 ) : R (I : R x 1 )) = (I k L : R L). It follows now from [73, Corollary 6.8.7] that (I k L : R L) = I k , and by virtue of (I k L : R L) ⊆ (I k L : R L), one has (I k L : R L) ⊆ I k . On account of I is normal, one derives I k = I k , and so (I k+1 : R I) ⊆ I k . Due to I k ⊆ (I k+1 : R I), one can conclude that (I k+1 : R I) = I k , as claimed.

Proposition 3 . 2 . 44 .Corollary 3 . 2 . 45 .

 32443245 Let S be a Noetherian integral domain, and I be an integrally closed ideal of S. Then (I k+1 : S I k ) = I for all positive integers k. Proof. Fix k ∈ N. The claim follows readily from [131, Proposition 1.58] by setting L := I k . Let R = K[x 1 , . . . , x n ] be a polynomial ring over a field K, and I be a square-free monomial ideal of R. Then (I k+1 : R I k ) = I for all positive integers k. In particular, (I 2 : R I) = I.

I k+1 1 I k+1 2 : R I 1 ) = I k+1 2 (I k+1 1 : R I 1 )I k+1 2 : R I 2 ) = I k 1 (I k+1 2 :

 1212112212 I 1 and I 2 have the strong persistence property, then I 1 ∩ I 2 has the strong persistence property. Proof. Assume that I 1 and I 2 have the strong persistence property. Since I 1 and I 2 are generated by disjoint sets of variables, [70, Lemma 1.1] yields that I a 1 ∩ I b 2 = I a 1 I b 2 for any positive integers a, b. Let k ≥ 1. By observing Lemma 2.1.12, one can deduce that (and (I k 1 R I 2 ).

Now, we calculate I d+1 n:

 d+1 I n for a fixed d ≥ 1. Note that I d+1 n : I n is a monomial ideal and I d n ⊆ (I d+1 n : I n ) so we seek monomials in I d+1 n : I n but not in I d n . If d ≥ n -2, then both I d n and I d+1 n are all of R j in all degrees j for which they are nonzero (namely j ≥ nd for I d n and j ≥ n(d + 1) for I d+1 n ) by Lemma 3.3.11(1). From this, it follows that (I d+1 n

3 .

 3 Then, by Lemma 3.3.11 part (3), (I d n ) nd+i has gaps of size 1 and (I d+1 n ) n(d+1)+i is equal to R n(d+1)+i . The monomials in these gaps of size 1 are mapped into (I d+1 n ) n(d+1)+i = R n(d+1)+i when multiplied by any minimal generator of I n (and indeed, by any element of R n ), hence lie in (I d+1 n

  14, which is what we have observed. Furthermore, the d + 1 = 3 subdivision intervals of (I 2 6 ) 12 are each of cardinality d + i + 1 = 2 + 0 + 1 = 3 and the subdivision intervals of (I 2 6 ) 13 are each of cardinality d + i + 1 = 2 + 1 + 1 = 4, both in agreement with Lemma 3.3.11 part (2), increasing by 1 each time we increase i by 1. There are d = 2 gaps in (I 2 6 ) 12 , namely {x 9 y 3 , x 8 y 4 } and {x 4 y 8 , x 3 y 9 }, each of cardinality n -d -2 -i = 6 -2 -2 -0 = 2 and the largest exponent of y in these gaps is nβ -β -1 = 4, 9 for β = 1, 2, all in agreement with Lemma 3.3.11 part (3). The gaps in (I 2 6 ) 13 , namely {x 9 y 4 } and {x 4 y 9 }, each of cardinality n-d-2-i = 6-2-2-1 = 1 and the largest exponent of y in these gaps is again nβ -β -1 = 4, 9 for β = 1, 2, also in agreement with

Case ( 1 ):

 1 . p ∈ Min(I k ). Because Min(I k ) = Min(I), one has p ∈ Min(I). Now, Proposition 3.3.17 yields that I p = p p , and we thus have I k p = p k p and I k+1 p = p k+1 p . On the other hand, Corollary 3.3.3 implies that (p k+1 : R p) = p k . Hence, one derives that (p k+1 p Rp p p ) = p k p , and so (I k+1 p

1 : R I 1 ) = I k 1 ,:Corollary 3 . 3 . 20 .

 1113320 and hence ((I 1 ) k+1 p : Rp (I 1 ) p ) = (I 1 ) k p . We thus have (I k+1 p Rp I p ) = I k p . This completes the proof. Let G = (V (G), E(G)) be a finite simple connected graph such that J(G) has the strong persistence property and T be a tree. Also, let L = (V (L), E(L)) be the finite simple graph such that V (L) := V (G) ∪ V (T ) and E(L) := E(G) ∪ E(T ). Then, under each of the following cases, J(L) has the strong persistence property.

(

  this implies that u = ab/gcd(a, b) for some a ∈ G(J(G)) and b ∈ G(J(H)). Hence, one may consider the following cases: Case 1. a = f and b = g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). Hence, one has u = f g. Case 2. a = x v f and b = g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). We thus have u = x v f g. Case 3. a = f and b = x v g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). Therefore, one obtains u = x v f g. Case 4. a = x v f and b = x v g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). Due to gcd(a, b) = x v , one can conclude that u = x v f g.

  3.19 and (3.3.1).To demonstrate Theorem 3.3.25, one needs to employ the following lemma. Lemma 3.3.24. Let G = (V (G), E(G)) be a finite simple connected graph and H be a triangle graph such that |V (G) ∩ V (H)| = 2 and |E(G) ∩ E(H)| = 1. Let L = (V (L), E(L)) be the finite simple graph such that V (L) :

  3.19 and (3.3.2).

Case 4 .

 4 a = x w f and b = x v g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). Similarly to Case 2, one has u = x v x w f g. Case 5. a = x w f and b = x w g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). It follows form gcd(a, b) = x w that u = x w f g. Case 6. a = x w f and b = x v x w g, where f (respectively, g) is a square-free monomial in the variables Γ (respectively, Λ). Since gcd(a, b) = x w , we thus have u = x v x w f g.

  3.19 and (3.3.3) to obtain the last claim.

Question 3 . 3 . 28 .

 3328 Let G = (V (G), E(G)) be a finite simple connected graph and K n the complete graph of order n such that |V (G)∩V (K n )| = 3 and |E

Definition 4 . 1 . 1 .

 411 [START_REF] Villarreal | Monomial algebras[END_REF] Definition 4.3.22]) Let I be an ideal of a ring R and p 1 , . . . , p r the minimal primes of I. Given an integer n ≥ 1, the n-th symbolic power of I is defined to be the ideal

Remark 4 . 1 . 3 .

 413 Let I be a radical ideal in a Noetherian ring R. Then the following basic properties hold:

  and only if a ≥ b; (4) If a ≥ b, then I (a) ⊆ I (b) ; (5) For all a, b ≥ 1, I (a) I (b) ⊆ I (a+b) ; (6) I n = I (n) if and only if I n has no embedded primes. Now, we recollect the definition of symbolic strong persistence property of ideals. Definition 4.1.4. ([118, Definition 13]) Let I be an ideal in a commutative Noetherian ring R. Then, we say that I has the symbolic strong persistence property if (I (i+1) : R I (1) ) = I (i) for each i.

Questions 4 . 1 . 8 .

 418 (i) Does the persistence property imply the symbolic strong persistence property?

Theorem 4 . 1 . 11 .

 4111 Every square-free monomial ideal has the symbolic strong persistence property. Proof. Let I be a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K with Ass R (R/I) = {p 1 , . . . , p r }. On account of Ass R (R/I) = Min(I), it follows from Lemma 4.1.10 and Corollary 3.3.3 that ( r i=1 p k+1 i : R r i=1 p i ) = 114 r i=1 p k i for all k ≥ 1. In addition, [60, Proposition 1.4.4] implies that I (k) = r i=1 p k i for all k ≥ 1. Therefore, we have (I (k+1) : R I (1) ) = I (k) for all k ≥ 1, that is, I has the symbolic strong persistence property, as desired.

Corollary 4 . 1 . 13 .

 4113 Every normally torsion-free square-free monomial ideal has the strong persistence property. Proof. Let I be a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K with Ass R (R/I) = {p 1 , . . . , p r }. In the light of Theorem 4.1.11, one has (I (k+1) : R I (1) ) = I (k) for all k ≥ 1. On the other hand, [60, Theorem 1.4.6] implies that I (k) = I k for all k ≥ 1. Cosequently, by Theorem 4.1.11, we get (I k+1 : R I) = I k for all k ≥ 1, that is, I has the strong persistence property.

Proposition 4 . 2 . 1 .

 421 Let I, J, and L be monomial ideals in a polynomial ring R = K[x 1 , . . . , x n ] over a field K, and w a weight over R. Then (I : R J) = L if and only if

Theorem 4 . 2 . 3 .

 423 Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K, and W a weight over R. Then I has the symbolic strong persistence property if and only if I W has the symbolic strong persistence property.Proof. We can combine together Proposition 4.2.1 and Lemma 4.2.2 to obtain the claim.

Theorem 4 . 2 . 7 .

 427 Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K such that I has the symbolic strong persistence property. Then I(p) has the strong persistence property for all p ∈ Min(I), and hence has the symbolic strong persistence property.Proof. Fix k ≥ 1 and p ∈ Min(I). The assumption implies that (I (k+1) : R I (1) ) = I(k) . This yields that (I (k+1) : R I(1) )(p) = I (k) (p). It follows now from Lemma 3.2.15(iv) that (I (k+1) (p) : R(p) I(1) (p)) = I (k) (p). Since p ∈ Min(I), by Lemma 3.2.15(vii), we obtain I (s) (p) = I s (p) for all s. Thus, we get (I k+1 (p) : R(p) I(p)) = I k (p). On account of Lemma 3.2.15(ii), we get ((I(p)) k+1 : R(p) I(p)) = (I(p)) k . Therefore, I(p) has the strong persistence property, and hence has the symbolic strong persistence property, as required.

x 5 ]

 5 over a field K. Then one can rapidly see that I is of clutter type. Note that I does not satisfy both the persistence property and strong persistence property since m = (x 1 , x 2 , x 3 , x 4 , x 5 ) ∈ Ass R (R/I) \ Ass R (R/I 2 ) and (I 2 : R I) = I. Question 4.2.10. Does every non-square-free monomial ideal of clutter type have the symbolic strong persistence property?

Theorem 4 . 2 . 11 .

 4211 Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. Then I has the symbolic strong persistence property if and only if I * has the symbolic strong persistence property.Proof. We first note that I = J if and only if I * = J * . To prove the forward implication, let I have the symbolic strong persistence property. Fix k ≥ 1. The assumption yields that (I (k+1) : R I (1) ) = I(k) . In view of[13, Lemma 1.1 and Corollary 1.4], one has the following equalities ((I * ) (k+1) : R * (I * ) (1) ) = ((I (k+1) ) * : R * (I (1) ) * ) = (I (k+1) : R I (1) ) * = (I (k) ) * = (I * ) (k) .

  of [13, Lemma 1.1 and Corollary 1.4], we get ((I * ) (k+1) : R * (I * )(1) ) = (I (k+1) : R I (1) ) * and (I * ) (k) = (I (k) ) * . This implies that (I (k+1) : R I (1) ) * = (I (k) ) * , and so (I (k+1) : R I (1) ) = I(k) . That is, I has the symbolic strong persistence property.

Corollary 4 . 3 . 3 .

 433 Let I be an ideal in a commutative Noetherian ring S. Let

Proof.

  To show the claim, it is sufficient to verify that (I k+1 : R I) ⊆ I k for all k. Fix k ≥ 1. Let Min(I) = {p 1 , . . . , p r }. Since Ass R (R/I s ) = Ass R (R/I) ∪ {q} and Min(I) = Ass R (R/I), we deduce from [133, Definition 4.3.22 and Proposition 4.3.25] and the assumption that

Lemma 4 . 3 . 5 .

 435 Let C 2n+1 be an odd cycle graph with V (C 2n+1 ) = {1, . . . , 2n + 1} and E(C) = {{i, i + 1} : i = 1, . . . , 2n + 1}, where 2n + 2 = 1. Then, for all s ≥ 2,

Theorem 4 . 3 . 8 .

 438 [START_REF] Hà | Symbolic powers of sums of ideals[END_REF] Theorem 3.4]

2 .

 2 ℓ ≤ b + 1. We thus have there exist u 1 ∈ G(I (ℓ) 1 ) and u 2 ∈ G(I (b+1-ℓ) 2 ) such that u 1 u 2 |m 1 m 2 . Consequently, u 1 |m 1 m 2 and u 2 |m 1 m 2 . Because gcd(u 1 , m 2 ) = 1 and gcd(u 2 , m 1 ) = 1, one has u 1 |m 1 and u 2 |m 2 . This implies that m 1 ∈ I (ℓ) 1 and m 2 ∈ I (b+1-ℓ) It follows from Remark 4.1.3, m 1 ∈ I (a 1 ) 1 \ I (a 1 +1) 1

Lemma 4 . 3 . 11 .

 4311 Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K, and h be a monomial in R. Also, let gcd(h, u) = 1 for all u ∈ G(I). Then I has the symbolic strong persistence property if and only if hI has the symbolic strong persistence property.

Definition 5 . 1 . 4 .

 514 Let R = K[x 1 , . . . , x n ] be a polynomial ring, and I be an ideal ofR. A monomial z in R is called an I-corner-element if z /∈ I and x i z ∈ I for each i = 1, . . . , n.

  or a path with length three such that V (T 1 ) = {f, v, w, g}, w ∈ L T , and E(T 1 ) = {(g, f ), (f, v), (v, w)}. Set G := T \ {w}. This gives that |V (G)| = |V (T )| -1. We now construct the finite simple hypergraph H corresponding to T , with V (H) = V (T ) and

  x 5 ]. One can easily see that J is the expansion of I with respect to the 5-tuple (2, 1, 2, 2, 3). Here and in the following, our aim is to show that I is normally torsion-free. To do this, consider the graph G, as shown in the figure below, on the vertex set V (G) := [5] = {1, 2, 3, 4, 5} and the following edge set E(G) := {{1, 2}, {1, 3}, {1, 4}, {3, 5}, {4, 5}}.

  Figure 5

Figure 6

 6 Figure 6

Theorem 5 . 3 . 3 .

 533 [START_REF] Simis | On the ideal theory of graphs[END_REF] Theorem 5.4]) Let R 1 and R 2 be polynomial rings over a field k, and let I 1 ⊂ R 1 and I 2 ⊂ R 2 be reduced ideals. Suppose the total rings of fractions of R 1 /I 1 and R 2 /I 2 are separably generated over k. If I 1 and I 2 are normally torsion-free, then I = (I 1 , I 2 ) is a reduced, normally torsion-free ideal of

  Hence, one can deduce that there exist monomials d 1 ∈ R 1 and d 2 ∈ R 2 such that p 1 = (I 1 : R 1 d 1 ) and p 2 = (I 2 : R 2 d 2 ). Thus, one gains the following equalities(I : R d 1 d 2 ) = (I 1 R + I 2 R : R d 1 d 2 ) = (I 1 : R d 1 d 2 ) + (I 2 : R d 1 d 2 ) = (I 1 : R 1 d 1 ) + (I 2 : R 2 d 2 ) = p 1 + p 2 = p.Therefore, p ∈ Ass R (R/I), and so I is normally torsion-free, as required.

3 . 10 . 7 . 5 . 3 . 6 .

 3107536 To demonstrate Proposition 5.3.8, one needs to apply Proposition 5.3.6 and Corollary 5.3.Proposition Let Q = (x ℓ 1 i 1 , . . . , x ℓr ir ) be an irreducible monomial ideal in a polymonial ring R = K[x 1 , . . . , x n ]. Then the following statements hold.

Lemma 5 . 3 . 9 . 1 ,

 5391 Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ], and W a weight over R. Then Ass R (R/I W ) = Ass R (R/I).Proof. Suppose that the weight W : {x 1 , . . . , x n } → N is given by w i = W (x i ) for each i = 1, . . . , n. Moreover, by virtue of [60, Theorem 1.3.1], assume that I = Q 1 ∩• • •∩Q s is the irredundant presentation of I. From now on, fix 1 ≤ i ≤ s, and let Q i = (x α 1 i 1 , . . . , x αr ir ) with i 1 , . . . , i r ∈ {1, . . . , n} and α 1 , . . . , α r be positive integers. It follows from the definition of weighted ideal that (Q i ) W = (x α 1 w 1 i . . . , x αrwr ir

Theorem 5 . 3 . 10 .

 5310 Let I be a monomial ideal of R, and W a weight over R. Then I is normally torsion-free if and only if I W is.

Lemma 5 . 3 . 11 .

 5311 Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] with G(I) = {u 1 , . . . , u m }, and h = x b 1 j 1

Theorem 5 . 3 . 15 .

 5315 then inspired by Lemma 3.2.20 and Theorem 5.3.12, one can easily conclude Theorem 5.3.15. Let I be a monomial ideal in R = K[x 1 , . . . , x n ], and 1 ≤ j ≤ n.

Proposition 5 . 4 . 3 .Theorem 5 . 4 . 4 .

 543544 Let G be a graph. Then DI(G) is the edge ideal of the transversal hypergraph of N (G).It is known that the neighborhood hypergraphs of trees are totally balanced, for example see [15, Example 3 on page 174]. We can now combine together Remark 5.4.2 and Proposition 5.4.3 to obtain the following theorem. In particular, it is an affirmative answer to [107, Question 2.8(i)] and[START_REF] Nasernejad | Dominating ideals and closed neighborhood ideals of graphs[END_REF] Question 2.20(i)]. The closed neighborhood ideals and dominating ideals of trees are normally torsion-free.

Theorem 5 . 4 . 8 .

 548 Let G be a strongly chordal graph. Then N (G) is totally balanced.

Theorem 5 . 4 . 9 .

 549 Let G be a strongly chordal graph. Then the following statements hold:(i) NI(G) is normally torsion-free. (ii) NI(G) is sequentially Cohen-Macaulay. (iii) DI(G) is normally torsion-free.

Theorem 5 . 4 . 10 .

 5410 Let I be a normally torsion-free square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] and v be a square-free monomial in R. LetG(I) = {u 1 , . . . , u m } such that gcd(v, u i ) divides gcd(v, u i+1 ) for all i = 1, . . . , m -1.Then the following statements hold: (i) I + vR is normally torsion-free.(ii) I + vR is nearly normally torsion-free. (iii) I + vR is normal. (iv) I + vR has the strong persistence property. (v) I + vR has the persistence property. (vi) I + vR has the symbolic strong persistence property. Proof. (i) To simplify the notation, set J :

  where j = 1, . . . , m}, and d := min{j : x λ | u j , where j = 1, . . . , m}.

Lemma 5 . 4 . 15 .

 5415 The dominating ideals of C 6 and C 9 are normally torsion-free.

17 for

 17 all s, where R

x 2 x 5

 5 ); I(m \ {x 4 }) = (x 7 , x 1 ) + (x 2 x 5 , x 2 x 6 , x 3 x 6 ); I(m \ {x 5 }) = (x 1 , x 2 ) + (x 3 x 6 , x 3 x 7 , x 4 x 7 ); I(m \ {x 6 }) = (x 1 , x 2 , x 3 ) + (x 4 x 7 );

1 = [ 1 , 3 ], A 2 = [ 4 , 6 ], and A 3 =

 1132463

Definition 6 . 3 . 10 .

 6310 t . To do this, we first recollect the definition of linear relation graph from [65, Definition 3.1]. Let I ⊂ R be a monomial ideal with G(I) = {u 1 , . . . , u m }. The linear relation graph Γ of I is the graph with the edge set E(Γ) = {{i, j} : there exist u k ,u l ∈ G(I) such that x i u k = x j u l }, and the vertex set V (Γ) = {i,j}∈E(Γ) {i, j}.

  and lim k→∞ depth(R(p)/I(p) k ) = 0.

Lemma 6 . 3 . 16 .

 6316 Let u = x a x b x n be a t-spread monomial andI = B t (u) ⊂ R = K[x i : x i ∈ supp(I)].Then the following statements hold.

  1, b], and c ′ ∈ A 3 = [2t + 1, n]. Since a > 1, we have |A 1 | > 1 and since b ≥ 2t + 1, we have |A 2 | > 1 and |A 3 | > 1.

4 . 3 . 6 . 4 . 4 .

 43644 CorollaryThe closed neighborhood ideals of star graphs are normally torsionfree.Proof. Proceed by induction on the number of vertices and use Lemma 6.4.3.

(

  iv) DI(H) has the strong persistence property. (v) DI(H) has the persistence property. (vi) Both DI(G) and DI(H) have the symbolic strong persistence property. Proof. Suppose that the cone H = C(G) is obtained by adding the new vertex w to G and joining every vertex of G to w. Using [122, Lemma 2.2] yields that DI(H) = DI(G) + (x w ). It follows now from [50, Lemma 3.4] that, for all s, (6.4.1) Ass(DI(H) s ) = {(p, x w ) : p ∈ Ass(DI(G) s )}. (i) Let DI(G) be normally torsion-free. Then the claim can be deduced from Theorem 5.3.4. Conversely, let DI(H) be normally torsion-free. By using Theorem 5.3.16, we obtain DI(G) is normally torsion-free. (ii) Based on (6.4.1), and by considering the fact that Min(DI(H)) = {(p, x w ) : p ∈ Min(DI(G))}, one can easily show this claim. (iii) One concludes this assertion by [2, Theorem 3.12]. (iv) By Definition 2.3.1, DI(H) is a unisplit monomial ideal. Hence, Theorems 2.3.10 and 2.3.17 imply that DI(H) has the strong persistence property.

Theorem 6 . 5 . 3 .

 653 [START_REF] Herzog | Persistence and stability properties of powers of ideals[END_REF] Theorem 3.3]) Let I ⊂ R = K[x 1 , . . . , x n ] be a monomial ideal generated in a single degree whose linear relation graph has r vertices and s connected components. Then

Remark 6 . 5 . 5 .

 655 Let n ≥ 4, and set I n := N I(C n ). Remark 6.5.4 leads us to the following:

Corollary 6 . 5 . 6 .

 656 Let n = 0(mod 3). Set I n := N I(C n ) ⊂ R = K[x 1 , . . . , x n ]. Then depth(R/I n-1 n ) = 0. In particular, m ∈ Ass(R/I n-1 n

Theorem 7 . 1 . 3 .

 713 Let I be a monomial ideal in the polynomial ring R = K[x 1 , . . . , x n ] with G(I) = {u 1 , . . . , u r }, v a square-free monomial in R such that gcd(v, u i ) = 1 for all i, and h ∈ I a monomial. If I is normal, then vI + hR is normal.

Theorem 7 . 1 . 7 .

 717 Let R = K[x 1 , . . . , x 2n+1 ] be a polynomial ring over a field K, C 2n+1 be the cycle graph on the vertex set {1, . . . , 2n + 1}, and I = J(C 2n+1 ) be the cover ideal associated to C 2n+1 . Then I is a normal square-free monomial ideal. Proof. First proof. Label the vertices of C 2n+1 in counterclockwise order with 1, . . . , 2n + 1, as shown in figure below.

  Fix t ∈ N, and assume that m

3 )

 3 By virtue of (7.1.2) and (7.1.3), along with Lemma 3.1.10, we obtain m ∈ I t . Hence, I t = I t , and therefore I is normal.Second proof. Similar to the first proof, label the vertices of C 2n+1 in counterclockwise order with 1, . . . , 2n + 1, as shown in figure above. Let X be the convex set in R 2n+1 defined by the inequalitiesl i + l i+1 ≥ d and Σ 2n+1 i=1 [(l i + l i+1 -d] ≥ d from Lemma 3.1.10, together with l i ≥ 0 for all i. Let Γ(I d ) be the exponent set of I d . Then Lemma 3.1.10 says that Γ(I d ) = N 2n+1 0 ∩ X. It is known [60, Corollary 1.4.3] that Γ(I d ) = N 2n+1 0 ∩ N P (I d ), where N P denotes the Newton polyhedron of an ideal. By definition, the Newton polyhedron of an ideal is the smallest convex set that contains the exponent set of the ideal. Therefore, N P (I d ) ⊆ X and we have Γ(I d ) = N 2n+1 0 ∩ N P (I d ) ⊂ N 2n+1 0 ∩ X = Γ(I d ), which implies that I d = I d .

Proof.

  It follows from I ⊆ I that I * ⊆ (I) * . Since I is integrally closed, the proof of Theorem 7.2.1 yields that (I) * is also integrally closed, and hence (I * ) ⊆ (I) * . Conversely, suppose that u ∈ (I) * for some monomial u in R * . Then π(u) ∈ I, and so (π(u)) k ∈ I k for some positive integer k. This implies that u k ∈ (I k ) * = (I * ) k , and thus u ∈ (I * ). Therefore, we get (I) * ⊆ (I * ), and the proof is done.

Theorem 7 . 2 . 4 .

 724 Let I be the integral closure of L 1 := (x a 1 1 , . . . , x an n ) in the polynomial ring R = K[x 1 , . . . , x n ]. Let J be the integral closure of the following monomial ideal

Corollary 7 . 3 . 5 .Corollary 7 . 3 . 6 .

 735736 Let R = K[x 1 , . . . , x 2n+1 ] be a polynomial ring over a field K, C 2n+1 be the cycle graph on the vertex set {1, . . . , 2n + 1}, and I = J(C 2n+1 ) ⊂ R be the cover ideal of C 2n+1 . Then I has the strong persistence property, and hence persistence property. Suppose that W 2n is a wheel graph of order 2n on the vertex set [2n]. Then J(W 2n ) has the strong persistence property, and hence persistence property.

Theorem 7 . 3 . 9 .

 739 Suppose that I is a square-free monomial ideal in the ring R = K[x 1 , . . . , x n ]. If I is normally torsion-free, then I \ x j is normally torsion-free square-free; hence, normal. Proof. It is enough to combine Theorem 5.3.16 and [60, Theorem 1.4.6].

Example 7 . 3 . 11 .

 7311 Consider the following monomial ideal

Using Theorem 7

 7 .3.1 and Remark 7.1.1, we get I \ x 5 and I \ x 6 are normal. Moreover, applying Theorem 7.1.3 verifies that I \ x 1 and I \ x 4 are normal. Finally, combining Theorem 7.1.3 and Remark 7.1.1 yields that I \ x 2 and I \ x 3 are normal.

Lemma 7 . 4 . 5 .Theorem 7 . 4 . 6 .

 745746 4.6 and 7.4.7. It is a straightforward application of[START_REF] Herzog | Monomial Ideals[END_REF] Theorem 1.4.2]. Suppose that I and J are two normal monomial deals in R such that gcd(u, v) = 1 for all u ∈ G(I) and v ∈ G(J). Then I ∩ J = IJ is normal. Let I be a normal monomial ideal of R, and J be a normal squarefree monomial ideal of R whose generators are pairwise relatively prime. Let h ∈ I a monomial and suppose gcd(u, v) = 1 for all u ∈ G(I) ∪ {h} and v ∈ G(J). Then L := JI + hR is normal.

Figure 7 . 4 . 1 : H 4

 7414 Figure 7.4.1: H 4

Theorem 7 . 4 . 13 .

 7413 Suppose that H 2n+1 with n ≥ 2 is a helm graph on the vertex set[4n+3]. Then J(H 2n+1 ) is normal.Therefore, it has the strong persistence property, and hence the persistence property.

Figure 7 . 4 . 2 :

 742 Figure 7.4.2: H 2n+1

Theorem 7 . 4 . 14 .

 7414 If n ≥ 4, then L 2 n is not integrally closed.

Theorem 7 . 4 . 16 .

 7416 If n ≥ 4, n even, or n > 5, n odd, then L i n , i ≥ 3 is not integrally closed.

Theorem 7 . 4 .

 74 19. L 2 5 : L 2 5 = (x 1 , . . . , x 6 ).

Figure 7 . 5 . 1 :

 751 Figure 7.5.1: Jahangir's graphs

Proposition 7 . 5 . 1 .

 751 The cover ideals of cycle graphs are normal.Proof. Let C n denote a cycle graph of order n. If n is even, then C n is bipartite. It follows from [48, Corollary 2.6] that J(C n ) is normally torsion-free. Now, [60, Theorem 1.4.6] yields that J(C n ) is normal.If n is odd, then C n is an odd cycle graph, and the claim can be deduce immediately from Theorem 7.1.7.

Figure 2

 2 Figure 2

Case 1 .:::

 1 p ∈ Min(L m ). Because Min(L m ) = Min(L), one can conclude from Proposition 3.3.17 that L m p = p m p and L m+1 p = p m+1 p , and so (L m+1 p Rp L p ) = (p m+1 p Rp p p ). In view of Proposition 4.3.2, p has the strong persistence property, that is, (p m+1 : R p) = p m , and hence (p m+1 p Rp p p ) = p m p . This yields that (L m+1 p : Rp

λλ

  j c t = c t .(7.6.4) Since ω j ∈ I t , then for each z ∈ W there is T z ⊆ S with i∈Tz ω j,2i ≤ ω j,2z+1 . We thus get j ω j,2z+1 = α 2z+1 .(7.6.5)

Corollary 7 . 7 . 2 .

 772 Let I ⊂ R = K[x 1 , . . . , x n ] be a monomial ideal generated in a single degree whose linear relation graph has n vertices and one connected component. Then (i) depth(R/I n-1 ) = 0. In particular, m = (x 1 , . . . , x n ) ∈ Ass(R/I n-1 ).(ii) If I satisfies the persistence property, then lim k→∞ depthR/I k = 0. In particular, m ∈ Ass(R/I k ) for all k ≥ n -1.

Theorem 7 . 7 . 3 .

 773 Let I and H be two normal monomial ideals in a polynomial ringR = K[x 1 , . . . , x n ] such that I + H is normal. Let h be a monomial in R with gcd(v, h) = 1 for all v ∈ G(I) ∪ G(H).Then the following statements hold: (i) L := I + hH is normal. (ii) L := I + hH has the strong persistence property. (iii) L := I + hH has the persistence property. (iv) L := I + hH has the symbolic persistence property. (v) If L := I + hH is generated in a single degree whose linear relation graph has n vertices and one connected component, then lim k→∞ depthR/L k = 0. In particular, m = (x 1 , . . . , x n ) ∈ Ass(R/L k ) for all k ≥ n -1.

Proposition 7 . 7 . 4 .Question 7 . 7 . 5 .

 774775 this completes the inductive step, and hence the claim has been shown by induction. Therefore, L = I + hH is normal, as claimed.(ii)-(iv) can be established similar to the proof of Corollary 6.2.3. (v) Corollary 7.7.2 together with (iii) yield the assertion.To show Theorem 7.9.7, we have to utilize the next proposition. Let I ⊆ H be two normal monomial ideals in a polynomial ringR = K[x 1 , . . . , x n ]. Let J be a monomial ideal of R such that gcd(u, v) = 1 for all v, u ∈ G(J) with u = v. Also, let gcd(u, v) = 1 for all u ∈ G(I) ∪ G(H) and v ∈ G(J).Then the following statements hold: (i) L := I + JH is normal. (ii) L := I + JH has the strong persistence property. (iii) L := I + JH has the persistence property.Proof. (i) Let G(J) = {u 1 , . . . , u s }. We proceed by induction on s. The case in which s = 1 is true according to Theorem 7.7.3(i). Now, assume that the claim holds for s -1. Set F := I + (u 1 , . . . , u s-1 )H. This implies that L = F + u s H. F + H = I + H = H is normal and also gcd(v, u s ) = 1 for all v ∈ G(F ) ∪ G(H), it follows readily from Theorem 7.7.3(i) that L is normal. This completes the inductive step, and so the claim has been shown by induction.(ii) and (iii) can be shown similar to the proof of Corollary 6.2.3. It has already been remained the following open question: (see Question 7.4.11) Let I be a normal square-free monomial ideal in R = K[x 1 , . . . , x n ] with G(I) ⊂ R. Then, in general, can one conclude that

Case 2 .

 2 |supp(J) ∩ supp(u s )| = 1. According to Corollary 7.7.7, one can derive that H is normal. Case 3. |supp(J) ∩ supp(u s )| = 2, say supp(J) ∩ supp(u s ) = {x a , x b }. In this case, one may consider the following subcases: Subcase 3.1. |a -b| = 1, say supp

Theorem 8 . 1 . 3 .

 813 Suppose that V = {1, . . . , n} is a non-empty set of n elements, A = {A 1 , . . . , A m } a non-empty family of m non-empty subsets of V , and 2 ≤ d ≤ n a positive integer. Then there exists a d-uniform hypergraph H with V (H) = V where A is the set whose elements are the minimal vertex cover sets of H if and only if the matrix M (A) = (a ij ) satisfies conditions (P1 ), (P2 )(d), and (P3 )(d).

Remark 8 . 1 . 4 .Example 8 . 1 . 5 .

 814815 Since there is a natural bijection between the set of simple hypergraphs H = (X , E) with the vertex set X = {1, . . . , n}, and the set of square-free monomial ideals I ⊂ S = K[x 1 , . . . , x n ], the algebraic interpretation of Theorem 8.1.3 is characterizing the possible prime ideals in the minimal primary decomposition of a square-free monomial ideal generated in degree d.The following examples will illuminate Theorem 8.1.3. Let V = {1, 2, 3, 4, 5, 6}, and A = {A 1 , A 2 , A 3 , A 4 , A 5 }, where A 1 := {3, 5}, A 2 := {2, 4, 5}, A 3 := {1, 3, 6}, A 4 := {2, 5}, and A 5 := {1, 4, 6}. This yields that the incidence matrix associated to A, i.e., matrix M (A) :

Example 8 . 1 . 8 .

 818 Let V = {1, 2, 3, 4, 5, 6}, d = 3, and A = {A i | i = 1, . . . , 7}, where A 1 := {3, 4}, A 2 := {1, 5, 6}, A 3 := {1, 2, 6}, A 4 := {2, 4}, A 5 := {1, 3, 5}, A 6 := {2, 3}, and A 7 := {1, 4}. The incidence matrix associated to A is the matrix:

  r and x z 1 • • • x zr are squarefree and in the same degree, we thus obtain that x θ s,1 • • • x θs,r = x z 1 • • • x zr , and therefore {z 1 , . . . , z r } = Γ s , as required.

6 4 =

 4 15 cases, there exist exactly six subsets Γ 1 := {1, 2, 4, 6}, Γ 2 := {1, 2, 3, 5}, Γ 3 := {2, 3, 4, 6}, Γ 4 := {1, 3, 4, 5}, Γ 5 := {2, 4, 5, 6}, and Γ 6 := {1, 3, 5, 6} which satisfy condition (iv).

Since x 1

 1 x 2 x 4 x 6 ∈ N I(G), one may consider the following cases: Case 1. N G (1) = {2, 4, 6}. Because x 1 x 2 x 3 x 5 ∈ N I(G), one can conclude that N G (2) = {1, 3, 5}. Furthermore, thanks to N G (2) = {1, 3, 5} and x 2 x 3 x 4 x 6 ∈ N I(G), this implies that N G (3) = {2, 4, 6}. Due to N G (3) = {2, 4, 6} and x 1 x 3 x 4 x 5 ∈ N I(G), we get N G (4) = {1, 3, 5}. It follows from N G (4) = {1, 3, 5} and x 2 x 4 x 5 x 6 ∈ N I(G) that N G (5) = {2, 4, 6}. Finally, since N G (5) = {2, 4, 6} and x 1 x 3 x 5 x 6 ∈ N I(G), this gives that N G (6) = {1, 3, 5}. Therefore, we deduce that G is the graph G 1 which has been shown in the figure below. Case 2. N G (2) = {1, 4, 6} and N G (4) = {2, 3, 6}. A similar argument yields that N G (1) = {2, 3, 5}, N G (3) = {1, 4, 5}, N G (5) = {1, 3, 6}, and N G (6) = {2, 4, 5}. We thus gain that G is isomorphic to G 2 which has been shown in the figure below. Case 3. N G (2) = {1, 4, 6} and N G (6) = {2, 3, 4}. Following a similar discussion, we get N G (1) = {2, 3, 5}, N G (3) = {1, 5, 6}, N G (4) = {2, 5, 6}, andN G (5) = {1, 3, 4}. Consequently, one has G is isomorphic to G 2 . Case 4. N G (4) = {1, 2, 6} and N G (3) = {1, 2, 5}. A similar discussion gives rise to N G (1) = {3, 4, 5}, N G (2) = {3, 4, 6}, N G (5) = {1, 3, 6}, and N G (6) = {2, 4, 5}. One can easily check that G is isomorphic to G 2 . Case 5. N G (4) = {1, 2, 6} and N G (5) = {1, 2, 3}. A similar argument implies that N G (1) = {3, 4, 5}, N G (2) = {4, 5, 6}, N G (3) = {1, 5, 6}, and N G (6) = {2, 3, 4}. It is not hard to investigate that G is isomorphic to G 2 .Case 6. N G (6) = {1, 2, 4} and N G (5) = {1, 2, 3}. By a similar discussion, we obtain that N G (1) = {3, 5, 6}, N G (2) = {4, 5, 6}, N G (3) = {1, 4, 5}, and N G (4) = {2, 3, 6}. One can rapidly check that G is isomorphic to G 2 . Case 7. N G (6) = {1, 2, 4} and N G (3) = {1, 2, 5}. A similar argument yields that N G (1) = {3, 5, 6}, N G (2) = {3, 4, 6}, N G (4) = {2, 5, 6}, and N G (5) = {1, 3, 4}.

1 Proof.

 1 We proceed by induction on n := |V (T )|. The case n = 1 is clear. Now, suppose, inductively, that n > 1 and that the result has been proved for all trees which have vertex sets with less than n elements. Let W be a minimal path cover set of a tree T and V (T ) = [n]. If W ⊆ L T and |L T | = d, then, in view of Proposition 9.1.7, it follows that |W | = d-1. Hence, assume that W ⊈ L T . Consider v ∈ W \L T and set ℓ := deg T v. Then T \ {v} has exactly ℓ connected components such as T 1 , .

Definition 9 . 1 . 11 .

 9111 Let G be a graph on the vertex set [n]. Then the ideal P I(G) :

(

  ii) It is clear by[START_REF] Herzog | Monomial Ideals[END_REF] Corollary 1.5.5].

Corollary 9 . 1 . 14 .

 9114 Let T be a tree. Then the following conditions are equivalent.(i) T is the path graph.(ii) P I(T ) is Cohen-Macaulay.(iii) P I(T ) is unmixed.

Definition 9 . 1 . 16 .

 9116 ([20, P. 1190]) Let G be a graph and let k be a positive integer. By the order of a path P we mean the number of vertices on P . A subset of vertices S ⊆ V (G) is called a k-path vertex cover if every path of order k in G contains at least one vertex from S. Also, we denote by ψ k (G) the minimum cardinality of a k-path vertex cover in G.

)

  It is straightforward by virtue of the definition of P C t (G), part (i) and [60, Corollary 1.5.5].

Theorem 9 . 1 . 19 .

 9119 Let T be a tree on the vertex set [n] and |L T | = d with d ≥ 3. Then the following conditions are equivalent. (i) T has a vertex of degree d.

  (ii) ht(P I(T )) = 1. (iii) T is starlike. (iv) If p ∈ Ass(P I(T )) and htp > 1, then htp = d -1.

Corollary 9 . 1 . 20 .

 9120 Let T be a tree on the vertex set [n] and |L T | = d. Then ht(P I(T )) = 1 if and only if T is a path graph or starlike. Proof. If d = 2, then T is a path graph and there is nothing to prove. We thus suppose that d ≥ 3. The claim is therefore an immediate consequence of Theorem 9.1.19.

Theorem 9 . 1 . 21 .

 9121 Let T be a tree on the vertex set [n] and |L T | = d. Then ht(P I(T )) = 2 if and only if there exist two starlike trees S 1 and S 2 such thatT = S 1 ∪ E ∪ S 2, where E satisfies one of the following conditions.

Theorem 9 . 2 . 1 .

 921 Let G be the union of a cycle C and a connected graph H such that v is the cut-point of this union. Suppose that W G is an arbitrary minimal path cover set of G and V (C) ∩ W G = {v ′ }. Then one of the following statements holds.

Theorem 9 . 2 . 5 .

 925 Let G be the union of a tree T and a connected graph H such that v is the cut-point of this union with deg T v = 1. Let W G be an arbitrary minimal path cover set of G and V (T ) ∩ W G = S. Suppose that, for each w = v in L T , there exist a path L ′ : v, . . . , z and a path graph L : z, . . . , w with degT z ≥ 3 (deg T z ≤ 2 when T is a path graph) in T such that V (L) ∩ W G = ∅.Then one of the following statements holds.

Proposition 9 . 2 . 8 .

 928 Let G 1 , . . . , G m be some complete subgraphs of a finite simple graph G, and m

Definition 9 . 2 . 10 .

 9210 The (m, n)-tadpole graph is the graph obtained by union a cycle C m to a path graph P n with a bridge.For example, the following figure is a (3, 4)-tadpole graph.

  Proof. Suppose that V (T ) = {1, . . . , λ} and R = K[x 1 , . . . , x λ ] is the polynomial ring over a field K. We proceed by induction on n := |L T |. According to the hypothesis, one has|L T | ≥ 4. If |L T | = 4,then T is a starlike graph which u is the center of the starlike with deg T u = 4. We thus can consider the vertex set of T as V (T ) = {i 1 , i 2 , i 3 , i 4 , i 5 , . . . , i s , u} with deg T i k = 1 for all k = 1, 2, 3, 4, and deg T i k = 2 for all k = 5, . . . , s, and also deg T u = 4. This yields that P I(T ) is given by P

  [n] = {1, . . . , n} and P ([n]) denote the set of all subsets of [n]. For any S ⊆ [n], the cardinality of S is denoted by |S|. A matroid M on the ground set [n] is a non-empty collection of subsets of [n] satisfying the following properties: 1. if A ∈ M and B ⊂ A, then B ∈ M; 2. if A, B ∈ M and |B| ≤ |A|, then there exists a ∈ A\B such that B ∪{a} ∈ M. A maximal element (with respect to inclusion) in M is called a base. Condition (2) can be used to show that all bases of M have the same cardinalities. If B be the set containing all the bases of M, then B is distinguished by the following "exchange property". (EP) For any A, B ∈ B, if a ∈ A \ B, then there exists b ∈ B \ A such that (A \ {a}) ∪ {b} ∈ B.

Theorem 9 . 3 . 4 . 1 i.

 9341 Let I ⊂ S = K[x 1 , . . . , x n ] be a polymatroidal ideal and 1 ≤ i ≤ n. Assume that I = d j=0 I j x j i , where d = max{deg x i u : u ∈ G(I)} andI j = (u/x j i : u ∈ G(I), deg x i u = j),for each j = 0, . . . , d. ThenI 0 ⊆ I 1 ⊆ • • • ⊆ I d .Proof. If d = 0, then I = I 0 and there is nothing to prove. Let d ≥ 1 and setJ 1 := d j=1 I j x j-It is easy to see that I = I 0 +x i J 1 and (I : x i ) = I 0 +J 1 . Following the proof of[12, Theorem 1.1], one can deduce that I 0 ⊆ J 1 , and hence (I : x i ) = J 1 is a polymatroidal ideal. Now, on account of I 0 = (u : u ∈ G(I), deg x i u = 0) andI 1 = (u/x i : u ∈ G(I), deg x i u = 1), we obtain I 0 ⊆ I 1 . If d = 1,then we are done. Otherwise, if d ≥ 2, we set J 2 := d j=2 I j x j-2 i

Proposition 9 . 3 . 5 .

 935 Let I ⊂ K[x 1 , . . . , x n ] be a fully supported matroidal ideal generated in degree d with gcd(I) = 1. Then the following statements hold.

  3.10. It is known from [65, Lemma 4.2 and Theorem 4.1] that if the linear relation graph of a matroidal ideal I has m vertices and s connected components, then (9.3.1) dstab(I) < ℓ(I) = m -s + 1, where ℓ(I) denotes the analytic spread of I, that is, the Krull dimension of the fiber ring R(I)/mR(I). Combining this with [66, Corollary 3.5], one can deduce that (9.3.2) lim k→∞ depth(S/I k ) = n -(m -s + 1).

(t - 1 )

 1 -path in G. It can be extended to a t-path by either joining the edge {y m+1 , x 1 }, or the edge {y m , x m+1 }. Hence, for u = y m+1 x 1 y1 x 2 y 2 • • • x m y m and v = x 1 y 1 x 2 y 2 • • • x m y m x m+1 , we have x m+1 u = y m+1 v, which gives {x m+1 , y m+1 } ∈ E(Γ).This shows that Γ is connected. In fact, by repeating the same argument as above, one can show that Γ is a complete graph on n vertices. Therefore, ℓ(I) = n-1+1 = n and by using the equality in (9.3.2), we obtain lim k→∞ depth(S/I t (G) k ) = 0 and one can conclude from (9.3.1) that dstab(I t (G)) < n. Again from Remark 9.3.13 we have depth(I t (G)) = t -1 > 0. Hence, dstab(I t (G)) > 1. This finishes the proof. Now, we consider the case in which G is an r-partite graph with r ≥ 3. Let d, a 1 , . . . , a n be positive integers with n i=1 a i ≥ d. Then the ideal generated by all monomials x b 1

Proposition 9 . 3 . 15 .

 9315 Let t ≥ 2 and G = K n 1 ,...,nr with |V (G)| = n and n i ≤ t/2 for all 1 ≤ i ≤ r. Then lim k→∞ depth(S/I t (G) k ) = 0 and dstab(I t (G)) = (n -1)/(n -t) .

3Lemma 9 . 3 . 16 .

 9316 ≤ n = n 1 + • • • + n r . First, we discuss the cases when n ∈ {3, 4}. The following lemma is a consequence of Proposition 9.3.15 and [133, Lemma 2.3.10]. Let G = K n 1 ,...,nr with r ≥ 3. Then the following statements hold. (i) If |V (G)| = 3, then I 3 (G) is a principal ideal and lim k→∞ depth(S/I 3 (G) k ) = 2 and dstab(I 3 (G)) = 1. (ii) If |V (G)| = 4, then G is either a complete graph on four vertices or isomorphic to K 2,1,1 . Moreover, lim k→∞ depth(S/I 3 (G) k ) = 0 and dstab(I 3 (G)) = 3.

Proposition 9 . 3 . 17 .

 9317 Let G = K n 1 ,...,nr with r ≥ 3. Then lim k→∞ depth(I(G) k ) = 0 and dstab(I(G)) = 2.

Theorem 9 . 3 . 18 .

 9318 Let G = K n 1 ,...,nr with r ≥ 3 and |V (G)| ≥ 5. Then lim k→∞ depth(I 3 (G) k ) = 0 and dstab(I 3 (G)) = 2.

First, assume that |V 1 |

 1 = • • • = |V r-1 | = 1. If |V r | = 1 or |V r | = 2,then the assertion follows from Proposition 9.3.15. More precisely, If |V r | = 1, then |V (G)| = r, and if |V r | = 2, then |V (G)| = r + 1. Since r ≥ 5, in both cases, we have dstab(I 3 (G)) = (s -1)/(s -3) = 2 for s = r, r + 1. Finally, if |V r | > 2, then take

Theorem 9 . 3 . 19 .

 9319 Let G = K n 1 ,...,nr with r ≥ 3 andn i ≥ t/2 for all 1 ≤ i ≤ r. Then lim k→∞ depth(I t (G) k ) = 0 and 1 < dstab(I t (G)) < n for t ≥ 3 . Proof. From Remark 9.3.13, we conclude that depth(I t (G)) = t -1 > 0. Let |V (G)| = n. Due to (9.3.1) and (9.3.2), it is enough to prove that the linear relation graph Γ of I t (G) has one connected component and |V (Γ)| = n.

Lemma 10 . 1 . 1 .

 1011 Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] with G(I) = {u 1 , . . . , u m }. If m = (x 1 , . . . , x n ) ∈ Ass R (R/I), then m ≥ n.

Proposition 10 . 1 . 2 .

 1012 Let I be a monomial ideal in a polynomial ring R = K[x, y] with G(I) = {u 1 , . . . , u m }. If m ≥ 2, then m = (x, y) ∈ Ass R (R/I).

Theorem 10 . 1 . 3 .

 1013 Suppose that I is a monomial ideal in a polynomial ring R =K[x 1 , . . . , x n ], m = (x 1 , . . . , x n ),andG(I) = {x r 1,1 1 • • • x r 1,n n , . . . , x r k,1 1 • • • x r k,n n }, with k ≥ n. Then m ∈ Ass R (R/I) ifand only if there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , k} such that the following conditions hold:

Lemma 10 . 1 . 5 . 1

 10151 Suppose that I ⊂ R = K[x 1 , . . . , x n ] is a monomial ideal, m = (x 1 , . . . , x n ), and G(I) = {x r

Corollary 10 . 1 . 6 . 1

 10161 Suppose that I is a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] with G(I) = {x r and m = (x 1 , . . . , x n ). Also, let C j = {i | r i,j =max{r 1,j , . . . , r n,j }} for each j = 1, . . . , n. Then m ∈ Ass R (R/I) if and only if the following conditions hold: (i) |C j | = 1 for each j = 1, . . . , n;

Proposition 10 . 2 . 1 .

 1021 Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K and p ∈ Ass(R/I). Then there exists a monomial v in R with supp(v) ⊆ u∈G(I) supp(u) such that p = (I : R v).Proof. By virtue of [60, Corollary 1.3.10], there exists a monomial w in R such that p = (I : R w). Without loss of generality, one may assume that u∈G(I) supp(u) = {x 1 , . . . , x m }. Certainly, one can write w = w 1 w 2 , where supp(w 1 ) ⊆ {x 1 , . . . , x m } and supp(w 2 ) ⊆ {x m+1 , . . . , x n }. It follows from Lemma 2.1.12 that (I : R w 2 ) = I, and so we get the following equalities p = (I : R w) = ((I : R w 2 ) : R w 1 ) = (I : R w 1 ).

Theorem 10 . 2 . 2 .

 1022 Let I 1 ⊂ R 1 = K[x 1 , . . . , x n ] and I 2 ⊂ R 2 = K[y 1 , . .. , y m ] be two monomial ideals in disjoint sets of variables. LetI = I 1 R + I 2 R ⊂ R = K[x 1 , . . . , x n , y 1 , . . . , y m ].Then p ∈ Ass(R/I) if and only if p = p 1 R + p 2 R, where p 1 ∈ Ass(R 1 /I 1 ) andp 2 ∈ Ass(R 2 /I 2 ). Proof. Let I 1 = Q 1 ∩ • • • ∩ Q s be a minimal primary decomposition of I 1 such that √ Q i = p i for all i = 1, . . . ,s. In view of [133, Exercise 6.1.23], we have I = s i=1

Definition 11 . 4 . 1 .

 1141 ([42, Definition 2.1]) Let u = x j 1 x j 2 • • • x j ℓ ∈ R = K[x 1 , . .

{x 1 ,

 1 . . . , x n } with edges e 1 , . . . , e r , we define the complement clutter, denoted by C c , as the clutter whose edges are {x 1 , . . . , x n } \ e i for each i = 1, . . . , r. In particular, we have (C c ) c = C. We denote the edge ideal of a clutter C by I(C).It is quite natural to want to compare the algebraic properties of clutters and complement clutters. The next question pursues this goal. Question 11.5.1. (i) Does I(C) have the strong persistence property if and only if I(C c ) have the strong persistence property? (ii) Does I(C) have the persistence property if and only if I(C c ) have the persistence property? (iii) Is I(C) normal if and only if I(C c ) is normal? (iv) Is I(C) normally torsion-free if and only if I(C c ) is normally torsion-free?

Example 11 . 6 . 1 .

 1161 Let G be the graph as shown in the following figure.

  The following example illuminates what happens in Corollary 2.2.14.

	Example 2.2.15. Suppose that T is the unrooted symmetric starlike tree, as shown
	in Figure 3, on the vertex set V (T ) = {z, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}.
	z
	1 2 3 4	5
	6 7 8 9	10
	11 12 13 14	15
	Figure 3

Definition 2.2.18. The

  2.2.19. centipede graph W n with n ∈ N, as shown in the figure below, is the graph on the vertex set {a 1 , . . . , a n } ∪ {b 1 , . . . , b n }. The set of edges of the centipede graph is given by

  5 , x 6 ]. Then I is not a separable monomial ideal. On the other hand, we obtain (I 3 : R I) = I 2 . Lemma 2.3.20 is essential for us in order to prove Theorem 2.3.22. To verify Lemma 2.3.20, one requires the subsequent lemma.

Lemma 2.3.19. ([90, Lemma 2.2]) Let I * denote the eventual stable value of the ascending chain (I 2

Lemma 2.3.30.

  Let I be a monomial ideal of R. Then I * has the persistence property if and only if I has. By virtue of Proposition 2.2.10, one has p ∈ Ass R (R/I n+1 ). This implies that I has the persistence property. Sufficiency follows from in a similar way, and the proof is done.Next proposition is necessary for us in order to clarify Example 2.3.32.

Proof. For necessity, consider p ∈ Ass R (R/I n ) for an arbitrary n ∈ N. According to Proposition 2.2.10, it follows that p * ∈ Ass R * (R * /I n * ). Due to Lemma 2.2.11(iii), we have I n * = (I * ) n . Since I * has the persistence property, one can conclude that p * ∈ Ass R * (R * /(I * ) n+1 ). Again, by Lemma 2.2.11(iii), we deduce that p * ∈ Ass R * (R * /I n+1 * ).

Proposition 2.3.31. Let

  Let n be an arbitrary positive integer. Due to Theorem 2.1.27, one has

I ⊂ R = K[x 1 , . . . , x n ] be a monomial ideal which has the persistence property, and let u = x a 1 i 1 • • • x ar ir be a monomial in R with a 1 , . . . , a r ∈ N. Then uI has the persistence property.

Proof.

  52 , x 11 x 21 x 3 31 x 41 x 51 , x 11 x 21 x 3 31 x 41 x 52 , x 11 x 21 x 3 31 x 42 x 51 , x 11 x 21 x 3 31 x 42 x 52 , x 12 x 21 x 3 31 x 41 x 51 , x 12 x 21 x 3 31 x 41 x 52 , x 12 x 21 x 3 31 x 42 x 51 , x 12 x 21 x 3 31 x 42 x 52 , x 21 x 2 31 x 2 51 x 61 x 71 , x 21 x 2 31 x 2 51 x 62 x 71 , x 21 x 2 31 x 2 51 x 63 x 71 , x 21 x 2 31 x 51 x 52 x 61 x 71 , x 21 x 2 31 x 51 x 52 x 62 x 71 , x 21 x 2 31 x 51 x 52 x 63 x 71 , x 21 x 2 31 x 2 52 x 61 x 71 , x 21 x 2 31 x 2 52 x 62 x 71 , x 21 x 2 31 x 2 52 x 63 x 71 ),

	Example 2.3.32. Let J be the following monomial ideal
	J := (x 11 x 2 21 x 2 31 x 41 x 51 , x 11 x 2 21 x 2 31 x 41 x 52 , x 11 x 2 21 x 2 31 x 42 x 51 , x 11 x 2 21 x 2 31 x 42 x 52 ,
	x 12 x 2 21 x 2 31 x 41 x 51 , x 12 x 2 21 x 2 31 x 41 x 52 , x 12 x 2 21 x 2 31 x 42 x 51 , x 12 x 2 21 x 2 31 x 42 x 52 ,
	x 2 21 x 3 31 x 51 , x 2 21 x 3 31 x

in the polynomial ring R = K[x 11 , x 12 , x 21 , x 31 , x 41 , x 42 , x 51 , x 52 , x 61 , x 62 , x 63 , x 71 ]. We claim that J has the persistence property. To do this, put p 1 := (x 11 , x 12 ), p 2 := (x 21 ), p 3 := (x 31 ), p 4 := (x 41 , x 42 ), p 5 := (x 51 , x 52 ), p 6 := (x 61 , x 62 , x 63 ), and p 7 := (x 71 ). It is routine to check that J = p 1 p 2 2 p 2 3 p 4 p 5 + p 2 2 p 3 3 p 5 + p 1 p 2 p 3 3 p 4 p 5 + p 2 p 2 3 p 2 5 p 6 p 7 .

Definition 3.1.5. ([45,

  sets such that for every E ∈ E, we have E ⊈ C i for all i = 1, . . . , d. (In the case of a graph G, this simply means that any two vertices connected by an edge receive different colors.) The C i s are called the color classes. Each color class C i is an independent set, meaning that C i does not contain any edge of the hypergraph. The chromatic number of H, denoted χ(H), is the minimal d such that H has a d-coloring.

	Definition 3.1.4. ([45, Definition 2.8]) A hypergraph H is called critically d-
	chromatic if χ(H) = d, but for every vertex x ∈ V, χ(H) < d, where H \ {x}
	denotes the hypergraph H with x and all edges containing x removed.

  Label the vertices of C 2n+1 in counterclockwise order with 1, . . . , 2n + 1. In order to simplify the notation, we set I := J(C 2n+1 ). Due to Theorem 3.1.11 and Proposition 3.1.12, one obtains the ideal I has the persistence property, and by virtue of m is the unique homogeneous maximal ideal of R, it is sufficient to verify that m ⊆ I 2 : R

	45, Corollary Proposition 3.1.13. Suppose that C 2n+1 is a cycle graph on the vertex set [2n + 1], 4.7]. R = K[x 1 , . . . , x 2n+1 ] is a polynomial ring over a field K, and m is the unique homogeneous maximal ideal of R. Then m ∈ Ass R (R/(J(C 2n+1 )) s ) for all s ≥ 2, and m / ∈ Ass R (R/(J(C 2n+1 ))). Proof. 2n+1 k=1

x ir ] which is obtained from I by applying the K-algebra homo- morphism R → R(p) with x j → 1 for all x j /

	∈ {x i 1 , . . . , x ir }. It is well-known that
	p ∈ Ass R (R/I) if and only if m p ∈ Ass(I(p)), where m p is the graded maximal ideal
	of R(p).

Theorem 3.1.20. Suppose that H 2n+1 with

		n ≥ 2 is a helm graph of order 2n + 1
	on the vertex set [4n + 3]. Then J(H 2n+1 ) satisfies the strong persistence property.
	Proof. Label the vertices of C 2n+1 in counterclockwise order with 1, . . . , 2n + 1, and
	the hub with 2n + 2, as shown in the figure below, such that we have
	E(H 2n+1 ) = E(W 2n+2 ) ∪ {{i, 2n + 2 + i} : i = 1, . . . , 2n + 1}.
		2n + 3
	2n + 4	1	4n + 3
	2	2n+2	2n + 1
	3		2n
	2n + 5		4n + 2
		H 2n+1

Proposition 3.1.21. Suppose

  that H 2n+1 with n ≥ 2 is a helm graph of order 2n + 1 on the vertex set [4n + 3], R = K[x 1 , . . . , x 4n+3 ] is a polynomial ring over a field

	K, and m = (x 1 , . . . , x 2n+2 ). Then m ∈ Ass R (R/(J(H 2n+1 )) s ) for all s ≥ 3, and
	m / ∈ Ass R (R/(J(H 2n+1 )) s ) for s = 1, 2.
	Proof. The result follows immediately from Proposition 3.1.18, Theorem 3.1.20, and
	the iteration of Theorem 3.1.6.
	We next give an example which explains Proposition 3.1.21.
	Example 3.1.22. Let H 5 be the helm graph, as shown in the figure below, on the
	vertex set [11] = {1, . . . , 11}, R = K[x 1 , . . . , x 11 ] be the polynomial ring over a field
	K, and m = (x 1 , . . . , x 6 ).
		7
	8	2	1 6	5	11
		3		4
	9			10
		H 5

Theorem 3.2.16. Let

  I be a monomial ideal in R = K[x 1 , . . . , x n ], and p ∈ V * (I).

	Then the following statements hold.
	(i) If I has the strong persistence property, then I(p) has the strong persistence
	property. In particular, I(p) has a superficial ideal.
	(ii) If I has the persistence property, then I(p) has the persistence property.
	Proof. (i) Assume that I has the strong persistence property. Hence, one can deduce
	that (I k+1 : R I) = I k for all positive integers k. Fix k ∈ N. It follows from
	(I k+1 : R I) = I k that (I k+1 : R I)(p) = (I k )(p). Now, parts (ii) and (iv) of Lemma
	3.2.15 yield that (I(p)) k+1 : R(p) I(p) = (I(p)) k . This means that I(p) has the
	strong persistence property. In particular, I(p) has a superficial ideal, as claimed.

  x 3 , x 4 , x 5 , x 6 , x 7 , x 8 ] over a field K. Takep := (x 1 , x 2 , x 3 , x 4 , x 5 ). It is easy to see that p ∈ V * (L) and L(p) = I. On account of (x 1 , x 2 , x 3 , x 4 , x 5 ) ∈ Ass R (R/I) \ Ass R (R/I 2 ), one can conclude that I does not satisfy the persistence property. It follows now from Corollary 3.2.17 that L does not satisfy the persistence property. On the other hand, it is routine to check that

  11 x 21 x 22 x 31 , x 21 x 31 x 32 x 41 , x 31 x 41 x 42 x 51 , x 41 x 51 x 52 x 11 , x 51 x 11 x 12 x 21 ), in the polynomial ring S = K[x 11 , x 12 , x 21 , x 22 , x 31 , x 32 , x 41 , x 42 , x 51 , x 52 ] over the field

  By direct computation one has I 2 6 = (x 12 , x 11 y, x 10 y 2 , x 7 y 5 , x 6 y 6 , x 5 y 7 , x 2 y 10 , xy 11 , y 12 ). monomials in (I 2 6 ) 12 are just the 9 minimal generators of I 2 6 and (I 2 6

	The

Example 3.3.14. Let us now look at I 2 6 and (I 3 6 : I 6 )\I 2 6 . This will illustrate the ideas in Lemma 3.3.11 and Proposition 3.3.13 and help ensure that we have gotten the notation straight.

  Definition 5.1.3. Let R = K[x 1 , . . . , x n ] be a polynomial ring. A parameter ideal of R is an ideal of the form (x a 1 1 , . . . , x an n ) with a 1 , . . . , a n are positive integers. If u = x a 1 1 • • • x an n with nonnegative integers a 1 , . . . , a n is a monomial in R, then we set P R (u) := (x a 1 +1 It is well-known that for a monomial z in R and monomial ideal I of R, one has z / ∈ I if and only if I ⊆ P R (z) (cf. [93, Exercise 6.1.12]).

	1	, . . . , x an+1

n

).

  y 1 , y 2 , y 3 , y 4 , y 5 , y 6 , y 7 },

	and the following edge set

  on page 207] states that transversal hypergraphs of simple balanced hypergraphs are also Mengerian. In addition, following [133, Theorem 6.3.39], we can rephrase [122, Lemma 2.2] as follows:

  We first assume that i = t. In this case, it follows from the definition thatI = (x 1 x t+1 , x 1 x t+2 , . . . , x 1 x n , x 2 x t+2 , . . . , x 2 x n , . . . , x t x 2t , x t x 2t+1 , . . . , x t x n ).It is routine to check that I is the edge ideal of a bipartite graph with the vertex set {1, 2, . . . , t} ∪ {t + 1, t + 2, . . . , n}. In addition,[START_REF] Villarreal | Monomial algebras[END_REF] Corollary 14.3.15] implies that I is normally torsion-free. Now, let i > t. One can conclude from the definition that the minimal generators of I are as follows:

3, we re-prove Proposition 4.4 in [9]. Proposition 6.3.4. Let u = x i x n be a t-spread monomial in R = K[x 1 , . . . , x n ] with i ≥ t. Then I = B t (u) is nearly normally torsion-free. Proof.

  and hence is normally torsion-free. In addition, by a similar argument, the ideal (x 1 , x 2 , . . . , x z-t ) is normally torsion-free. Furthermore, since (x 1 , x 2 , . . . , x z-t ) and (x α x β : α = z -t + 1, . . . , i, β = z + 1, . . . , n, β -α ≥ t) have no common variables, we are able to derive from Theorem 5.3.4 that I(m \ {x z }) is normally torsion-free too. Inspired by Corollary 6.1.3, we conclude that I is nearly normally torsion-free, as claimed. Let u = x 4 x 7 and t = 3.If v = x a x b ∈ G(B 3 (u)), then a ∈ [1, 4]and b ∈[4, 7]. The complete list of minimal generators of B 3 (x 4 x 7

	We illustrate the statement of Proposition 6.3.4 through the following example.
	Example 6.3.5.

is the edge ideal of a bipartite graph with the following vertex set {z -t + 1, . . . , i} ∪ {z + 1, . . . , n},

  It is known from[34, Lemma 5.2] that if I is a monomial ideal generated in degree d and the first syzygy of I is generated in degree d + 1, then

	(6.3.2)
	it can
	be concluded from [10, Theorem 2.3] that the first syzygy of B t (u) is generated in
	degree d + 1. Hence, [34, Lemma 5.2] gives a way to compute the analytic spread
	of B t (u). Before proving the other main result of this section, we first analyze the
	linear relation graph of B t (u).

ℓ(I) = r -s + 1,

where r is the number of vertices and s is the number of connected components of the linear relation graph of I. If u is a t-spread monomial of degree d, then Lemma 6.3.11.

Let u = x i 1 x i 2 • • • x i d-1 x i d and A k = [(k -1)t + 1, i k ] for each k = 1, . . . d.

Moreover, let Γ be the linear relation graph of I = B t (u). Then the following statements hold:

  d. Since I is not principal, it follows from Remark 6.3.8(4) that |A for some k, then A k ⊆ V (Γ) and the induced subgraph on A k is a complete graph. This shows that V (Γ) is the union of all A i 's for which |A i | ≥ 2. Moreover, each such A i determines a connected component in Γ. Hence, Γ has only one connected components if and only if |A d | ≥ 2, and |A i | = 1 for all i = 1, . . . , d -1. In this case,

k | ≥ 2 for some 1 ≤ k ≤ d. Since i d-1 ≤ (d -1)t,

one can deduce from Lemma 6.3.11 that we do not have any edge in Γ of the form {f, h} with f and h lie in different A k 's and if |A k | ≥ 2,

  iii) if a > 1 and b ≥ 2t + 1, then I is not nearly normally torsion-free.

Proof. (i) This claim can be shown by virtue of Theorem 6.3.13. (ii) Let u ∈ G(I). Then u = x 1 x i 2 x i 3 , where i 2 ∈ {t + 1, . . . , b} and i 3 ∈ {2t + 1, . . . , n}. It can be easily seen that I = x 1 J, where J = B t (x b x n ). It follows from Proposition 6.3.4 that B t (x b x n ) is nearly normally torsion-free. Then we get the required result by using Lemma 6.1.5. (iii) We will prove the assertion by constructing two monomial prime ideals p 1 and p 2 that belong to Ass

  is normally torsion-free, then DI(H) is normally torsion-free. Proof. Let DI(G) be normally torsion-free. It follows from [122, Lemma 2.2] that

  x n+1 ] be a polynomial ring over a field K. Let us simplify the notation by L := J(H), I := J(G), and g := n i=1 x i . In the following, our goal is to apply Theorem 7.1.3. By definition L = IR ∩ n i=1 (x i , x n+1 ). It follows from [133, Exercises 6.1.22 and 6.1.23] that n i=1 (x i , x n+1 ) = ( n i=1 x i , x n+1 ), and so one has L = IR ∩ ( n i=1 x i , x n+1 ). By virtue of g = n i=1

  First assume that I is a normal monomial ideal of R. Also, define the Kalgebra homomorphism π : R * → R given by π(x ij ) = x i for all i and j. One can easily observe that if u is a monomial in R * , then π(u) ∈ I if and only if u ∈ I * . Fix t ∈ N, and pick an arbitrary monomial u in (I * ) t . Hence,[START_REF] Herzog | Monomial Ideals[END_REF] Theorem 1.4.2] implies that there exists a positive integer k such that u k ∈ ((I * ) t ) k , and so u k ∈ (I * ) tk . On account of [13, Lemma 1.1(iii)], one has (I * ) tk = (I tk ) * . Consequently, we can derive u k ∈ (I tk ) * , and thus π(u k ) ∈ I tk . Because π(u k ) = (π(u)) k , one gains that (π(u)) k ∈ (I t ) k . This yields that π(u) ∈ I t . Thanks to I is normal, one can conclude that π(u) ∈ I t , and so u ∈ (I t ) * . Once again, [13, Lemma 1.1(iii)] gives that (I t ) * = (I * ) t . Therefore, u ∈ (I * ) t ; hence, I * is a normal monomial ideal in R * .Conversely, let I * be normal. Fix t ∈ N, and consider a monomial v ∈ I t . Then v k ∈ (I t ) k for some k ∈ N. Let u be a monomial in R

Theorem 7.2.1. Let I be a monomial ideal of R. Then I is normal if and only if I * is normal. 200 Proof. * such that π(u) = v. Then (π(u)) k ∈ I tk , and so u k ∈ (I tk ) * = (I * ) tk . This gives that u ∈ (I * ) t . Since I * is normal, one has u ∈ (I * ) t = (I t ) * . Therefore, v ∈ I t , and hence I is a normal monomial ideal. Proposition 7.2.2. Let I be a monomial ideal of R. Then (I) * = (I * ).

  K[x 11 , . . . , x 1m , x 2 , . . . , x n ] as follows 11 , . . . , x 1m ) a j . Remark 7.1.1 and the discussion above give

	Proof. Note (x	a j 11 , . . . , x	a j 1m ) = (x
			N =	s j=1	(x	a j 11 , . . . , x	a j 1m )u ′ j .
	Then L 201

*

and N have the same integral closure, that is, L * = N .

  necessarily imply that the weighted ideal is also normal. But the converse is always true. The following theorem asserts that if the weighted ideal of a monomial ideal is normal, then the original ideal is also normal. An application of this theorem will be appeared in Chapter 11. Let I be a monomial ideal in R = K[x 1 , . . . , x n ], and W a weight over R. If I W is normal, then I is normal.Proof. We use the notations that have been introduced in Definition 3.2.11, in particular, for a monomial u ∈ R, we denote h(u) = u W . Let I W be normal. Fix t ∈ N, and pick a monomial u ∈ I t . Hence, [60, Theorem 1.4.2] implies that there exists a positive integer k such that u k ∈ (I t ) k . We thus gain (u k ) W ∈ (I tk ) W . It follows from Lemma 3.2.12(ii) that (I tk ) W = (I W ) tk . Consequently, one can conclude that

	To state Theorem 7.2.5, one first should recall the definition and properties of weighted ideals, see Definition 3.2.11 and Lemma 3.2.12. Theorem 7.2.1 gives that the ideal M = (x 1 , x 2 , . . . , x Theorem 7.2.5.

n ) is normal in R = K[x 1 , . . . , x n ]. But if we consider the weight W := {x 1 , . . . , x n } → N over R with W (x i ) ≥ 2

for all i, then it is obvious that x i x j ∈ M W \M W ; hence, M W is not integrally closed. This shows that if the original monomial ideal is normal, then it does not

  is normal by[START_REF] Villarreal | Monomial algebras[END_REF] Proposition 12.2.3]. Let I be a square-free monomial ideal in R = K[x 1 , . . . , x n ] and p ∈ V * (I). If I is normally torsion-free, then I(p) is square-free normally torsionfree; hence, normal.Proof. It is enough to combine Theorem 5.3.12 and[START_REF] Herzog | Monomial Ideals[END_REF] Theorem 1.4.6].The subsequent corollary is an immediate consequence of Theorem 7.3.2. Let I be a square-free monomial ideal in R 1 = K[x 1 , . . . , x r ] with

	Theorem 7.3.2. Corollary 7.3.3.

  2.19. By virtue of Theorem 7.3.2, one can conclude the following result.

Corollary 7.3.7. Suppose that I is a square-free monomial ideal in the polynomial ring R = K[x 1 , . . . , x n ]. If I is normally torsion-free, then I/x j is normally torsionfree square-free; hence, normal.

  x 1 ). It is easy to see that I is the edge ideal associated to C 5 , and J is the cover ideal associated to C 5 . Based on Theorem 7.1.7, the ideal J is normal. By virtue of[START_REF] Villarreal | Monomial algebras[END_REF] Corollary 10.5.9], one derives I is normal. Moreover, it follows from J ⊆ I that (L : R x 6 ) = I, and hence (L : R x 6 ) is normal. Now, put α := x 1 x 2 x 3 x 4 x 5 x 6 x 7 . Direct computation gives that

	Now, let C 5 be the odd cycle graph on the vertex set V (G) = {1, 2, 3, 4, 5} and the
	edge set E(C 5 ) = {{1, 4}, {4, 3}, {3, 2}, {2, 5}, {5, 1}}.

  and V (H) = {y 1 , y 2 , y 3 , y 4 , y 5 } with E(H) = {{y i , y i+1 }} 5 i=1 , where x 6 represents x 1 , and y 6 represents y 1 . It follows from Theorem 7.1.7 that I and J are normal monomial ideals. Set α := 5 i=1 x i 5 i=1 y i in the monomial ideal Q := IK[x 1 , . . . , x 5 , y 1 , . . . , y 5 ] + JK[x 1 , . . . , x 5 , y 1 , . . . , y 5 ].

  n}, where x 2n+1 represents x 1 . For any t ≥ 1, it has already been shown in Lemma 3.2.39 that x ℓ 1 1 • • • x ℓ 2n 2n ∈ G(I t ) if and only if (i) n i=1 ℓ 2i = t and (ii) ℓ 2i+1 = ℓ 2i +ℓ 2i+2 +ℓ 2i+4 for each i = 1, . . . , n. By Theorem 3.2.40 the ideal I is normal. But this can now be deduced directly from Theorem 7.6.14 by setting c t = t, S = {1, . . . , n}, W = {1, . . . , n}, and

  follows from (7.7.2) that δ ∈ (I + H) t . Thanks to I + H is normal, one has (I + H) t = (I + H) t , and so δ ∈ (I + H) t . Thus, we get δ ∈ I l H z for some l and z with l + z = t, in particular, δ k ∈ I lk H zk . One can derive from the minimality of p that p ≤ lk. On account of δ ∈ I l H z , we can write

		s
	(7.7.3)	δ =
		i=1

  Theorem 7.7.8. Let I ⊂ R = K[x 1 , . . . , x n , x n+1 ] be a normal square-free monomial ideal such that I ∩ (x n ) + (I : R x n+1 ) is normal. Then L := I ∩ (x n , x n+1 ) is normal.Proof. It is easy to check I ∩ (x n ) = x n (I : R x n ) and I ∩ (x n+1 ) = x n+1 (I : R x n+1 ). Hence, in view of[START_REF] Villarreal | Monomial algebras[END_REF] Exercise 6.1.23], we get L = x n (I : R x n ) + x n+1 (I : R x n+1 ). To simplify our notation, put F := (I : R x n ) and G := (I : R x n+1 ). Thus, L =x n F + x n+1 G. Let G(I) = {u 1 , . . . , u s }, and define f i = u i if x n ∤ u i , and f i = u i /x n if x n | u i .By the square-free assumption on I, none of the f i is divisible by x n . Similarly, define a set of elements g i by considering divisibility by x n+1 , i.e.,g i = u i if x n+1 ∤ u i and g i = u i /x n+1 if x n+1 | u i .Again, by the square-free assumption, none of the g i is divisible by x n+1 . One can readily see that F = (f 1 , . . . , f s ) and G = (g 1 , . . . , g s ). By[START_REF] Villarreal | Monomial algebras[END_REF] Proposition 12.2.3], F and G are normal, and so x n F and x n+1 G are also normal. By hypothesis, x n F + G is normal. Accordingly, we obtain (x n F + G) t = (x n F + G) t for all t ≥ 1. One has to prove that L t = L t for all t ≥ 1.

  2 , x 5 }. Let C n be a cycle graph with the vertex set V (C n ) = {1, . . . , n} and the edge set E(C n ) = {{x i , x i+1 } : i = 1, . . . , n}, where x 0 (respectively, x n+1 ) represents x n (respectively, x 1 ). Let

	Lemma 7.9.3. H :=	(x j-1 , x j , x j+1 ).
		j∈{1,...,n}\{ℓ 1 ,...,ℓ k }

  2m+1}\{ℓ 1 ,...,ℓ k } (x j-1 , x j , x j+1 ).Set I := DI(R G ) and H := j∈{1,...,2m+1}\{ℓ 1 ,...,ℓ k } (x j-1 , x j , x j+1 ). Hence, one has DI(G) = I + JH. By virtue of Lemma 7.9.3, we obtain H is normal. Moreover, Corollary 7.9.4 yields that I is normal. Because I ⊆ H, gcd(u, v) = 1 for all

	one can conclude immediately from Proposition 7.7.4(i) that DI(G) is normal, as
	desired.
	(ii) and (iii) can be verified similar to the proof of Corollary 6.2.3.

v, u ∈ G(J) with u = v, and also gcd(u, v) = 1 for all u ∈ G(I) ∪ G(H) and v ∈ G(J),

  there exists a positive integer1 ≤ j ≤ n such that a i 1 ,j > a i 2 ,j . (P2 )(d) when, for each d -1 distinct positive integers 1 ≤ j 1 , . . . , j d-1 ≤ n, there exists a positive integer 1 ≤ i ≤ m such that a i,j 1 + • • • + a i,j d-1 = 0. (P3 )(d) when, for each ℓ ≥ d distinct positive integers 1 ≤ j 1 , . . . , j ℓ ≤ n such that a i,j 1 + • • • + a i,j ℓ ≥ 1 for all i = 1, . . . ,m, then there exist at least d distinct integers j α 1 , . . . , j α d ∈ {j 1 , . . . , j ℓ } such that a i,jα 1 + • • • + a i,jα d ≥ 1 for all i = 1, . . . , m.

  m, if and only if {j 1 , . . . , j ℓ } ∩ A i = ∅ for all i = 1, . . . , m, if and only if {x j 1 , . . . , x j ℓ } ∩ p i = ∅ for all i = 1, . . . , m, if and only ifx j 1 • • • x j ℓ ∈ p i for all i = 1, . . . , m. This gives x j 1 • • • x j ℓ ∈ m i=1 p i , and hence x j 1 • • • x j ℓ ∈ I(H).Because H is a d-uniform hypergraph, we deduce that there exists an element u ∈ G(I(H))

  1 for all i = 1, . . . , m, then there exist at least r distinct integers j α 1 , . . . , j αr ∈ {j 1 , . . . , j ℓ } such that a i,jα 1 + • • • + a i,jα r ≥ 1 for all , . . . , θ t,r } ⊆ {1, . . . , n} for any integer 1 ≤ t ≤ n such that we have(a) a i,θ t,1 + • • • + a i,θt,r ≥ 1 for i = 1, . . . , m and t = 1, . . . , n. Furthermore, if z 1 , . . . , z r are r distinct integers such that a i,z 1 + • • • + a i,zr ≥ 1 for i = 1, . . . , m, then {z 1 , . . . , z r } = Γ s for some 1 ≤ s ≤ n;(b) for any integer 1 ≤ t ≤ n there exist r sets Γ i 1 , . . . , Γ ir with t ∈ Γ i k for all k = 1, . . . , r; ∈ Γ t if and only if t ∈ Γ θ t,i for i = 1, . . . , r and t = 1, . . . , n, which ensures us that the resulting graph is not directed. Suppose that V = {1, . . . , n} is a non-empty set of n elements, and let S = {S 1 , . . . , S m } be a non-empty set of m subsets of V . We define the if {x j 1 , . . . , x j ℓ } ∩ p i = ∅ for all i = 1, . . . , m, if and only if x j 1 • • • x j ℓ ∈ p i for all i = 1, . . . , m. This leads to x j 1 • • • x j ℓ ∈ ∩ m i=1 p i . On account of N I(G) = ∩ m i=1 p i and G is an (r-1)-regular graph, this implies that there exists an element u ∈ G(N I(G))

	i = 1, . . . , m.
	(iv) when, there exist n subsets (possibly some of them are the same) Γ t :=
	{θ t,1 (c) t ∈ Γ t for t = 1, . . . , n;
	(d) θ t,i Definition 8.2.2.

  , . . . , N G [n] such that, for each t = 1, . . . , n, N G [t] as the closed neighborhood of the vertex t ∈ V (G) has r elements. Set N G [t] := {θ t,1 , . . . , θ t,r } for each t = 1, . . . , n. It follows from the definition of closed neighborhood ideal of G that N I

  r. Since G is undirected, this is equivalent to this statement that θ t,i ∈ N G [t] if and only if t ∈ N G [θ t,i ] for i = 1, . . . , r and t = 1, . . . , n. This yields that M (S) satisfies condition (iv).

  • • • x θt,r : t = 1, . .. , n).In what follows, we prove that every element of S is a minimal dominating set of G. To see this, we first show that S 1 , . . . , S m are dominating sets of G. To do this, assume that N G [t] = {θ t,1 , . . . , θ t,r } is the closed neighborhood of the vertex t with 1 ≤ t ≤ n. By condition (iv)(a), we have that a i,θ t,1 + • • • + a i,θt,r ≥ 1 for all i = 1, . . . , m. Since, for all i = 1, . . . , m, the i-th row of the matrix M (S) is associated to S i and a i,θ t,1 + • • • + a i,θt,r ≥ 1, we obtain {θ t,1 , . . . , θ t,r } ∩ S i = ∅ for all i = 1, . . . , m. Consequently, N G [t] ∩ S i = ∅, and so S i is a dominating set of G for all i = 1, . . . , m.

  similar to Case 1, one can conclude that W i is a minimal path cover set ofT ′ i . Case 4. deg T w i = 1. Then V (T i ) = {w i }, and so W i = ∅. Set d i := |L T ∩ V (T i )| for all i = 1, . . . , ℓ and suppose that Case 1 occurs for T 1 , . . . , T t , Case 2 occurs for T t+1 , . . . , T k , Case 3 occurs for T k+1 , . . . , T m and finally Case 4 occurs for T m+1 , . . . , T ℓ

  there exists a unique vertex w i ∈ V (T i ) such that {z 1 , w i } ∈ E(T ). Coming back to the proof of theorem 9.1.8, without loss of generality, we can assume that for each deg T w i (2 ≤ i ≤ ℓ 1 ), Case 2 or Case 4 of the proof of Theorem 9.1.8 happens and for deg T w 1 , Case 1 or Case 3 happens. Hence, z 1 is the center of a starlike subgraph S 1 of T and for T 1 one of the following two cases happens:Case 1. deg T w 1 ≥ 3. In this case, since {z 2 } is a minimal path cover set of T 1 , T 1 is a path graph or a starlike graph. Hence, we getT = S 1 ∪ E ∪ S 2, where S 2 is also a starlike graph whose center belongs to V (T 1 ) and E satisfies one of the conditions (i) or (ii).Case 2. deg T w 1 = 2 and T 1 is not a path graph. Then T 1 is the union of a path P 1 and a tree T ′ 1 such that if v ′ 1 is the cut point of this union, then deg T v ′ 1 ≥ 3 and {z 2 } is a minimal path cover set of T ′ 1 . Therefore, T ′ 1 is a path graph or a starlike graph and again we conclude that T = S 1 ∪ E ∪ S 2 , where S 2 is also a starlike graph whose center belongs to V (T ′ 1

  The example below illuminates what happens in Theorem 9.2.1. Assume that G 1 , G 2 , and G 3 are the graphs as shown in the fol-

	Example 9.2.2.

  Suppose that G 1 , . . . , G k are path graphs, G k+1 , . . . , G t are cycle graphs, G t+1 , . . . , G r are complete graphs, and G r+1 , . . . , G s are graphs which each of them is the union of a cycle graph and a (3, n)-tadpole graph for some natural number n such that satisfies the condition of Proposition 9.2.11. Suppose that G = s i=1

	6		5	5				6	5
	9			4	3	2	1	9		4	3	2	1
	8	G	7	7	T			8	7	G ∪ T
	Theorem 9.2.12.

  that is, all z i are distinct. This implies that deg T z i ≥ 3 for each i = 1, . . . , m, and so m i=1 degT z i ≥ 3m. It is clear that m i=1 deg T w i = m. Note that V (T ) \ {w 1 , .. . , w m , z 1 , . . . , z m } = ∅, and also for any v ∈ V (T ) \ {w 1 , . . . , w m , z 1 , . . . , z m }, we have deg T v ≥ 2. We therefore deduce that Thus, we get v∈V (T ) deg T v ≥ 2n. On the other hand, it is well-known that v∈V

	m	m
	deg T w i +	deg T z i +	deg T v ≥ m + 3m + 2(n -2m).
	i=1	i=1	v∈V (T )\{w 1 ,...,wm,z 1 ,...,zm}

  1. deg T f k ≥ 3. In view of the proof of Theorem 9.1.8, B 1,k and B 2,k are two minimal path cover sets of T k . Due to |S T k | is less than n, it follows from the inductive hypothesis that B 1,k = B 2,k . Case 2. deg T f k = 2 and T k is not a path graph. Hence, T k is the union of a path Q and a tree T ′ such that if z is the cut-point of this union, then deg T z ≥ 3. In view of the proof of Theorem 9.1.8, we get B 1,k and B 2,k are two minimal path cover sets of T ′ . According to |S T ′ | is less than n, the inductive hypothesis yields that B 1,k = B 2,k . Case 3. deg T f k = 2 and T k is a path graph. These imply that B 1,k

  It is routine to check that, in the polynomial ring R = K[x 1 , . . . , x 12 ] over a field K, we haveP I(T ) = (x 1 x 2 x 3 , x 5 x 6 x 7 , x 1 x 2 x 4 x 5 x 7 , x 2 x 3 x 4 x 5 x 6 , x 2 x 3 x 4 x 5 x 7 , x 1 x 2 x 4 x 8 x 9 , x 1 x 2 x 4 x 8 x 10 ,x 1 x 2 x 4 x 8 x 11 x 12 , x 2 x 3 x 4 x 8 x 9 , x 2 x 3 x 4 x 8 x 10 , x 2 x 3 x 4 x 8 x 11 x 12 , x 4 x 5 x 6 x 8 x 9 , x 4 x 5 x 6 x 8 x 10 , x 4 x 5 x 6 x 8 x 11 x 12 , x 4 x 5 x 7 x 8 x 9 , x 4 x 5 x 7 x 8 x 10 , x 4 x 5 x 7 x 8 x 11 x 12 , x 1 x 2 x 4 x 5 x 6 , x 8 x 9 x 10 , x 8 x 9 x 11 x 12 , x 8 x 10 x 11 x 12 ).

	2	4	8	9 10
		5	11	12
	6		7

  Example 9.3.2. Let t = 4 and K 1,2,3 be as shown in figure below with partition sets {x 1 }, {x 2 , x 3 } and {x 4 , x 5 , x 6 }.

		x 2
	x 3		x 1
	x 4	x 5	x 6
		K 1,2,3

  x t+1 2 +1 , x 2 , x t+1 2 +2 , . . . , x t-1 -path in G, by virtue of consecutive vertices in the above sequence belong to different V i 's. Similarly, if t is even, then x 1 , x t 2 +1 , x 2 , x t 2 +2 , . . . , x t 2 , x t is a t-path in G. This completes the proof. 9.3.3 limit depth of I t (K n 1 ,...,n r )

	2	, x t , x t+1 2	,
	is a t

  1 1 • • • x sn n with nonnegative integers s 1 , . . . , s n . It follows from m = (I : R h) that h / ∈ I, and also hx i ∈ I for each i = 1, . . . , n. Fix 1 ≤ t ≤ n. Since hx t ∈ I, one can conclude that there exists a positive integer 1 ≤ i t ≤ k such that x Thus, r it,j ≤ s j for each j ∈ {1, . . . , t -1, t + 1, . . . , n}, and also r it,t ≤ s t + 1. On the other hand, h / ∈ I implies that x ∤ x s 1 1 • • • x sn n for each i = 1, . . . , k. Therefore, we gain x

	r i,1 1 • • • x	r i,n n
	r i t ,1
	1
	r i t ,1 1	• • • x	r i t ,n n	|x s 1 1 • • • x	s t-1 t-1 x st+1 t	x	s t+1 t+1 • • • x sn n .
	306

Sseparable monomial ideal , 35 simplicial complex, 7 special cycle, 147 split generator, 35 spoke, 219 square-free monomial ideal, 4 square-free Veronese ideal of degree d, 299 stable set, 3 starlike tree, 25 strong elimination ordering, 149 strong persistence property, 3 strongly chordal graphs, 149 strongly superficial element, 32 subgraph, 4 superficial element, 32 superficial ideal, 3 support, 4 suspension, 90 symbolic strong persistence property, 3 symmetric starlike tree, 25

Acknowledgments

First and foremost, I would like to express my gratitude and appreciation for Professor André Leroy, my mentor, whose guidance, support, and encouragement have been invaluable throughout this study. I also would like to state my most heartfelt thanks to Professor Ahmad Laghribi the director of the Department of Mathematics at Artois University for his academic support. Finally, I would like to thank my wife for her patience and encouragement during the preparation of this project.

Theorem 5.3.16. Let I be a square-free monomial ideal in R = K[x 1 , . . . , x n ], and 1 ≤ j ≤ n. If I is normally torsion-free, then I \ x j is so.

Proof. Assume that I is normally torsion-free. Fix k ∈ N, and suppose that Ass R (R/I) = {p 1 , . . . , p r }, i.e., we have I = p 1 ∩ • • • ∩ p r . Since I is a square-free monomial ideal, [START_REF] Villarreal | Monomial algebras[END_REF]Proposition 3.3.26] yields that I k = I (k) , where I (k) denotes the k-th symbolic power of I. On the other hand, it follows from [START_REF] Villarreal | Monomial algebras[END_REF]Exercise 5.1.29] that

Consequently, one has

Thanks to parts (ii) and (iii) of Lemma 3.2.22, one can conclude the following equalities

Furthermore, part (v) of Lemma 3.2.22 deduces that

is a primary decomposition of (I \ x j ) k . Suppose that Ass R\x j ((R \ x j )/(I \ x j )) = {p i 1 \ x j , . . . , p it \ x j },

where {i 1 , . . . , i t } ⊆ {1, . . . , r}. Accordingly, one derives that

is the minimal primary decomposition of I \ x j . Due to p i 1 \ x j , . . . , p it \ x j are minimal associated primes of I \ x j , we get

Once again, by virtue of [START_REF] Villarreal | Monomial algebras[END_REF]Exercise 5.1.29] yields the following equality

Therefore, we obtain (I \x j ) (k) = (I \x j ) k . Now, [START_REF] Villarreal | Monomial algebras[END_REF]Proposition 3.3.26] implies that I \ x j is normally torsion-free, as claimed.

Proof. We first note that the assumption concludes that the monomial ideal I must have the following form

for some positive integers d 1 , . . . , d n . This gives rise to Ass R (R/I) = Min(I). In addition, since (m \ {x i }) d i , for each i = 1, . . . , n, is normally torsion-free, the claim can be promptly deduced from Lemma 6.1.2.

The next lemma says that a monomial ideal is nearly normally torsion-free if and only if its monomial multiple is nearly normally torsion-free under certain conditions. It is an updated version of [START_REF] Herzog | On the associated prime ideals and the depth of powers of squarefree principal Borel ideals[END_REF]Lemma 3.5] and Lemma 5.3.11. Lemma 6.1.5. Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ], and h be a monomial in R such that gcd(h, u) = 1 for all u ∈ G(I). Then I is nearly normally torsion-free if and only if hI is nearly normally torsion-free.

Proof. (⇒) Assume that I is nearly normally torsion-free. Let h = x b 1 j 1 • • • x bs js with j 1 , . . . , j s ∈ {1, . . . , n}. On account of Theorem 2.1.27, we obtain, for all ℓ, (6.1.1)

Ass R (R/(hI) ℓ ) = Ass R (R/I ℓ ) ∪ {(x j 1 ), . . . , (x js )}.

Due to gcd(h, u) = 1 for all u ∈ G(I), it is routine to check that (6.1.2) Min(hI) = Min(I) ∪ {(x j 1 ), . . . , (x js )}.

Since I is nearly normally torsion-free, there exist a positive integer k and a monomial prime ideal p such that Ass R (R/I m ) = Min(I) for all 1 ≤ m ≤ k, and

Ass R (R/I m ) ⊆ Min(I) ∪ {p} for all m ≥ k + 1. Select an arbitrary element q ∈ Ass R (R/(hI) m ). Then (6.1.1) implies that q ∈ Ass R (R/I m ) ∪ {(x j 1 ), . . . , (x js )}.

If 1 ≤ m ≤ k, then (6.1.2) yields that q ∈ Min(hI). Hence, let m ≥ k + 1. The claim follows readily from (6.1.1), (6.1.2), and Ass R (R/I m ) ⊆ Min(I) ∪ {p}.

(⇐) Let hI be nearly normally torsion-free. This means that there exist a positive integer k and a monomial prime ideal p such that Ass R (R/(hI) m ) = Min(hI) for all 1 ≤ m ≤ k, and Ass R (R/(hI) m ) ⊆ Min(hI) ∪ {p} for all m ≥ k + 1. Take an arbitrary element q ∈ Ass R (R/I m ). Because q ∈ Ass R (R/I m ), we get q / ∈ {(x j 1 ), . . . , (x js )}. If 1 ≤ m ≤ k, then (6.1.1) gives that q ∈ Ass R (R/(hI) m ), and so q ∈ Min(hI). As q / ∈ {(x j 1 ), . . . , (x js )}, we gain q ∈ Min(I). We thus assume that m ≥ k +1. One can derive the assertion according to the facts q / ∈ {(x j 1 ), . . . , (x js )}, (6.1.1), (6.1.2), and Ass R (R/(hI) m ) ⊆ Min(hI) ∪ {p}.

We conclude this section by observing that nearly normally torsion-freeness does not imply the strong persistence property. It is not known if nearly normally torsionfreeness implies the persistence property. Direct computations show that A 1 = {6, 7, 8}, A 4 = {9, 10}, A 2 = A 3 = A 5 = ∅, and B {1,5} = {11, 12, 13, 14}, B {2,3} = {15, 16, 17}, B {1,2} = B {3,4} = B {4,5} = ∅.

The following theorem is the first main result of this section. Theorem 6.2.1. Assume that G = (V (G), E(G)) is a finite simple connected graph, and J(G) denotes the cover ideal of G. Then J(G) is nearly normally torsion-free if and only if G is either a bipartite graph or an almost bipartite graph.

Proof. To show the forward implication, let J(G) be nearly normally torsion-free. Suppose, on the contrary, that G is neither bipartite nor almost bipartite. This gives that G has at least two induced odd cycle subgraphs, say C and C ′ . It follows from Proposition 3.1.15 that p = (x j : j ∈ V (C)) ∈ Ass(J(C) s ) and p ′ = (x j : j ∈ V (C ′ )) ∈ Ass(J(C ′ ) s ) for all s ≥ 2. Since G p = C and G p ′ = C ′ , we can deduce from [START_REF] Francisco | Colorings of hypergraphs, perfect graphs and associated primes of powers of monomial ideals[END_REF]Lemma 2.11] that p, p ′ ∈ Ass R (R/J(G) s ) for all s ≥ 2. This contradicts the assumption that J(G) is nearly normally torsion-free.

Conversely, if G is bipartite, then on account of [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6], one has J(G) is normally torsion-free, and so J(G) is nearly normally torsion-free. Next, we assume that G is an almost bipartite graph, and let C be its unique induced odd cycle subgraph. Put p = (x j : j ∈ V (C)). We claim that Ass(J(G) s ) = Min(J(G)) ∪ {p} for all s ≥ 2. Fix s ≥ 2. For any i ∈ V (C) and e ∈ E(C), assume that A i and B e are the vertex subsets of G as defined in the discussion above. Without loss of generality, suppose that A i = ∅ for all i = 1, . . . , r and B e j = ∅ for all j = 1, . . . , t. Set H i := A i ∪ {i} for all i = 1, . . . , r, and L j := B e j ∪ {α j , β j } for all j = 1, . . . , t, where e j = {α j , β j }. Accordingly, we get |V (C) ∩ V (H i )| = 1 for all i = 1, . . . , r, |V (C) ∩ V (L j )| = 2 and |E(C) ∩ E(L j )| = 1 for all j = 1, . . . , t.

On the other hand, it should be noted that all H i and L j are bipartite, and so [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6] yields that Ass(J(H i ) s ) = Min(J(H i )) and Ass(J(L j ) s ) = Min(J(L j )) for all i = 1, . . . , r and j = 1, . . . , t. Now, repeated applications of Theorems 3.3.23 and 3.3.25 give that Ass(J(G) s ) = Ass(J(C) s ) ∪ Min(J(H i )) ∪ Min(J(L j )) for all i = 1, . . . , r and j = 1, . . . , t. By virtue of Proposition 3.1.15, one obtains Ass(J(C) s ) = {p} ∪ Min(J(C)). We thus have Ass(J(G) s ) = Min(J(G)) ∪ {p}, as claimed. This shows that J(G) is nearly normally torsion-free, and the proof is done.

Now, we focus on cover ideals of hypergraphs. For this purpose, we state the following theorem. To do this, one has to recollect Definitions 3.1.3, 3.1.4, and 3.1.5, which will be necessary for understanding Theorem 6.2.2.

Theorem 6.2.2. Assume that

Then the Alexander dual of I is given by

Then the edge ideal and cover ideal of the hypergraph H are given by

It is easy to see that J(H) = I, as claimed. We are in a position to provide the second main result of this section in the following corollary. Corollary 6.2.3. Let I be a normally torsion-free square-free monomial ideal in

(ii) L is nearly normally torsion-free.

(iii) L is normal.

(iv) L has the strong persistence proeprty.

(v) L has the persistence property.

(vi) L has the symbolic strong persistence property.

Proof. (i) In the light of the argument which has been stated above, we can assume that I = J(H) such that the hypergraph H corresponds to I ∨ , where I ∨ denotes the Alexander dual of I. Fix t ≥ 1. It follows now from Theorem 6.2.2 that

Since I is normally torsion-free, one can deduce that Ass R (R/J(H) t ) = Min(J(H)), and so Ass S (S/L t ) = Min(J(H)) ∪ {(x n , x n+1 )}. Therefore, Ass S (S/L t ) = Min(L). This means that L is normally torsion-free, as desired.

(ii) It is obvious from the definition of nearly normally torsion-freeness and (i). x i ).

Since i∈V (G) x i ∈ N I(G), this implies that N I(H) = x w N I(G).

(i) This claim can be deduced from Lemma 5. Proof. We first note that one may assume gcd(v, h) = 1. To see this, let d := gcd(v, h). This yields that v = dv 1 and h = dh 1 for some monomials v 1 and h 1 . Consequently, one derives vI

Also, it follows from h ∈ I that there exists an element u ∈ G(I) such that u|h, and so u|h 1 d. By virtue of gcd(u, v) = 1, one has gcd(u, d) = 1, and hence u|h 1 , that is, h 1 ∈ I. Furthermore, since gcd(v, u) = 1 for every u ∈ G(I), one can conclude that gcd(v 1 , u) = 1 for every u ∈ G(I). Accordingly, if we show that v 1 I + h 1 R is normal, then Remark 7.1.1 implies that vI + hR is also normal. Thus, one may assume that gcd(v, h) = 1.

Assume α is a monomial in (vI + hR) t , thus α k ∈ (vI + hR) tk for some k. Hence, α k ∈ (vI) p (hR) q for some p and q with p + q = tk. Choose q to be the minimal according to this membership. If q = 0, then α k ∈ (vI) tk ; thus, α ∈ (vI) t . Since I is normal, then vI is also normal by Remark 7.1.1. Therefore, α ∈ (vI) t ⊆ (vI + hR) t , and the proof is done. Therefore, assume q > 0. As G(I) = {u 1 , . . . , u r } and α k ∈ (vI) p (hR) q , we thus get (7.1.1)

As h ∈ I and I is a monomial ideal, then by [START_REF] Herzog | Monomial Ideals[END_REF]Proposition 1.1.5], h is divisible by some element of G(I), say u r . Thus, we obtain h = u r h ′ for some monomial h ′ . Now, if v divides β, say β = vβ ′ , then (7.1.1) can be rewritten as

contradicting the minimality of q. Therefore, we may assume in (7.1.1) that v ∤ β,

This finishes the proof. Remark 7.1.4. We show that the hypotheses in the above theorem are best possible.

(i) The square-freeness of the monomial v in the above theorem is necessary. For instance, let 2 , and therefore, the ideal vI + hR is not integrally closed, hence not normal.

(ii) The condition that gcd(u, v) = 1 for every u ∈ G(I) is necessary. For instance, let

Throughout this section, if G is a finite simple graph, then N I(G) (respectively, DI(G)) stands for the closed neighborhood ideal (respectively, dominating ideal) of G.

Some criteria for normality of monomial ideals

This subsection is devoted to presenting some criteria which help us to detect the normality of monomial ideals. In particular, our results will be used in studying closed neighborhood ideals and dominating ideals of some classes of graphs.

The following theorem is essential for us to verify Theorem 7.7.3 and Proposition 7.7.6. It is necessary to observe that Theorem 7.7.1 is an updated version of Theorem 6.5.1. In fact, in Theorem 7.7.1, we focus on the monomial ideals, while in Theorem 6.5.1, the authors argued on square-free monomial ideals. Theorem 7.7.1. Let I and H be two normal monomial ideals in a polynomial ring

To establish the normality of L, one has to prove that L t = L t for all integers t ≥ 1. Obviously, L t ⊆ L t , so it is enough to show that L t ⊆ L t . For this it suffices to show that an arbitrary monomial α ∈ L t is in L t . Write α = x b 1 δ with x 1 ∤ δ and δ ∈ R. In view of [START_REF] Herzog | Monomial Ideals[END_REF]Theorem 1.4.2], one has α k ∈ L tk for some integer k ≥ 1. Hence, one can conclude that α k ∈ I p (x c 1 H) q for some p and q with p + q = tk. If p = 0 (respectively, q = 0), then x bk

, and the proof is complete. Accordingly, one can assume that p > 0 and q > 0. Then, we can write (7.7.1)

In this case, we can consider the following subcases: Subcase 4.1. There exists exactly one integer α ∈ {1, . . . , s -1} such that |supp(u α ) ∩ supp(u s )| = 2, say supp(u α ) ∩ supp(u s ) = {x is-1 , x is }. Also, there exists unique β ∈ {1, . . . , s-1}\{α} such that x is+1 ∈ u β . Assume that u β = (x is+1 , x r , x t ). Set A := j∈{1,...,s-1}\{β} u j . We thus get the following equalities

It follows from the inductive hypothesis that A is normal. Furthermore, by mimicking the argument in Case 3, one can derive that A ∩ (x is-1 , x is ) ∩ (x r , x t ) is normal. We therefore conclude from Theorem 7.7.3(i) that H is normal.

Subcase 4.2. There exist exactly two distinct integers

. Put B := j∈{1,...,s-1}\{α,β} u j . This yields the following equalities

One can deduce from the inductive hypothesis that B is normal. By repeating the argument in Case 3, we can deduce that B ∩ (x is-1 , x is+1 ) ∩ (x θ 1 , x θ 2 ) ∩ (x λ 1 , x λ 2 ) is normal as well. Thanks to Theorem 7.7.3(i), we obtain H is normal.

This completes the inductive step, and hence the claim has been proved by induction.

(ii) and (iii) can be shown similar to the proof of Corollary 6.2.3.

The next corollary which is an immediate consequence of Lemma 7.9.3, says that the dominating ideals of cycle graphs are normal and also satisfy the (strong) persistence property. Moreover, this corollary will be used in proving Theorem 7.9.7. Corollary 7.9.4. Let C n be a cycle graph with the vertex set V (C n ) = {1, . . . , n} and the edge set E(C n ) = {{x i , x i+1 } : i = 1, . . . , n}, where x n+1 represents x 1 , and DI(C n ) be its dominating ideal. Then the following statements hold:

(ii) DI(C n ) has the strong persistence property.

(iii) DI(C n ) has the persistence property. incidence matrix associated to S, denoted by M (S), as the binary matrix M (S) = (a ij ) with m rows and n columns such that a ij = 0 if j / ∈ S i , and a ij = 1 if j ∈ S i for all i = 1, . . . , m and j = 1, . . . , n.

We are ready to state the main result of this paper in the following theorem. Theorem 8.2.3. Suppose that V = {1, . . . , n} is a non-empty set of n elements, S = {S 1 , . . . , S m } a non-empty set of m non-empty subsets of V , and 2 ≤ r ≤ n a positive integer. Then there exists an undirected (r -1)-regular graph G with V (G) = V , where S is the set whose elements are the minimal dominating sets of G if and only if the matrix M (S) = (a ij ) satisfies conditions (i)-(iv).

If there exists such (r -1)-regular graph G, then the closed neighborhood ideal of G is given by

with j 1 , . . . , j r are distinct positive integers).

Proof. To establish the forward implication, assume there exists an (r -1)-regular graph G with V (G) = V , where S is the set whose elements are the minimal dominating sets of G. We show that M (S) satisfies condition (i). On the contrary, assume that condition (i) does not hold. So there exist 1 ≤ i 1 , i 2 ≤ m with i 1 = i 2 such that a i 1 ,j ≤ a i 2 ,j for each j = 1, . . . , n. Hence, we get S i 1 ⊆ S i 2 , which contradicts the minimality of S i 2 . Accordingly, we conclude that M (S) satisfies condition (i).

We prove that M (S) satisfies condition (ii). On the contrary, assume that there exist r-1 distinct positive integers 1 ≤ j 1 , . . . , j r-1 ≤ n, such that a i,j 1 +• • •+a i,j r-1 = 0 for each 1 ≤ i ≤ m. To simplify the notation, put p i := (x w : w ∈ S i ) for each i = 1, . . . , m. Due to S 1 , . . . , S m are the minimal dominating sets of G, [122, Lemma 2.2] implies that N I(G) = ∩ m i=1 p i , where N I(G) denotes the closed neighborhood ideal of G. The assumption gives that {j 1 , . . . , j r-1 } ∩ S i = ∅ for each i = 1, . . . , m. Hence, one can conclude that {x j 1 , . . . , x j r-1 } ∩ p i = ∅ for each i = 1, . . . , m. This yields that x j 1 • • • x j r-1 ∈ p i for each i = 1, . . . , m, and so

On the other hand, since G is an (r -1)-regular graph, one obtains degu = r, which is a contradiction.

Here, we demonstrate that M (S) satisfies condition (iii). Suppose that for each

Want to show that there are at least r distinct integers j α 1 , . . . , j αr ∈ {j 1 , . . . , j ℓ } such that a i,jα

, m, if and only

Chapter 9

On the max-path ideals and matroidal path ideals

Maximal paths and path cover sets

Monomial ideals play a fundamental role in studying the connection between commutative algebra and combinatorics. The relation between these two subjects allows us to use techniques and methods in commutative algebra to solve combinatorial problems, and vice versa. So, commutative algebraists have started exploring the properties of finite simple graphs through monomial ideals, see [START_REF] Miller | Combinatorial commutative algebra[END_REF] for more information.

In the following text, our major goal is to introduce two new notions which are related to paths in a finite simple graph. In fact, similar to edge ideals and cover ideals, we define the notions max-path ideals and path cover ideals associated to simple finite graphs. It should be noted that the results of this section can be found in [START_REF] Sharifan | Minimal path cover sets and monomial ideals[END_REF].

Let G be a graph on the vertex set [n] = {1, . . . , n}. Then the sequence of distinct vertices i 1 , . . . , i r is called a maximal path when i 1 , . . . , i r is a path and furthermore, for all j ∈ {1, . . . , n} \ {i 1 , . . . , i r }, none of j, i 1 , . . . , i r and i 1 , . . . , i r , j is a path.

Then W 1 := {3, 5} and W 2 := {1, 4} are some path cover sets of G. Definition 9.1.4. Suppose that G is a graph and set

Clearly, if G is a cycle or a complete graph or a path graph, then I G = 1.

Example 9.1.5. Assume that G is the graph of Example 9.1.3. Then we can easily see that I

In this section, we focus on the case that G is a tree. Furthermore, from now on, we consider the graphs K n , C n , and P n , as the complete, the cycle, and the path graphs on the vertex set [n] = {1, . . . , n}, respectively.

In the next proposition we present some elementary properties of maximal paths in a tree. Proposition 9.1.6. Suppose that T is a tree. Then (i) The path i 1 , . . . , i t is a maximal path if and only if both i 1 and i t belong to L T .

(iii) If i 1 , . . . , i t and j 1 , . . . , j s are two distinct maximal paths, then we have

The following proposition is an immediate consequence of Proposition 9.1.6(i) which presents an upper bound for the index of covering of a tree.

In the next theorem we present an upper bound for the number of elements of any minimal path cover set of a tree. Theorem 9.1.8. Suppose that T is a tree and

is called the max-path ideal of G, and also

is called the path cover ideal of G.

We now focus on max-path ideals and path cover ideals of trees. Note that two different graphs on the vertex set [n] may have the same max-path ideal. For instance, consider the graphs K n , C n , and

In fact, in the graph K n we have n!/2 different maximal paths that all of them are corresponded to the generator

As another example, let G be a graph on [5] = {1, 2, 3, 4, 5} with

Then

is the set of all maximal paths in G,

and

In the following proposition we study some properties of max-path and path cover ideals. Proposition 9.1.12. Let G be a graph on the vertex set [n].

is the minimal primary decomposition of P I(G).

(ii) P C(G) = P I(G) ∨

, where I ∨ means the Alexander dual ideal of I.

Proof. (i) By using the definition of the max-path ideal of G, we only need to show that

To understand the next result, one has to recollect the definitions of starlike trees and symmetric starlike trees, see Definition 2.2.4.

In the following corollary, we investigate the relation between max-path ideals and path ideals. Corollary 9.1.18. Let T be a tree on the vertex set [n] which is not a path graph. Then the following conditions are equivalent.

(i) P I(T ) = P t (T ) for some integer t.

(iv) T is a symmetric starlike tree.

Proof. (i) =⇒ (ii). Let P I(T ) = P t (T ) for some integer t. By Proposition 9.1.17(iii), we have the following equalities

Now, in view of Proposition 9.1.17(v), it follows that P C(T ) is unmixed.

(ii) =⇒ (iii). Let u, v ∈ L T . Then there exists a path between them in T such as u, i 1 , . . . , i r , v. Hence,

According to (iii), it is enough to prove that T is a starlike tree. By contrary assume that there exist

and the path u ′ , u, u ′′ to a maximal path j 1 , . . . , u ′ , u, u ′′ , . . . , j t+1 as the following figure. It is routine to check that

). On the other hand, one can easily see that

) is a minimal path cover set of the tree which the vertex v = 5 is the cut-point of the union of the tree and cycle.

In the light of Theorem 9.2.1, we provide the following proposition.

Proposition 9.2.3. Let G be the union of a cycle C and a tree T such that v is the cut-point of this union. Suppose that W G is an arbitrary minimal path cover set of

Due to Theorem 9.2.1, we have the following two cases:

We finally get

Let G be the union of a path graph L and a connected graph H such that v is the cut-point of this union. Also, let W G be an arbitrary minimal path cover set of G and V (L) ∩ W G = S. It follows now from the minimality of W G that, if S is non-empty, then S is singleton. The following theorem discusses the minimal path cover set of H with respect to W G . Theorem 9.2.4. Let G be the union of a path graph L : v, r 1 , . . . , r s , w with w ∈ L G and a connected graph H such that v is the cut-point of this union. Suppose that W G is an arbitrary minimal path cover set of G and V (L) ∩ W G = {t}. Then one of the following statements holds.

Proof. We use induction on n := |L T |. Since T is a tree and deg T v = 1, one has

then T is a path graph, and the assertion follows from Theorem 9.2.4. Now, suppose, inductively, that n ≥ 3 and that the result has been proved for any graph which satisfies the hypothesis of our theorem with the size of the set of leaves of the tree is less than n. Let G be the union of a tree T and a connected graph H such that v is the cut-point of this union with deg T v = 1 and |L T | = n. Let also W G be an arbitrary minimal path cover set of G and V (T ) ∩ W G = S. Suppose that, for each w = v in L T , there exist a path L ′ : v, . . . , z and a path graph L : z, . . . , w

Then one can consider the graph G as the union of a connected graph G ′ and the path graph L : z, . . . , w with deg T z ≥ 3 such that z is the cut-point of this union. Set S ′ := V (L) ∩ W G . By our hypothesis, we get S ′ = ∅. Now, according to Theorem 9.2.4, we can deduce that one of the following statements holds.

This implies that G ′ is the union of the tree T ′ and graph H such that v is the cut-point of this union. Set

according to the inductive hypothesis, we can conclude that one of the following statements holds.

(1) W G is a minimal path cover set of H.

Due to (9), by virtue of {z} \ S 3 = ∅, one can conclude that

Hence, one has (iii). This completes the inductive step, and therefore the theorem has been proved by induction.

We are now ready to present the main result of this section in the following theorem.

Theorem 9.2.6. Let G be the union of trees T 1 , . . . , T n and a connected graph H such that v is the cut-point of this union with deg

Let also W G be an arbitrary minimal path cover set of G, and for each w = v in L T i , there exist a path L ′ i : v, . . . , z and a path graph L i : z, . . . , w with deg

Proof. We proceed by induction on n. By Theorem 9.2.5, the assertion is true for n = 1. Now, suppose, inductively, that n > 1 and that the result has been proved for all graphs which are the union of some trees and a connected graph with a cutpoint, such that the number of trees is less than n and satisfy the condition has been stated in our theorem. Let G be the union of trees T 1 , . . . , T n and a connected graph H such that v is the cut-point of this union. Let also W G be an arbitrary minimal path cover set of G, and the tree T i satifies the condition has been stated in our theorem for all i = 1, . . . , n.

Assume that T is the union of the trees T 1 , . . . , T n-1 with v is the cut-point of this union. Now, suppose that G ′ is the union of the tree T and the connected graph H such that v is the cut-point of this union. Therefore, G is the union of the graph G ′ and the tree T n such that v is the cut-point of this union. Put S ′ := V (T n ) ∩ W G . By the inductive hypothesis, it follows that one of the following statements holds.

Once again, according to the inductive hypothesis, we can deduce that one of the following statements holds.

(1) W G is a minimal path cover set of H.

(2) W G \ S 1 is a minimal path cover set of H. 1) holds, then we get (i).

Assume that statement (2) holds. It follows that

)), and one has

is a minimal path cover set of H, we deduce that W G \ S 1 ⊆ V (H), and hence

and so v ∈ S 1 , which is a contradiction. Thus, S ′ \ S 1 = ∅. This implies that W G \ S 1 = W G \ S, and so we obtain (ii).

In the statement (3), based on the argument has been stated in (2), we have

∪ {v}, and so we have (iii).

Assume that statement (4) holds. One can easily check that

Let I(G) be not a principal ideal. If gcd(I(G)) = x i , for some variable x i , then G is a complete bipartite graph with r-partition V 1 = {x i } and V 2 = V (G) \ {x i }, and again the assertion holds in this case. Finally, we assume that gcd(I) = 1. Note that for each i = 1, . . . , n, we have

, where N G (x i ) denotes the neighborhood of x i in G.

Then from Proposition 9.3.5(ii), we conclude that if

We can partition vertices of G into disjoint sets, say V 1 , . . . , V r such that for all k = 1, . . . , r, the vertices in V k have the same neighborhood. Since G is a simple graph, it is clear that each of the V k is an independent set of G. We claim that

To prove our claim, it only remains to verify the following: if

. By virtue of Proposition 9.3.5(ii), we have

Then {x i , x k } and {x j , x ℓ } are disjoint edges in G. Since I(G) is matroidal, then by employing the exchange property on x i x k and x j x ℓ , we conclude that {x i , x j } ∈ E(G), as claimed. This completes the proof.

To establish the main result of this section, we first need the following auxiliary lemma. Lemma 9.3.7. Let G = K n 1 ,...,nr and t ≥ 3. Furthermore, let P 1 : x i 1 , . . . , x i t-1 be a (t -1)-path in G and P 2 : x j 1 , . . . , x jt be a t-path in G. Then there exists

) such that one of the following statements is satisfied.

(3) t ≥ 4 and there exists 2

..,nr . We may assume that

If (1) is not true, then every

If

, and hence we obtain a t-path P : x i 1 , . . . , x i t-1 , x j k in G, as given in (2). Note that if t = 3, then in P 1 : x i 1 , x i 2 , both vertices cannot be in G 1 because G is a complete r-partite graphs. Hence, either (1) or (2) must be true for t = 3.

If both (1) and ( 2) do not hold, then t ≥ 4. In this case, we try to construct a path as given in (3). The negation of (1) and (2) gives us that

Similarly, one defines the square-free counterpart of a Veronese ideal. The monomial ideal I is called square-free Veronese ideal of degree d in S if I is generated by all square-free monomials of degree d in S. It follows from [START_REF] Herzog | Cohen-Macaulay polymatroidal ideals[END_REF]Theorem 4.2] that a polymatroidal ideal I is Cohen-Macaulay if and only if I is (i) a principal ideal, or (ii) a Veronese ideal, or (iii) a square-free Veronese ideal.

In particular, if I is a matroidal ideal, then it is Cohen-Macaulay if and only if it is a principal ideal or square-free Veronese ideal. From Remark 9.3.3(v) it follows that

and hence in this case it can be viewed as the square-free Veronese ideal of degree t = n in S. Therefore, for a given t, to be able to characterize n 1 , . . . , n r such that I t (K n 1 ,...,nr ) is Cohen-Macaulay, one only needs to check that when I t (K n 1 ,...,nr ) is square-free Veronese of degree t in S. More precisely, one needs to check that for which n 1 , . . . , n r with

. , x n ],

that is, for every 1 ≤ i = j ≤ n, we have x i x j ∈ I(G). Moreover, x i x j ∈ I(G) for every 1 ≤ i = j ≤ n, if and only if G is a complete graph. By using Theorem 9.3.6, the above proposition can be rephrased as follows:

We are in a position to state the main result of this section in which we discuss the general case when t ≥ 2. Theorem 9.3.11. The t-path ideal I t (K n 1 ,...,nr ) = 0 is Cohen-Macaulay if and only if n i ≤ t/2 for each i = 1, . . . , r. [START_REF] Herzog | Cohen-Macaulay polymatroidal ideals[END_REF]Theorem 4.2] and Remark 9.3.3(v), one can conclude that if I t (G) ⊂ S = K[x 1 , . . . , x n ] is Cohen-Macaulay, then it is a square-free Veronese ideal of degree t, and hence x i 1 • • • x it ∈ I t (G) for every Remark 9.3.12. In [61, Proposition 2.1], it is shown that for any graded ideal I, the depth(S/I k ) is a non-increasing function of k if all powers of I have a linear resolution. This is indeed the case for matroidal ideals. Therefore, one concludes the following: if limit depth of a matroidal ideal is s and the k is the smallest integer for which depth(S/I k ) = s, then dstab(I) = k. Remark 9.3.13. It follows from [28,Theorem 2.5] that if I is a fully supported matroidal ideal generated in degree d, then depth(I) = d -1. It is shown in Remark 9.3.3(iv) that if I t (K n 1 ,...,nr ) = 0, then it is a fully supported ideal. This shows that whenever I t (K n 1 ,...,nr ) = 0, we have depth(I t (K n 1 ,...,nr )) = t -1.

At first, we will compute limit depth of t-path ideals of complete bipartite graphs. Theorem 9.3.14. Let G = K p,q with n = p + q and t ≥ 2. Then the following statements hold.

(i) if p = t/2 and q = t/2 , then lim k→∞ depth(S/I t (G) k ) = n -1 and

(ii) if p = t/2 and q > t/2 , then lim k→∞ depth(S/I t (G) k ) = t/2 and dstab(I t (G)) = (q -1)/(q -t/2 ) .

(iii) if p > t/2 and q > t/2 , then lim k→∞ depth(S/I t (G) k ) = 0 and 1 < dstab(I t (G)) < n if t is odd, and lim k→∞ depth(S/I t (G) k ) = 1 and 1 ≤ dstab(I t (G)) < n -1 if t is even.

Proof. Let V 1 = {x 1 , . . . , x p } and V 2 = {y 1 , . . . , y q } be the 2-partition of G.

(i) In this case n = p + q = t and I t (G) is a principal ideal and the assertion follows trivially.

(ii) If p = t/2 and q > t/2 , then it is evident from Remark 9.3.3(i) that all vertices of V 1 appear in every t-path in G. Then, we deduce from the definition of

and Γ is a complete graph on q vertices, and has only one connected component. We conclude ℓ(I) = q and use the equality in (9.3.2) to compute lim k→∞ depth(S/I t (G) k ) = n -q = p = t/2 .

To prove the assertion about dstab(I t (G)), we first observe that I t (G) = JL, where J = (x 1 • • • x p ) and L is generated by all monomials of degree t/2 in q variables. Indeed, as discussed before, all vertices of V 1 appear in every t-path in G, equivalently, x 1 • • • x p divides every generator of I t (G). Moreover, for every subset {y i 1 , . . . , y im } of V 2 of size m = t/2 , there is a t-path P in G with V (P ) =

Hence, J is a principal ideal and L is the square-free Veronese ideal of degree t/2 in variables y 1 , . . . , y q . Let S 1 = K[x 1 , . . . , x p ] and S 2 = K[y 1 , . . . , y q ]. Then depth(S 1 /J k ) = p -1 and it follows Example 10.1.4. Let R = K[x 1 , x 2 , x 3 ] be the polynomial ring over a field K, and

x 3 , and u 5 := x 5 1 x 3 2 x 2 3 . After selecting i 1 := 4, i 2 := 2, and i 3 := 3, one can easily check that r it,1 < r i 1 ,1 for all t ∈ {2, 3}, r it,2 < r i 2 ,2 for all t ∈ {1, 3}, and r it,3 < r i 3 ,3 for all t ∈ {1, 2}. That is, the condition (i) in Theorem 10.1.3 holds. On the other hand, since r 1,3 ≥ r i 3 ,3 and r 5,1 ≥ r i 1 ,1 , we conclude that the condition (ii) in Theorem 10.1.3 holds. This implies that m = (x 1 , x 2 , x 3 ) ∈ Ass R (R/I). Theorem 10.1.7. Suppose that I is a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ], m = (x 1 , . . . , x n ), and

Then m ∈ Ass R (R/I) if and only if there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , k} such that the following conditions hold: (i) |C j | = 1, where C j = {i t | r it,j = max{r i 1 ,j , . . . , r in,j }} for all j = 1, . . . , n;

Proof. To prove the forward implication, let m ∈ Ass R (R/I). It follows from Theorem 10.1.3 that there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , k} such that r it,j < r i j ,j for each t = j, where j = 1, . . . , n. Now, Lemma 10.1.5 yields that m ∈ Ass R (R/J), where J = (x

). Accordingly, Corollary 10.1.6 implies that the conditions (i) and (ii) hold. Moreover, Theorem 10.1.3 gives that for each i ∈ {1, . . . , k} \ {i 1 , . . . , i n }, there exists a positive integer 1 ≤ t ≤ n such that r i,t ≥ r it,t , and so r i,t > r it,t -1. Thus, x

for each i ∈ {1, . . . , k} \ {i 1 , . . . , i n }. This means that the condition (iii) is proved.

Conversely, assume that there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , k} such that the conditions (i)-(iii) hold. Put J := (x

). On account of Corollary 10.1.6, one can conclude that m ∈ Ass R (R/J). Now, Lemma 10.1.5 implies that there exist distinct integers s 1 , . . . , s n ∈ {i 1 , . . . , i n } such that r st,j < r s j ,j for each t = j, where j = 1, . . . , n. Since {s 1 , . . . , s n } = {i 1 , . . . , i n }, after rearranging, one may assume that i j = s j for each j = 1, . . . , n. This shows that the condition (i) in Theorem 10.1.3 holds. In addition, it follows from the condition (iii) that for each i ∈ {1, . . . , k} \ {i 1 , . . . , i n }, there exists a positive integer 1 ≤ t ≤ n such that r i,t > r it,t -1, and hence r i,t ≥ r it,t . Therefore, the condition (ii) in Theorem 10.1.3 is proved. Consequently, we get m ∈ Ass R (R/I), as desired.

We are now ready to present an example which illustrates Theorem 10.1.7.

Example 10.1.8. Consider the following monomial ideal

To achieve this, it is sufficient to apply Theorem 10.1.7.

To see this, set u

, and u 5 := x 3 1 x 2 2 x 3 3 . By choosing i 1 = 1, i 2 = 4, and i 3 = 5, one has C 1 = {5}, C 2 = {1}, and C 3 = {4}. Thus, |C 1 |=|C 2 |=|C 3 |=1, and also C i ∩ C j = ∅ for all i = j, that is, the conditions (i) and (ii) are proved. Moreover, since r 5,1 = 3, r 1,2 = 4, and r 4,3 = 4, and by virtue of u i ∤ x 2 1 x 3 2 x 3 3 for each i ∈ {1, 2, 3, 4, 5} \ {1, 4, 5}, one can conclude that the condition (iii) holds. Therefore, m = (x 1 , x 2 , x 3 ) ∈ Ass R (R/I). In general, there are 5 3 = 10 cases for choosing i 1 , i 2 , and i 3 . Direct computations show that we can also consider the following cases:

(1) i 1 = 1, i 2 = 2, and i 3 = 4.

(2) i 1 = 2, i 2 = 3, and i 3 = 5.

The following corollary is a well-known result, which can be shown by the minimal primary decomposition technique. But we re-prove it as an application of Theorem 10.1.7.

Corollary 10.1.9. Suppose that

Proof. We first assume that m ∈ Ass R (R/I). In addition, for each j = 1, . . . , n, let C j = {i t | r it,j = max{r i 1 ,j , . . . , r in,j }}. Thanks to Theorem 10.1.7, one can derive that there exist distinct integers i 1 , . . . , i n ∈ {1, . . . , k} such that |C j | = 1 for all j = 1, . . . , n, and C i ∩ C j = ∅ for all i = j. Since I is a square-free monomial ideal, one has r i,j ∈ {0, 1} for all i and j. This yields that r i j ,j = 1 and r it,j = 0 for all t = j, where j = 1, . . . , n. Accordingly, one must have n = k, and so {x i 1 , . . . , x in } = {x 1 , . . . , x n }. Due to x i 1 , . . . , x in ∈ I, one gains m ⊆ I, and thus I = m. The converse is obvious based on the fact that Ass R (R/m) = {m}.

On the embedded associated primes and cornerelements

In this section, our first motivation is to give a large class of square-free monomial ideals which satisfy normality, normally torsion-freeness, and the (symbolic) (strong) persistence property. Moreover, our second aim is to present some results on the corner-elements of monomial ideals. It should be noted that the results of this section can be found in [START_REF] Sayedsadeghi | On the embedded associated primes of monomial ideals[END_REF].

Here, we introduce some necessary definitions which will be used in the following subsections.

Let I be a square-free monomial ideal and Γ ⊆ G(I). We say that Γ is an independent set in I if gcd(f, g) = 1 for each f, g ∈ Γ with f = g. We denote the maximum cardinality of an independent set in I by β 1 (I).

Also, a König ideal is a square-free monomial ideal I with G(I) = {u 1 , . . . , u r } such that the maximum number of pairwise disjoint monomials u 1 , . . . , u r is equal given by θ(r + I) = (r + Q 1 + I 2 , . . . , r + Q s + I 2 ) for all r ∈ R. This implies that

Assume that

Consequently, there exists the following R-monomorphism

given by ϕ(r

for all i = 1, . . . , s and j = 1, . . . , t. Therefore, Ass(R/I) ⊆ {p | p = p i + p ′ j for all i = 1, . . . , s and j = 1, . . . , t}.

To establish the reverse inclusion, pick arbitrary elements p ∈ Ass(R/I 1 ) and p ′ ∈ Ass(R/I 2 ). Then there exist monomials v and v ′ in R such that p = (I 1 : R v) and p ′ = (I 2 : R v ′ ). Based on Proposition 10.2.1, we may assume that (10.2.1)

supp(g).

In the light of (10.2.1) and Lemma 2.1.12, one can deduce that (I 1 : R vv ′ ) = (I 1 : R v) and (I 2 : R vv ′ ) = (I 2 : R v ′ ). Accordingly, we obtain the following equalities

This implies that p + p ′ ∈ Ass R (R/I). Thus, the reverse inclusion holds.

To prove Corollary 10.2.4, we have to employ the following crucial theorem. Proof. We first assume that m ∈ Ass(R/(I t : s i=1 y i )). Based on [133, Exercise 2.1.62], we get the following short exact sequence

where ψ(r) = r s i=1 y i and ϕ(r) = r. In addition, by virtue of [START_REF] Sharp | Steps in commutative algebra[END_REF]Exercise 9.42], one can derive that (10.2.2)

Since m ∈ Ass(R/(I t : s i=1 y i )), we get m ∈ Ass(R/I t ). To establish the converse implication, let m ∈ Ass(R/I t ). Our aim is to prove that m ∈ Ass(R/(I t : s i=1 y i )). Proceed by induction on s. Let J := I \y 1 . Because the generators of J t are precisely the generators of I t that are not divisible by y 1 , this implies that (I t , y 1 ) = (J t , y 1 ).

By assumption, one has m \ y 1 / ∈ Ass(R/J t ). Also, it follows from Theorem 10.2.2 that p ∈ Ass(R/(J t , y 1 )) if and only if p = (p 1 , y 1 ), where p 1 ∈ Ass(R/J t ). We thus have m / ∈ Ass(R/(J t , y 1 )), and so m / ∈ Ass(R/(I t , y 1 )). Also, (10.2.2) yields that Ass(R/I t ) ⊆ Ass(R/(I t : y 1 )) ∪ Ass(R/(I t , y 1 )). Accordingly, we obtain m ∈ Ass(R/(I t : y 1 )). Hence, the claim is true for the case in which s = 1. Now, suppose that the assertion has been shown for a product of s -1 variables, and that m ∈ Ass(R/I t ). Set M := s-1 i=1 y i . One can deduce from the induction that m ∈ Ass(R/(I t : M )). Moreover, (10.2.3) Ass(R/(I t : M )) ⊆ Ass(R/((I t : M ) : y s )) ∪ Ass(R/((I t : M ), y s )).

Let K := I\y s . It remains to check that ((I t : M ), y s ) = ((K t : M ), y s ). Since K ⊆ I, one can easily conclude that ((K t : M ), y s ) ⊆ ((I t : M ), y s ). To prove the reverse inclusion, take a monomial u in ((

and the argument is done. If y s ∤ u, then u ∈ (I t : M ), and hence uM ∈ I t . This yields that there exists a monomial f ∈ G(I t ) such that f | uM . Because y s ∤ u and y s ∤ M , this implies that y s ∤ f , and so f ∈ K t . We therefore get u ∈ (K t : M ), and thus ((I t : M ), y s ) ⊆ ((K t : M ), y s ). It also follows from our assumption that m \ y s / ∈ Ass(R/K t ). As Ass(R/(K t : M )) ⊆ Ass(R/K t ), one can conclude that m \ y s / ∈ Ass(R/(K t : M )). Due to Theorem 10.2.2, we obtain p ∈ Ass(R/((K t : M ), y s )) if and only if p = (p 1 , y s ), where p 1 ∈ Ass(R/(K t : M )).

This gives rise to m / ∈ Ass(R/((K t : M ), y s )), and so m / ∈ Ass(R/((I t : M ), y s )). In the light of (10.2.3), one has m ∈ Ass(R/((I t : M ) : y s )) = Ass(R/(I t : s i=1 y i )). This completes the inductive step, and therefore the claim has been proved by induction.

We are ready to express one of the main results of this section in the subsequent corollary.

Corollary 10.2.4. Let I ⊂ R = K[x 1 , . . . , x n ] be a square-free monomial ideal, m = (x 1 , . . . , x n ), and {u 1 , . . . , u β 1 (I) } be a maximal independent set of minimal generators of

i=1 u i and some positive integer t, where I \ x i denotes the deletion of I at

Proof. To simplify the notation, set ℓ := β 1 (I). Let m ∈ Ass(R/I t ). Without loss of generality, one may assume that ℓ i=1 u i = r i=1 x i . In view of Theorem 10.2.3, we can conclude that m ∈ Ass(R/(I t : r i=1 x i )). Since ℓ i=1 u i ∈ I ℓ , we get

. This implies that t ≥ ℓ + 1, as required.

Applying Corollary 10.2.4 enables us to study König ideals in the next theorem.

Theorem 10.2.5. Let I be an unmixed König ideal in the polynomial ring R = K[x 1 , . . . , x n ] over a field K, m = (x 1 , . . . , x n ), and {u 1 , . . . , u β 1 (I) } be a maximal independent set of minimal generators of I such that I \ x i is normally torsion-free for all x i | β 1 (I) i=1 u i . Then the following statements hold:

(i) I is normally torsion-free.

(ii) I is normal.

(iii) I has the strong persistence proeprty.

(iv) I has the persistence property.

(v) I has the symbolic strong persistence property.

Proof. (i) For convenience of notation, put ℓ := β 1 (I). Suppose, on the contrary, that I is not normally torsion-free. Thus, there exists a positive integer t such that I t has embedded primes. We consider t minimal with respect to this property. Let p be an embedded prime of I t . In view of [START_REF] Sharp | Steps in commutative algebra[END_REF]Lemma 9.38], we get p ∈ Ass(R/I t ) if and only if pR p ∈ Ass(R p /(IR p ) t ). This enables us to assume that p = m. Without loss of generality, one may assume that ℓ i=1 u i = r i=1 x i . In the light of Theorem 10.2.3, one has m ∈ Ass(R/(I t : r i=1 x i )), and so m ∈ Ass(R/(I t : ℓ i=1 u i )). On the other hand, since m \ x i / ∈ Ass(R/(I \ x i ) t ) for each i = 1, . . . , r, we deduce 314 from Corollary 10.2.4 that t ≥ ℓ + 1. Because t is minimal, I t-ℓ has no embedded primes, and hence I t-ℓ = I (t-ℓ) . It follows now from [50, Proposition 3.8] that

We therefore obtain m ∈ Ass(R/I t-ℓ ), which contradicts the fact that I t-ℓ has no embedded primes. Consequently, I is a normally torsion-free square-free monomial ideal.

(ii)-(v) can be shown similar to the proof of Corollary 6.2.3.

We are in a position to state the other main result of this section in the following theorem.

Theorem 10.2.6. Let I be a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K and m = (x 1 , . . . , x n ). If there exists a square-free monomial v ∈ I such that v ∈ p \ p 2 for any p ∈ Min(I), and I \ x i is normally torsion-free for all x i ∈ supp(v), then the following statements hold:

(i) I is normally torsion-free.

(ii) I is normal.

(iii) I has the strong persistence proeprty.

(iv) I has the persistence property.

(v) I has the symbolic strong persistence property.

Proof. (i) Suppose, on the contrary, that I is not normally torsion-free. Let t be minimal such that I t has embedded prime ideals. Since I is a square-free monomial ideal, this implies that Ass(R/I) = Min(I), and hence t ≥ 2. Take an arbitrary p ∈ Min(I). In the following, we show that (I t : v) = I t-1 . Because v ∈ I, one must have I t-1 ⊆ (I t : v). To prove the reverse inclusion, select a monomial f in (I t : v). Thus, f v ∈ I t , and so there exist some monomials g 1 , . . . , g t ∈ G(I) and some monomial

one can conclude that p contains exactly one variable that divides v. This gives that f ∈ p t-1 . Because p is arbitrary, one obtains f ∈ p∈Min(I) p t-1 . Applying [START_REF] Villarreal | Monomial algebras[END_REF]Proposition 4.3.25] leads to f ∈ I (t-1) . Since t is minimal such that I t has embedded prime ideals, this yields that I t-1 has no embedded prime ideals, and so I (t-1) = I t-1 . We thus gain f ∈ I t-1 . Consequently, (I t : v) ⊆ I t-1 , and hence (I t : v) = I t-1 . Let q be an embedded prime ideal of I t . It follows from [START_REF] Sharp | Steps in commutative algebra[END_REF]Lemma 9.38] that q ∈ Ass(R/I t ) if and only if qR q ∈ Ass(R q /(IR q ) t ). We therefore may assume that q = m. Hence, m ∈ Ass(R/I t ). In view of Theorem 10.2.3, one derives that m ∈ Ass(R/(I t : v)). We thus get m ∈ Ass(R/I t-1 ). This contradicts our assumption. Accordingly, the square-free monomial ideal I is normally torsion-free.

(ii)-(v) can be shown similar to the proof of Corollary 6.2.3.

To illustrate Theorem 10.2.6, one has to review the definition of t-spread monomial ideals, see Definition 6.3.1.

Example 10.2.7. Assume that

] is a square-free monomial ideal with K is a field and I = (x 1 x 3 x 6 , x 1 x 3 x 7 , x 1 x 4 x 6 , x 1 x 4 x 7 , x 1 x 5 x 7 , x 2 x 4 x 7 , x 2 x 5 x 7 ). Direct computations show that I is a 2-spread monomial ideal. Using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], one can deduce that Min(I) = {(x 1 , x 2 ), (x 1 , x 7 ), (x 6 , x 7 ), (x 1 , x 4 , x 5 ), (x 3 , x 4 , x 7 ), (x 3 , x 4 , x 5 )}. Also, note that I is not unmixed. Set v := x 1 x 3 x 6 . It is straightforward to check that v ∈ p \ p 2 for each p ∈ Min(I). To complete our argument, one needs to demonstrate that I \ x i is normally torsion-free for all i ∈ {1, 3, 6}. Since I \ x 1 = (x 2 x 4 x 7 , x 2 x 5 x 7 ) = x 2 x 7 (x 4 , x 5 ), it follows from Lemma 5.3.11 that I \x 1 is normally torsion-free. Note that F := I \ x 3 = (x 1 x 4 x 6 , x 1 x 4 x 7 , x 1 x 5 x 7 , x 2 x 4 x 7 , x 2 x 5 x 7 ). In addition, one has Ass(R/(I \ x 3 )) = {(x 1 , x 2 ), (x 4 , x 5 ), (x 4 , x 7 ), (x 6 , x 7 ), (x 1 , x 7 )}. Let G = (V (G), E(G)) be the graph with the vertex set V (G) = {1, 2, 4, 5, 6, 7} and the edge set E(G) = {{1, 2}, {4, 5}, {4, 7}, {6, 7}, {1, 7}}. It is routine to check that G is a tree, and F is its cover ideal. By virtue of [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6], F is normally torsion-free; thus, I \ x 3 is normally torsion-free. Finally, notice that

Put L := (x 1 x 3 , x 1 x 4 , x 1 x 5 , x 2 x 4 , x 2 x 5 ). Assume that H = (V (H), E(H)) is the graph with the vertex set V (H) = {1, 2, 3, 4, 5} and the following edge set

Thanks to H has no odd cycle subgraph, we get H is bipartite. In addition, L is the edge ideal of H, and so [START_REF] Simis | On the ideal theory of graphs[END_REF]Theorem 5.9] yields that L is normally torsion-free. Due to Lemma 5.3.11, one has I \ x 6 = x 7 L is normally torsion-free as well. It follows now from Theorem 10.2.6 that I is normally torsion-free, as desired.

We terminate this section by giving an application of Theorem 10.2.6. In fact, we re-prove the fact that every square-free transversal polymatroidal ideal is normally torsion-free, see [START_REF] Herzog | The stable set of associated prime ideals of a polymatroidal ideal[END_REF]Corollary 3.6]. To accomplish this, one has to recall the following definition from [66, Page 298]. Definition 10.2.8. Let F be a non-empty subset of [n] = {1, . . . , n}. We denote by p F the monomial prime ideal (x j : j ∈ F ). A transversal polymatroidal ideal is an ideal I of the form I = p F 1 • • • p Fr , where F 1 , . . . , F r is a collection of non-empty subsets of [n] with r ≥ 1.

Theorem 10.2.9. Every square-free transversal polymatroidal ideal is normally torsion-free.

Proof. Let I ⊂ R = K[x 1 , . . . , x n ] be a square-free transversal polymatroidal ideal. Thus, we can write I = p F 1 • • • p Fr , where F 1 , . . . , F r is a collection of non-empty subsets of [n] with r ≥ 1 and F i ∩ F j = ∅ for all 1 ≤ i = j ≤ r. Without loss of generality, one may assume that u∈G(I) supp(u) = {x 1 , . . . , x n }. We proceed by induction on n. If n = 1, then I = (x 1 ) which is normally torsion-free. Now, suppose that n > 1, and that the claim has been proved for n -1. Without loss of generality, we may assume that x i ∈ p F i for each i = 1, . . . , r.

follows from the inductive hypothesis that I \ x i is normally torsion-free. By virtue of Theorem 10.2.6, we deduce that I is normally torsion-free.

Some results on the corner-elements of monomial ideals

Let I be a monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K and p ∈ Ass(R/I). It is well-known by [START_REF] Herzog | Monomial Ideals[END_REF]Corollary 1.3.10] that p = (I : h) for some monomial h in R. Moreover, it has been proved that if

is a square-free monomial ideal and p ∈ Ass(R/I), then there exists a square-free monomial h in R with supp(h) ⊆ u∈G(I) supp(u) such that p = (I : R h), see Theorem 2.1.15. In this subsection, we are going to detect which variables must divide h. To accomplish this, we begin with the following proposition. Proposition 10.2.10. Let I ⊂ R = K[x 1 , . . . , x n ] be a monomial ideal. Let p, q ∈ Ass(R/I t ) with p = q such that p = (I t : f ) and q = (I t : g) for some positive integer t and some monomials f and g in R. Then f ∤ g and g ∤ f .

Proof. We need only show that f ∤ g, since the other claim can be proved by a similar argument. Suppose, on the contrary, that f | g. That is, g = f v for some monomial v in R. This gives that q = (I t : f v). Since (I t : f v) = ((I t : f ) : v), one gains that q = (p : v). It is easy to see that either (p : v) = R or (p : v) = p. We thus have either q = R or q = p. This leads to a contradiction. Therefore, we get f ∤ g, as claimed.

To establish Proposition 10.2.11 and Corollary 10.2.14, one has to apply the notion of corner-elements which has been introduced in Definition 5.1.4. Proposition 10.2.11. Let I ⊂ R = K[x 1 , . . . , x n ] be a monomial ideal and m = (x 1 , . . . , x n ). Let f and g be two I t -corner-elements for some positive integer t. Then f ∤ g and g ∤ f . Question 10.2.15. (see Question 3.2.23) Suppose that I is a square-free monomial ideal in R = K[x 1 , . . . , x n ], G(I) = {u 1 , . . . , u m }, m i=1 supp(u i ) = {x 1 , . . . , x n }, and m = (x 1 , . . . , x n ) is the graded maximal ideal of R. If there exists a positive integer 1 ≤ j ≤ n such that m \ x j ∈ Ass(R/(I \ x j ) k ) for some positive integer k, then can we deduce that m ∈ Ass R (R/I k )?

We provide a counterexample. To do this, let G = (V (G), E(G)) be the graph with the vertex set V (G) = {1, 2, 3, 4, 5, 6}, and the following edge set

Assume that I is the cover ideal of G, that is,

Using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF], one can derive that (x 1 , x 2 , x 3 , x 4 , x 5 ) ∈ Ass(R/(I \ x 6 ) 2 ), while (x 1 , x 2 , x 3 , x 4 , x 5 , x 6 ) / ∈ Ass(R/I 2 ).

The proposition below says that if p = (I s : h) ∈ Ass(R/I s ) for some positive integer s, then we can always find an upper bound for deg x i h for each i. Proof. To verify the claim, it is enough for us to show that h|( n i=1 x i ) s-1 . On the contrary, assume that h ∤ ( n i=1 x i ) s-1 . Thus, there exists some 1 ≤ i ≤ n such that deg x i h ≥ s. This implies that the exponent of x i in hx i is at least s + 1. By p = (I s : h), one has hx i ∈ I s , and so there exist monomials u 1 , . . . , u s ∈ G(I) such that hx i = u 1 • • • u s f for some monomial f in R. Since I is a square-free monomial ideal, then for each j = 1, . . . , s, u j is a square-free monomial, one has the exponent of x i in each u j is at most one. Accordingly, we get x i |f , and hence h ∈ I s , which is a contradiction. We therefore have h|( n i=1 x i ) s-1 , as desired.

We close our argument with the subsequent proposition which focuses on the minimal prime ideals associated to square-free monomial ideals. Proposition 10.2.17. Let I be a square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. If p ∈ Min(I), then p = (I : h), where h is the product of the variables that are not in p.

Proof. Let p ∈ Min(I). We first demonstrate that (I : h) ⊆ p. To see this, pick a monomial u in (I : h). Hence, uh ∈ I, and so uh ∈ p. We thus have x j | uh for some x j ∈ p. Since gcd(x j , h) = 1, one can conclude that x j | u, and so u ∈ p. Therefore,

Chapter 11

Some perspectives for future works

The major point of this chapter, which has been divided into seven parts, is to give some open questions and ideas for future work. To do this, we will recall and repeat any needed notions and definitions from commutative algebra or combinatorics in each part.

Relations among NTF, NNTF, and SPP

Let R be a commutative Noetherian ring and I be an ideal of R. The ideal I is said to have the persistence property if Ass(R/I k ) ⊆ Ass(R/I k+1 ) for all positive integers k. Moreover, the ideal I satisfies the strong persistence property (SP P for short) if (I k+1 : I) = I k for all positive integers k. Furthermore, the ideal I is called normally torsion-free (N T F for short) if Ass R (R/I k ) ⊆ Ass R (R/I) for all positive integers k. In addition, a monomial ideal I in a polynomial ring R = K[x 1 , . . . , x n ] over a field K is called nearly normally torsion-free (N N T F for short) if there exist a positive integer k and a monomial prime ideal p such that Ass R (R/I m ) = Min(I) for all 1 ≤ m ≤ k, and Ass R (R/I m ) ⊆ Min(I) ∪ {p} for all m ≥ k + 1.

Based on Example 6.1.6, we know that, in general, nearly normally torsionfreeness does not imply the strong persistence property. In fact, let R = K[x 1 , x 2 , x 3 ] be the polynomial ring over a field K and I

This means that I has the persistence property and is also a nearly normally torsion-free ideal. On the other hand, using Macaulay2 [START_REF] Grayson | Macaulay2, a software system for research in algebraic geometry[END_REF] shows that (I 2 : R I) = I. We therefore deduce that I does not satisfy the strong persistence property.

In this direction, one can ask the following question. In fact, up to now, the author could not find any counterexample for it. Question 11.1.1. Let I be a nearly normally torsion-free square-free monomial ideal in a polynomial ring R = K[x 1 , . . . , x n ] over a field K. Then can we conclude that I has the strong persistence property?

To see a large class in this theme, we know that all cover ideals of odd cycle graphs are nearly normally torsion-free, and also satisfy the strong persistence property.

We finish this section with the following open questions.

Question 11.1.2. Let I be a square-free monomial ideal in R = K[x 1 , . . . , x n ]. Let I be minimally not normally torsion-free, i.e., I is not normally torsion-free but all its minors are. Find the least power t ≥ 1 such that Ass R (R/I t ) = Ass R (R/I).

To see an example, consider the odd cycle graph

Then the cover ideal associated to C 5 is given by

Since the cover ideal of any bipartite graph is normally torsion-free and also by remembering this fact that the intersection of two normally torsion-free square-free monomial ideals which have no common variable is again normally torsion-free, we can deduce that any minor of L is normally torsion-free. On the other hand, we know that Ass(L s ) = Min(L) ∪ {(x 1 , x 2 , x 3 , x 4 , x 5 )} for all s ≥ 2. Therefore, L is not normally torsion-free, but all its minors are normally torsion-free. In other words, L is minimally not normally torsion-free. Note that in this example, we have t = 2.

Question 11.1.3. Let R = K[x 1 , . . . , x n ] be a polynomial ring over a field K and

be the set of square-free monomial ideals in R such that satisfy (respectively, do not satisfy) the strong persistence property. Then, does the following limit exist? can be 0?

where |A| denotes the cardinality of A.

we have to verify that I has both of the persistence property and strong persistence property. Our strategy is to use [START_REF] Rajaee | Superficial ideals for monomial ideals[END_REF]Corollary 6.4], which says that every normal monomial ideal has a superficial ideal. Note that a monomial ideal has the strong persistence property if and only if it has a superficial ideal. A fast approach is to refer to Normaliz software, but we utilize a recursive method to demonstrate the normality of I. We establish this claim in four steps:

Step 1. Put A 1 := (x 4 , x 2 5 ) and B 1 := (x 4 x 5 , x 2 ). Since A 1 is integrally closed, we can deduce from [START_REF] Villarreal | Cohen-Macaulay graphs[END_REF]Theorem 12.1.10]

Step 3. Put A 3 := x 4 x 5 (x 2 , x 3 ) and B 3 := (x 2 x 3 , x 2 x 5 , x 3 x 4 ). It follows from [4, Theorem 3.4 and Remark 1.2] that A 3 is normal. To show the normality of [START_REF] Gitler | Blowup algebras of ideals of vertex covers of bipartite graphs[END_REF]Corollary 2.6] yields that B 3 is normally torsion-free, and by [60, Theorem 1.4.6], we deduce that B 3 is normal. Also, note that A 3 + B 3 = B 3 is normal. Here, Theorem 7.7.3 implies L := A 3 + x 6 B 3 is normal.

Step 4. Thanks to I = L + x 1 H, and according to Steps 2 and 3, H and L are normal, it is enough for us to prove the normality of

), similar to the argument which has been presented in Steps 2 and 3, we conclude that C, D, and C + D are normal, and so H + L is normal. In view of Theorem 7.7.3, I has the (strong) persistence property. This completes the discussion.

Along with this argument, we are ready to provide the following open questions. It should be noted that when t = 1, the construction above is simply the cover ideal of a finite simple graph G. In fact, Bhat, Biermann, and Van Tuyl, in [17,Lemma 2.4], proved the following lemma:

t-spread principal Borel ideals

Here, we turn our attention to vector-spread monomials of type t. In fact, in [108, Theorem 5.13 and Lemma 5.15], the authors characterized all normally torsion-free t-spread principal Borel ideals. We review the needed definitions as follows. monomial ideal. Observe that, for a hypergraph H, the deletion H \ x at a vertex x ∈ V (H) has the effect of setting x = 0 in the edge ideal I(H), and the contraction H/x has the effect of setting x = 1 in the edge ideal I(H). We call an ideal I ′ a minor of a square-free monomial ideal I if I ′ can be obtained from I by a sequence of taking quotients (=deletion) and localizations (=contraction) at the variables. In this direction, one can ask the following questions.

Question 11.7.1. Let I be a square-free monomial ideal in R = K[x 1 , . . . , x n ]. Also, assume that I is minimally not normally torsion-free, that is, I is not normally torsion-free, but all its minors are normally torsion-free. Then find the least power t such that Ass R (R/I t ) = Ass R (R/I). Question 11.7.2. Let I be a square-free monomial ideal in R = K[x 1 , . . . , x n ]. Then find the least integer ℓ (may depend on I) such that if I (t) = I t for 0 ≤ t ≤ ℓ, then I (t) = I t for all t ≥ 1, that is, I is normally torsion-free.

Let H denote a hypergraph over d vertices {x 1 , . . . , x d } and R = K[x 1 , . . . , x d ] will be the corresponding polynomial ring. A vertex x ∈ V (H) is called an isolated vertex if {x} ∈ E(H). It follows from the definition that if x is an isolated vertex of H, then {x} is the only edge in H that contains x. A vertex cover of H is a set of vertices that has non-empty intersection with all of the edges of H. The minimum cardinality of a vertex cover of H is denoted by α 0 (H). It should be noted that by the correspondence between minimal primes and minimal vertex covers, α 0 (I) is also the height of I. We say the generators of a square-free monomial ideal I as being independent if the corresponding edges of the associated hypergraph are pairwise disjoint; that is, the generators have disjoint support. We denote the maximum cardinality of an independent set in I by β 1 (I). Furthermore, note that a subset of the monomial generators of I is independent if and only if it forms a regular sequence. Thus, β 1 (I) is equal to the monomial grade of I, denoted by m-grade(I), where the monomial grade of an ideal I is the maximum length of a regular sequence of monomials in I. It not hard to see that α 0 (I) ≥ β 1 (I). In connection with combinatorial applications, we express an algebraic reformulation of the Conforti-Cornuéjols conjecture as follows.

Conjecture 11.7.7. Let I be a square-free monomial ideal such that I and all of its minors satisfy the property that their heights are the same as their m-grades. Then gr I := q≥0 I q /I q+1 is reduced; or equivalently, I k = I (k) for all k ≥ 1; or equivalently, I is normally torsion-free.

In broadly speaking, finding a counterexample for this conjecture is too hard, and the famous result in this theme is due to Hà