Thomas Bauer
email: tbauer@mathematik.uni-marburg.de

Sandra Di

David Schmitz

Tomasz Szemberg

Justyna Szpond

On the postulation of lines and a fat line

Keywords: postulation problems, fat flats, Hilbert functions, Serre vanishing Mathematics Subject Classification (2010) 14C20, 14F17, 13D40, 14N05

In the present note we show that the union of r general lines and one fat line in P 3 imposes independent conditions on forms of sufficiently high degree d, where the bound is independent of the number of lines. This extends former results of Hartshorne and Hirschowitz on unions of general lines, and of Aladpoosh on unions of general lines and one double line.

Introduction

Let X ⊂ P n be a closed subscheme defined over an algebraically closed field K of characteristic zero. The Hilbert function of X encodes a number of properties of X and has been classically an object of vivid research in algebraic geometry and commutative algebra. We first recall the definition. Definition 1.1 (Hilbert function). The Hilbert function of a scheme X ⊂ P n (K) is

HF X : Z d → dim K [S(X)] d ∈ Z,
where S(X) denotes the graded homogeneous coordinate ring of X.

It is well known that the Hilbert function becomes eventually (i.e., for large d) a polynomial. We denote the Hilbert polynomial of X by HP X . Whereas the Hilbert polynomial can be (in principle) computed algorithmically, the Hilbert function is more difficult to compute. For some varieties, like P n , the Hilbert function is equal to the Hilbert polynomial, but this behaviour is rare. The next simplest behaviour occurs for subschemes with bipolynomial Hilbert function. Definition 1.2 (Bipolynomial Hilbert function). Following [START_REF] Carlini | Bipolynomial Hilbert functions[END_REF] we say that X has a bipolynomial Hilbert function if HF X (d) = min {HP P n (d), HP X (d)} [START_REF] Aladpoosh | Postulation of generic lines and one double line in P n in view of generic lines and one multiple linear space[END_REF] for all d 1.

In other words, X has a bipolynomial Hilbert function if X ⊂ P n imposes the expected number of conditions on forms of arbitrary degree d 1. Essentially by definition, a scheme consisting of general points in P n has bipolynomial Hilbert function. An analogous result for X consisting of r general lines in P n with n 3 has been proved by Hartshorne and Hirschowitz in [START_REF] Hartshorne | Droites en position générale dans l'espace projectif[END_REF]Theorem 0.1]. A new proof has been announced recently by Aladpoosh and Catalisano [START_REF] Aladpoosh | On the Hartshorne-Hirschowitz theorem[END_REF]. Recently Carlini, Catalisano and Geramita [START_REF] Carlini | On the Hilbert function of lines union one non-reduced point[END_REF] showed that if X consists of r general lines and one general fat point, then, up to a short list of exceptions in P 3 , X has bipolynomial Hilbert function, see also [START_REF] Aladpoosh | Postulation of disjoint unions of lines and a multiple point[END_REF] and [START_REF] Ballico | Postulation of disjoint unions of lines and a multiple point[END_REF].

Aladpoosh in [START_REF] Aladpoosh | Postulation of generic lines and one double line in P n in view of generic lines and one multiple linear space[END_REF] has proved recently that also schemes consisting of general lines and one double line have bipolynomial Hilbert function, with the exception of one double line and two simple lines in P 4 imposing dependent conditions on forms of degree 2. She also conjectured [START_REF] Aladpoosh | Postulation of generic lines and one double line in P n in view of generic lines and one multiple linear space[END_REF]Conjecture 1.2] that the same holds true for r general lines and one fat flat of arbitrary dimension. In the present note we provide evidence supporting this conjecture for a fat line of arbitrary multiplicity m. Our main result is the following.

Main Theorem. Let m 1 be a fixed integer. Then for d d 0 (m) := 3 m+1

3

, the Hilbert function of a subscheme X ⊂ P 3 consisting of r 0 general lines and one line of multiplicity m (i.e. defined by the m-th power of the ideal of a line) satisfies formula [START_REF] Aladpoosh | Postulation of generic lines and one double line in P n in view of generic lines and one multiple linear space[END_REF].

In other words, a general fat line and an arbitrary number of general lines with multiplicity 1 impose independent conditions on forms of degree d d 0 (m) (see Theorem 4.1).

It follows from the Serre Vanishing Theorem [11, Theorem 1.2.6] that for any subscheme X ⊂ P n , there exists a bound d 0 (X) such that X imposes independent conditions on forms of degree d d 0 (X). The point here is that we obtain an explicit bound that depends only on the multiplicity of the fat line and is independent of the number of reduced lines.

We will set up the proof in a way which employs the general strategy of Hartshorne and Hirschowitz [START_REF] Hartshorne | Droites en position générale dans l'espace projectif[END_REF] and Carlini, Catalisano and Geramita [START_REF] Carlini | On the Hilbert function of lines union one non-reduced point[END_REF]. This amounts to work inductively by constructing a suitable sequence of generic subschemes Z 0 , Z 1 , . . . , along with suitable specializations Z i of Z i . The starting scheme Z 0 consists of the lines in the theorem plus a number of generic points. The essential difficulty in this strategy lies in the choice of the intermediate schemes Z i and their specializations Z i . In our approach this is achieved by using intermediate schemes that contain, apart from disjoint lines and points, also crosses and so-called zig-zags (see Def. 2.4).

Preliminaries and auxiliary results

We begin by recalling a formula for the number of conditions, c(n, m, d), imposing the vanishing of a form of degree d m to the order m along a line in P n :

c(n, m, d) = m(nd + 2n + m -mn -1) n(n -1) n + m -2 m . (2)
For a proof see e.g. [START_REF] Dumnicki | Linear subspaces, symbolic powers and Nagata type conjectures[END_REF]Lemma 2.1]. Note that

c(n, 1, d) = d + 1
for all n 1.

In the next Lemma we present a useful formula relating some of numbers c(n, m, d).

Lemma 2.1. For all positive integers n, m, d we have

c(n, m, d) = c(n, m -1, d -1) + c(n -1, m, d).
Proof. This is a straightforward computation.

In P 3 the formula (2) reduces to

c(d, m) = c(3, m, d) = 1 6 m(m + 1)(3d + 5 -2m).
Our approach to the Main Theorem uses the specialization method. This employs the semicontinuity Theorem [9, Theorem III.12.8] in the following way:

Let f : X → B be a projective morphism of noetherian schemes and let F be a coherent sheaf on X, flat over B. The vanishing h 0 (X b , F b ) = 0 for some b implies then the vanishing h 0 (X b , F b ) = 0 for all b in a neighborhood of b.

In our situation, with B = K this means concretely that if h 0 (P n , O P n (d) ⊗ I Z b ) = 0 for a (special) subscheme Z b , then h 0 (P n , O P n (d)⊗I Z b ) = 0 for a (general) subscheme Z b such that Z b and Z b vary in a flat family over K.

We are going to use and generalize the notion of sundials following the ideas of Carlini, Catalisano and Geramita, see [START_REF] Carlini | 3-dimensional sundials[END_REF]Sections 2,[START_REF] Aladpoosh | On the Hartshorne-Hirschowitz theorem[END_REF] for definitions and motivations.

Definition 2.2 (Sundials and crosses). A sundial in P n is the limiting subscheme obtained by a collision of two skew lines (hence spanning a P 3 ⊂ P n ) moving in a one-dimensional family. It has a nonreduced structure in the collision point which can be thought of as a vector generating the P 3 mentioned above together with the plane spanned by the two intersecting lines.

A union of two lines in P n intersecting in a single point is called a cross. A cross is hence a sundial with the reduced structure.

Carlini, Catalisano and Geramita proved in [START_REF] Carlini | Bipolynomial Hilbert functions[END_REF]Lemma 2.5] that there exists a flat family g : W → K of schemes in P n , with n 3 such that a general member W t ⊂ W is a union of two disjoint lines, whereas the special fiber W 0 is a sundial. As this property is central to our argument, we recall briefly how W can be obtained. It is enough to consider it in the projective space of dimension 3.

Construction 2.3. In P 3 with homogeneous coordinates (x : y : z : w), let M be the line defined by the ideal I(M ) = z, w and let L t be lines given by the ideals I(L t ) = y, z -tx . For t = 0 let W t = M ∪ L t . Intersecting the defining ideals we get J t = yw, txw -zw, yz, txz -z 2 as the ideal of W t . Thus the limiting subscheme W 0 is given by the ideal J 0 = yw, zw, yz, z 2 . The primary decomposition of J 0 is J 0 = z, w ∩ y, z ∩ y, z 2 , w , so that the scheme structure at the intersection point P = (1 : 0 : 0 : 0) of the lines M and L 0 is non-reduced.

It is easy to see that J 0 can be also written down as

J 0 = I(M ) ∩ I(L 0 ) ∩ I(P ) 2 ,
which is in accordance with [START_REF] Carlini | 3-dimensional sundials[END_REF]Definition 3.7].

It is a crucial point in our proof of the Main Theorem to use the following generalization of the sundial idea. Definition 2.4 (Zig-zag). A zig-zag of length z is the limiting subscheme obtained by a collision of an ordered set of z general lines L 1 , L 2 , . . . , L z in the following way:

• the line L 2 intersects L 1 ; • the line L 3 intersects L 2 in a point different from L 2 ∩ L 1 ;
• and so on, i.e., L i+1 intersects L i in a point different from

L i ∩ L i-1 , for i z -1.
We assume that there are no other intersection points among the lines but those listed above. A zig-zag of length z has thus (z -1) singular points.

A reduced zig-zag is a zig-zag with reduced structure, i.e., it has no embedded points.

Thus a sundial is just a zig-zag of length 2 and a cross is a reduced zig-zag of length 2.

Figure 1 shows a zig-zag of length 7. Note that the lines in the figure are all skew but the ones indicated in the figure, which means there are no other intersection points but those indicated in this figure. The intersection points are embedded points as explained in Construction 2.3.

L 1 L 2 L 3 L 4 L 5 L 6 L 7
Figure 1: A zig-zag of length 7

Lemma 2.5. For an integer z 2, there exists a flat family {X λ } λ∈K of schemes in P n , with n 3 such that a general member of {X λ } λ∈K is a union of z disjoint lines and the special fiber is a zig-zag of length z.

Proof. The proof is a straightforward generalization of the argument in [5, Lemma 2.5].

Zig-zags are useful in our approach because of the following fact.

Lemma 2.6. Let S be a zig-zag of length z in P 3 formed by lines L 1 , . . . , L z . Let Q be a smooth quadric in P 3 such that all singular points of S lie on Q but none of the lines in S is contained in Q. Then the colon ideal

J = I S : I Q defines the reduced zig-zag V (J) = L 1 ∪ . . . ∪ L z .
Proof. Again straightforward.

Apart from the semicontinuity, the residual exact sequence and Castelnuovo's inequality are key ingredients in the proof. We discuss them now. One has the following residual exact sequence

0 -→ I Z (-Y ) -→ I Z -→ I Z ,Y -→ 0 , (3)
where I W is the sheafification of the ideal I W . Twisting (3) by

O P n (d) we get 0 -→ O P n (d -e) ⊗ I Res Y (Z) -→ O P n (d) ⊗ I Z -→ O Y (d) ⊗ I Tr Y (Z) -→ 0. ( 4
)
Taking then the long cohomology sequence of ( 4) we obtain the following statement, which is called Castelnuovo's inequality, see e.g. [6, Lemma 3.3].

Lemma 2.8 (Castelnuovo's inequality). Let Y ⊂ P n be a divisor of degree e and let d e be an integer. Let Z ⊂ P n be a closed subscheme. Then

h 0 (P n , O P n (d) ⊗ I Z ) h 0 (P n , O P n (d -e) ⊗ I Res Y (Z) ) + h 0 (Y, O Y (d) ⊗ I Tr Y (Z) ). (5)
We call the space

H 0 (P n , O P n (d -e) ⊗ I Res Y (Z) ) the residual linear system of H 0 (P n , O P n (d) ⊗ I Z ) with respect to Y and H 0 (Y, O Y (d) ⊗ I Tr Y (Z) ) the trace linear system of H 0 (P n , O P n (d) ⊗ I Z ) on Y .
3 Nonspeciality of certain linear series on P 1 × P 1

In the proof of the Main Theorem we will consider trace linear systems on a smooth quadric in P 3 . This section serves as a preparation of relevant results on linear systems on a smooth quadric in P 3 identified with P 1 × P 1 . Special linear systems with general points of multiplicity at most 3 on P 1 × P 1 have been classified by Lenarcik in [START_REF] Lenarcik | Linear systems over P 1 × P 1 with base points of multiplicity bounded by three[END_REF]. Here we recall a part of [START_REF] Lenarcik | Linear systems over P 1 × P 1 with base points of multiplicity bounded by three[END_REF]Theorem 2] relevant in our situation. Lemma 3.1. Let Z be the fat point scheme in P 1 × P 1 defined by the ideal

I Z = I(P 1 ) ∩ . . . ∩ I(P p ) ∩ I(Q 1 ) 2 ∩ . . . ∩ I(Q q ) 2 ,
where P 1 , . . . , P p , Q 1 , . . . , Q q are general points in P 1 × P 1 . Let 0 a b be nonnegative integers. The linear system

H 0 (P 1 × P 1 , O P 1 ×P 1 (a, b) ⊗ I Z )
is special if and only if one of the following cases holds

• a = 0, p + 2q b and q 1,
• a = 2, p = 0, b = q -1 and q is odd.

Using this result, we prove now an auxiliary postulation statement for higher multiplicities: Lemma 3.2. Given an integer m 2 let k be an integer with k m+1 3

. Then a scheme composed by 2 general points P 1 , P 2 taken with multiplicity m imposes independent conditions on linear systems on P 1 × P 1 of bidegree (a, b) if a b, a k -1, a > 0 and b 3k.

Proof. For m = 2 the assertion for arbitrary k m+1 3

follows from Lemma 3.1. Indeed, the case a = 0 is excluded by our assumption and the case a = 2 is also excluded, since in our situation q = 2.

We proceed by induction on m and k. Let m > 2 and k m+1 3

be fixed and assume that the assertion holds for all m < m and all k m +1 3

. Let s = (a + 1)(b + 1) -2 m+1 2 and let Q 1 , . . . , Q s be s general points on P 1 × P 1 . It is enough to show that there is no divisor of bidegree (a, b) which passes through Q 1 , . . . , Q s and additionally through the points P 1 , P 2 with multiplicity m. It suffices to prove this claim for a particular position of points Q 1 , . . . , Q s .

To this end let C be a smooth curve of bidegree (1, 1) passing through P 1 and P 2 . Thus C is a smooth rational curve. Let t = a+b-2m+1. By above assumptions this is a non-negative integer. We specialize now the points Q 1 , . . . , Q t onto the curve C leaving the points Q t+1 , . . . , Q s as general points on P 1 × P 1 , hence not on C. Assume to the contrary that there is a divisor Γ such that mult P i Γ m for i = 1, 2 and mult Q j Γ 1 for j = 1, . . . , s. Then C must be a component of Γ, because (Γ • C) = a + b. But the trace of Γ on C has at least 2 points of multiplicity m and other t reduced points with 2m + t = a + b + 1. The residual divisor Γ = Γ -C has bidegree (a -1, b -1), passes through the points P 1 and P 2 with multiplicity m -1 and also passes through the points Q t+1 , . . . , Q s . Since s -t = ab -2 m 2 , the existence of Γ is excluded by our induction assumption. This concludes the proof.

The proof of the Main Theorem

In this section we will prove the Main Theorem, which is equivalent to the following statement.

Theorem 4.1 (Maximal rank property). For a subscheme X ⊂ P 3 consisting of a general line of multiplicity m and an arbitrary number r of general lines, the restriction map

H 0 (P 3 , O P 3 (d)) → H 0 (X, O X (d))
has maximal rank for all d d 0 (m) = 3 m+1

3

.

As pointed out in the introduction, we will employ the general strategy of Hartshorne and Hirschowitz [10, Theorem 1.1]. Specifically, we will proceed inductively along a suitable sequence of subschemes Z 0 , Z 1 , . . . , for which we choose suitable specializations Z 0 , Z 1 , . . . . Subschemes consisting of general lines, a fat line and additional points are unfortunately too simple for the specialization process. Our idea is to allow instead some intermediate schemes Z = Z(m, r, s, q, z) consisting of one general line of multiplicity m, r general lines, s general crosses, q general points and a reduced zig-zag of length z (along with particular specializations Z of Z, which will be introduced in Definition 4.4).

We now set up some notation that will be useful for the remainder of the paper. We denote by L(k, ε; m, r, s, q, z) = L(d; Z) = H 0 (P 3 , O P 3 (d) ⊗ I Z ) the linear system of polynomials in P 3 of degree d = 3k + ε, with ε ∈ {0, 1, 2} vanishing along the subscheme Z. One defines the virtual dimension of L(d; Z) to be:

vdim(L(d; Z)) = d + 3 3 -c(d, m) -r(d + 1) -s(2d + 1) -q -(zd + 1).
Similarly we will write

HP P 3 (d) = c(d, m) + r(d, m)(d + 1) + q(d, m). (6)
So HP P 3 (d) is the virtual number of conditions that one m-fold line, r(d, m) generic ordinary lines, and q(d, m) generic points impose. In particular,

• for d = 3k r(d, m) = 3 2 k 2 + 5 2 k + 1 - m + 1 2 and q(d, m) = 2 m + 1 3 , • for d = 3k + 1 r(d, m) = 3 2 k 2 + 7 2 k + 2 - m + 1 2 and q(d, m) = 2 m + 1 3 , • for d = 3k + 2 r(d, m) = 3 2 k 2 + 9 2 k + 3 - m + 1 2 and q(d, m) = k + 1 + 2 m + 1 3 .
The following theorem (to be proved in Subsection 4.1) implies the Main Theorem. ,r(d,m) + 1, 0, 0, 0). Let further Q be some smooth quadric. Then there exists a sequence Z = Z 0 , Z 1 , . . . , Z u of schemes Z i = Z(m i , r i , s i , q i , z i ) together with specializations Z i such that the following statements hold for each i = 0, . . . , u -1

(1) Z i+1 = Res Q (Z i ); (2) h 0 (Q, O Q (d -2i) ⊗ I Tr Q (Z i ) ) = 0,
and such that Z u satisfies the conditions

(i) Z u = Z(m u , r(d -2u, m u ), 0, q(d -2u, m u ), 0), or Z u = Z(m u , r(d -2u, m u ) + 1, 0, 0, 0); (ii) d -2u d 0 (m u ); (iii) m u ∈ {m -1, m -2, 1, 0}.
Proof of Theorem 4.1. We proceed by induction on m. The base case m = 1 has been proved for all d 0 = d 0 (1) in [START_REF] Hartshorne | Droites en position générale dans l'espace projectif[END_REF] and the base case m = 2 by Aladpoosh [START_REF] Aladpoosh | Postulation of generic lines and one double line in P n in view of generic lines and one multiple linear space[END_REF] for all d 3 = d 0 (2). Consider now m 3. For d d 0 (m) 4 it suffices to prove the bijectivity of the restriction map in the case of schemes Z = Z(m, r(d, m), 0, q(d, m), 0), and the injectivity in the case of schemes Z = Z(m, r(d, m) + 1, 0, 0, 0). This amounts in either case to proving the identity dim(L(d; Z)) = 0 .

Z u )) + u-1 i=0 h 0 (Q, O Q (d -2i) ⊗ I Tr Q (Z i ) ) = dim(L(d -2u; Z u )) .
The latter must be zero since Z u satisfies the induction hypothesis.

Proof of Theorem 4.3

In order to prove Theorem 4.3, we will need the next lemma describing which schemes result from certain specializations.

Definition 4.4. Let Q be a smooth quadric in P 3 . We denote by R(δ, , s , z , t, t s , t z ) the specialization Z of Z = Z(m, r, s, q, z) given by assuming the following lines to be disjoint lines belonging to the same ruling of Q:

• δ m-fold lines (here δ will be either 0 or 1);

• ordinary lines;

• s lines from s crosses (one line from each cross);

• z = z 2 lines from the reduced zig-zag of length z, and assuming furthermore

• t among the q points to be general points on Q,

• 2t s of the r lines to form t s sundials whose intersection with Q is a zerodimensional scheme containing the singular points of the sundials,

• t z + 1 of the lines to form one zig-zag whose zero-dimensional intersection with Q contains all t z singular points.

Lemma 4.5. Let Z be the specialization R(δ, , s , z , t, t s , t z ) of the scheme Z = Z(m, r, s, q, z). Then

Res Q (Z ) = Z(m -δ, r -+ s + (z -z ) -2t s -(t z + 1), s -s + t s , q -t, t z + 1)
and

Tr Q (Z ) = D + Ω(2r -2 -2 z -3 s -2t s -2t z + t + 4s + z + γ, t s + t z , 2 -2δ, m),
where D is a divisor on Q consisting of δ lines of multiplicity m, with δ ∈ {0, 1}, and + s + z reduced lines, all contained in the same ruling on Q.

Here γ = 0, if z = 0, 1, if z > 0.
Proof. The argument amounts just in counting the various objects to which the specialization applies and which it results. We omit the simple details. Now we turn to the proof of Theorem 4.3. The particular sequence of subschemes differs according to the divisibility of d by 3. In order to simplify the notation we denote the relevant linear series by B(k, ε, m) = L(k, ε; m, r(3k + ε, m), 0, q(3k + ε, m), 0);

I(k, ε, m) = L(k, ε; m, r(3k + ε, m) + 1, 0, 0, 0).
As the proof is technically quite involved, we start by outlining the overall course of the argument: The proof consists in a number of cases, in each of which an initial system L 0 will yield a residual system L u through a sequence of reduction steps with intermediate systems

L 0 , L 1 , . . . , L u ,
according to the plan shown in Table 1.

The intermediate systems L i will be constructed as residual systems from suitably chosen schemes Z i as

L i+1 := L(d -2i -2; Res Q (Z i )) ,
giving on the quadric the trace systems

Λ i+1 := Λ((d -2i, d -2i); Tr Q (Z i )) .

The bijective cases

With d = 3k + ε, the initial system in every case is

L(k, ε; m, r(3k + ε, m), 0, q(3k + ε, m), 0).
which is, by Lemma 3.1, a zero-dimensional system. Then we set Z 1 = R(0, 1, 2k, 0, 0, 0, 0) and obtain the residual system

L 2 = L(k -1, 0; m -1, r(3(k -1), m -1), 0, q(3(k -1), m -1), 0) = B(k -1, 0, m -1),
and the trace system

Λ 2 = Λ((k -2, 3k -1); 3k 2 -3k -m 2 + m, 0, 2, m -1).
Applying Lemma 3.2, we obtain dim(Λ 2 ) = 0.

Case B(k, 2, m). In this case we use the specialization

Z 0 = R(1, 2k + 2 -m, 0, 0, k + 1 + m(m -1), 0, 0).
We obtain

L 1 = L(k, 0; m -1, r(3k, m -1), 0, q(3k, m -1), 0) = B(k, 0, m -1)
and Λ 1 = Λ((k, 3k + 2); 3k 2 + 6k + 3, 0, 0, m) which is of dimension 0.

The injective cases

With d = 3k + ε, the initial state in every case now is L 0 = L(k, ε; m, r(3k + ε, m) + 1, 0, 0, 0).

Case I(k, 0, m). We have L 0 = L(k, 0; m, r(3k, m) + 1, 0, 0, 0) so that

vdim(L 0 ) = -3k -1 + 1 3 m(m -1)(m + 1) < 0 for d = 3k d 0 (m) = 3 m+1

3

. We apply the specialization

Z 0 = R(1, 2k + 1 -m, 0, 0, 0, 0, m(m -1) -2).
By Lemma 4.5 the trace system is

Λ 1 = Λ((3k, k -1); 2(r(3k, m) + 1 -(2k + 1 -m) -m(m -1) + 2), m(m -1) -2, 0, m).
It is easy to see that Λ 1 has non-positive virtual dimensions for d d 0 (m), and thus the actual dimension is zero by Lemma 3.1. Hence the quadric is a component of L 0 and the residual system is

L 1 = L(k -1, 1; m -1, 3 2 k 2 + 1 2 k + 2 - 3 2 m 2 + 3 2 m, 0, 0, m 2 -m -1).
The second specialization is

Z 1 = R(1, 2k + 1 -m -( 1 2 m(m -1) -1), 0, 1 2 m(m -1) -1, 0, 0, 0).
Therefore, Lemma 4.5 implies that the trace system is now Λ 2 = Λ((3k -2, k -2); 3k 2 -3k + 2 -2m 2 + 4m, 0, 0, m -1).

By Lemma 3.1 again, its dimension is zero, so that the quadric turns out to be also a fixed component of L 1 . Note that we have the identity

r(3k, m) + 1 -(2k + 1 -m) -(2k + 1 -m) = r(3(k -2) + 2, m -2) + 1.
Hence the final residual system is

L 2 = L(k -2, 2; m -2, r(3(k -2) + 2, m -2) + 1, 0, 0, 0) = I(k -2, 2, m -2).
Case I(k, 1, m). Here L 0 = L(k, 1; m, r(3k + 1, m) + 1, 0, 0, 0), which has virtual dimension vdim(L 0 ) = -3k -2 + 1 3 m(m -1)(m + 1) < 0.

We apply the specialization

Z 0 = R(1, 2k + 2 -m, 0, 0, 0, 0, 0)
which by the identity

r(3k + 1, m) + 1 -(2k + 2 -m) = r(3(k -1) + 2, m -1) + 1 yields L 1 = L(k -1, 2; m -1, r(3(k -1) + 2, m -1) + 1, 0, 0, 0) = I(k -1, 2, m -1)
as the residual system and Λ 1 = Λ((k -1, 3k + 1); 3k 2 + 3k + 2 -m 2 + m, 0, 0, m) as the trace system. Its virtual dimension is

vdim(Λ 1 ) = -k -2 + m 2 -m < 0, so dim(Λ 1 ) = 0.
Case I(k, 2, m). In this case we will employ the novel machinery introduced in Sect. 2, because specializing to lines on a conic and sundials seems not to suffice. Apart from using zig-zags this case is also considerably more complicated in that it decomposes into three subcases, each of which in turn requires a number of reduction steps. The division into subcases is determined by the divisibility of m by 3, which is a direct consequence of the chosen specializations. The particular choice of reductions is far from obvious and can in fact be considered to be the core innovation of this note. We have L 0 = L(k, 2; m, r(3k + 2, m) + 1, 0, 0, 0) and

vdim(L 0 ) = -2k -2 + 1 3 m(m -1)(m + 1) < 0.
In each case the first specialization is

Z 0 = R(1, 2k + 2 -m, 0, 0, 0, 0, k + m(m -1) -1).
It results in

L 1 = L(k, 0; m -1, 3 2 (k 2 + k -m 2 + m) + 2, 0, 0, k + m 2 -m).
In subsequent steps we work, for p = 2, . . . , m -1, with the specializations

Z p-1 = R(1, 2k + 2 -m - p -1 3 - t z p-1 + 1 2 , 0, t z p-1 + 1 2 , 0, 0, t zp ),
where the number t zp depends on the divisibility of p by 3:

t zp = k + pm(m -p) + 1 3 p(p -1)(p + 1) -2p + 1 if p ≡ 1, 2 pm(m -p) + 1 3 p(p -1)(p + 1) -2p + 2 p 3 if p ≡ 0 (mod 3).
The definition of the numbers t zp is governed by the requirement that the trace system on Q is of the virtual (and hence by Lemma 3.1 also the actual) dimension 0.

We conclude the proof by explaining in detail the final reductions in each of the subcases.

Subcase I(k, 2, m = 3 ). In this case we consider the sequence Z 0 , Z 1 , . . . , Z m-2 defined above and use as a final step Z m-1 = Res Q (Z ) for

Z = R(1, 2k + 2 -m -( -1) - t z m-2 + 1 2 + 1, 0, t z m-2 + 1 2 , 0, 0, 0).
The final residual system is

L m-1 = L(k -2 + 1, 1; 1, r(3k + 2, 3 ) + 21 2 2 - 23 2 + 3 -6k + 2k, 0, 0, 0). Since r(3k + 2, 3 ) + 21 2 2 - 23 2 + 3 -6k + 2k = r(3(k -2 + 1) + 1, 1)
and q(3(k -2 + 1) + 1, 1) = 0 we have and q(3(k -2 ), 1) = 0, so we have L m-1 = B(k -2 , 0, 1). The final trace system is Λ m-1 = Λ((k -2 -1, 3k -6 +2); 2r(3k +2, 3 +2)-12k -4k -9 3 +3 2 -2 , 0, 0, 2).

L m-1 = B(k -2 + 1, 1, 1). The final trace system is Λ m-1 = Λ((k -2 , 3k -6 +6); 2r(3k +2, 3 )-12k -16 +21 2 +3k +3-9 3 , 0, 0, 2), which has virtual dimension -2k -2 + 1 3 m(m -
Its dimension is zero since vdim(Λ m-1 ) = -2k -2 + 1 3 m(m -1)(m + 1) < 0.

Final remarks

We have developed a software to handle calculations necessary here. See [13] for a Maple core file and a file containing an explicit example explaining how to use our program. The software has proved indispensable in order to manipulate sets of data and to discover general patterns leading to suitable specializations. Using this software we were not able to find any systems in the range d < d 0 (m) for which the maximal rank statement in Theorem 4.1 would fail. We therefore expect that the statement holds in these cases as well: We hope that with some modifications, the software mentioned above might prove useful in similar situations, in particular might help to advance towards the proof of Aladpoosh's Conjecture. We also expect that our results can be generalized to projective spaces of arbitrary dimension. This is a subject of ongoing research.

Definition 2 . 7 (

 27 Trace and residual scheme). Let Y be a divisor of degree e in P n and let Z ⊂ P n be a closed subscheme. Then the subscheme Z = Tr Y (Z) defined in Y by the ideal I Z ,Y = (I Y + I Z ) /I Y ⊂ O Y is the trace of Z on Y . The colon ideal I Z = (I Z : I Y ) ⊂ O P n defines Z = Res Y (Z), the residual scheme of Z with respect to Y .

Λ

  ((a, b); p, p d , p m , m) = Λ((a, b); Ω) = H 0 (P 1 × P 1 , O P 1 ×P 1 (a, b) ⊗ I Ω ) to indicate the linear system on P 1 × P 1 of polynomials of bidegree (a, b) vanishing along the subscheme Ω = Ω(p, p d , p m , m) consisting of p general points, p d general double points and p m general points of multiplicity m. In our considerations p m is either 0 or 2, depending on whether we specialize the fat line onto the quadric or not. We define the virtual dimension of Λ((a, b); Ω) as vdim(Λ((a, b); Ω)) = (a + 1)(b + 1) -p -3p d -m + 1 2 p m . Given m 1 and d d 0 (m) = 3 m+1 3 there exist unique integers r(d, m) 0 and 0 q(d, m) d such that

Remark 4 . 2 .

 42 Concretely+ 1)(3d + 5 -2m) -(d + 1)r(d, m).

Theorem 4 . 3 .

 43 Let d d 0 (m) = 3 m+1 3 and let Z = Z(m, r(d, m), 0, q(d, m), 0), or Z = Z(m

Theorem 4 .

 4 3 together with Castelnuovo's inequality yields dim(L(d; Z)) dim(L(d -2u;

1 )(m + 1 ) 2 - 1 2 - 1 - 1 - 2 -

 1122112 < 0.Subcase I(k, 2, m = 3 + 1). Consider the sequence Z 0 , Z 1 , . . . , Z m-1 defined above and use as a final step Z m = Res Q (Z ) forZ = R(1, 2k + 2 -m --t z m-1 final residual system is L m = L(k -2 , 0; 0, r(3k + 2, 3 + 1) + 21 2 6k -2k, 0, 0, 0) 6k -2k = r(3(k -2 ), 0)and q(3(k -2 ), 0) = 0 equals the system B(k -2 , 0, 0), as required. The final trace system is Λ m = Λ((k-2 -1, 3k-6 +2); 2r(3k+2, 3 +1)-12k -4k-9 3 +12 2 + -2, 0, 0, 1).Also in this case we havevdim(Λ m ) = -2k -2 + 1 3 m(m -1)(m + 1) < 0.Subcase I(k, 2, m = 3 +2). Use as in the first subcase the sequence Z 0 , Z 1 , . . . , Z m-2 defined above and use as a final step Z m-1 = Res Q (Z ) forZ = (1, 2k + 2 -m --t z m-2 + 1 2 + 1, 0, t z m-2 + 1 2 , 0, 0, 0).The final residual system is L m-1 = L(k -2 , 0; 1, r(3k + 2, 3 + 26k -2k = r(3(k -2 ), 1)

Conjecture 5 . 1 (

 51 Maximal Rank Conjecture). The restriction maps in Theorem 4.1 have maximal rank for all d 1.

Acknowledgement. This project has been started during the mini-workshop "Arrangements of Subvarieties, and their Applications" held in Mathematischem Forschungsinstitut Oberwolfach February 29 -March 5, 2016. We are grateful to MFO for providing perfect working conditions. Part of the paper has been written up during the visit of the last two authors at KTH. We thank KTH for providing financial support and stimulating working atmosphere. The final reductions have been discovered at the Karma coffee place in Kraków. It is a pleasure to acknowledge support and understanding of the Karma team.

We would like to thank the referees for making a number of valuable suggestions that improved the presentation of this note.

Research of DS was supported by DFG research fellowship SCHM 3223/1-1. Research of TS and JS was partially supported by National Science Centre, Poland, grant 2014/15/B/ST1/02197. Research of SDR was supported by her VR grant [NT:2014-4763].

The system yields through the system L 0 a sequence of length

B(k, 2, m) 1 B(k, 0, m -1)

Table 1: Initial systems and systems resulting from reduction steps Case B(k, 0, m). We specialize only once, and we pick

By Lemma 4.5, we obtain the trace system

which is of virtual dimension

As only reduced general points are involved, this system is non-special, so its actual dimension is also zero. This shows that condition (2) in Theorem 4.3 is fulfilled.

The residual system is

by Lemma 4.5. Note that the subscheme

Case B(k, 1, m). In this case we use two specializations. First set

resulting in the residual system

and the trace system


Joydip Saha
email: saha.joydip56@gmail.com

AND Indranath Sengupta
email: indranathsg@iitgn.ac.in

Gaurab Tripathi

IDEALS OF THE FORM I 1 (XY )

Keywords: 2010 Mathematics Subject Classification. Primary 13P10; Secondary 13C40, 13D02 Gröbner basis, Betti numbers, determinantal ideals, completely irreducible systems

In this paper we compute Gröbner bases for determinantal ideals of the form I 1 (XY ), where X and Y are both matrices whose entries are indeterminates over a field K. We use the Gröbner basis structure to determine Betti numbers for such ideals.

INTRODUCTION

Let K be a field and {x ij ; 1 ≤ i ≤ m, 1 ≤ j ≤ n}, {y j ; 1 ≤ j ≤ n} be indeterminates over K. Let K[x ij ] and K[x ij , y j ] denote the polynomial algebras over K. Let X denote an m × n matrix such that its entries belong to the ideal {x ij ; 1 ≤ i ≤ m, 1 ≤ j ≤ n} . Let Y = (y j ) n×1 be the generic n × 1 column matrix. Let I 1 (XY ) denote the ideal generated by the 1 × 1 minors or the entries of the m × 1 matrix XY . Ideals of the form I 1 (XY ) appeared in the work of J. Herzog [START_REF] Herzog | Certain Complexes Associated to a Sequence and a Matrix[END_REF] in 1974. These ideals are closely related to the notion of Buchsbaum-Eisenbud variety of complexes. A characteristic free study of these varieties can be found in [START_REF] De Concini | On the variety of complexes[END_REF], where the defining equations of these varieties have been described as minors of matrices using combinatorial structure of multitableux. It has also been proved that the varieties are Cohen-Macaulay and Normal. The ideal I 1 (XY ) is a special case of the defining ideal of a variety of complexes, when n 0 = m, n 1 = n, n 2 = 1, in the notation of [START_REF] De Concini | On the variety of complexes[END_REF]. These ideals feature once again in [START_REF] Tchernev | Universal complexes and the generic structure of free resolutions[END_REF], in the study of the structure of a universal ring of a universal pair defined by Hochster. It has been proved in [START_REF] Tchernev | Universal complexes and the generic structure of free resolutions[END_REF] that the set of standard monomials form a free basis for the universal ring. The initial ideal of the defining ideal is given by the set of all nonstandard monomials, which form a monomial ideal. A combination of Gröbner basis techniques and representation theory techniques yield the results in [START_REF] Tchernev | Universal complexes and the generic structure of free resolutions[END_REF]. We were not aware of this work when we computed a Gröbner basis for the ideal I 1 (XY ) using very elementary techniques. Our technique uses nothing more than the Buchberger's criterion and the description of Gröbner bases for the ideals of minors of matrices from [START_REF] Conca | Gröbner basis of Ideals of Minors of a Symmetric Matrix[END_REF] and [START_REF] Sturmfels | Gröbner bases and Stanley decompositions of determinantal rings[END_REF].

Given determinantal ideals I and J, the sum ideal I + J is often difficult to understand and they appear in various contexts. Ideals I 1 (XY ) + J are special in the sense that they occur in several geometric considerations like linkage and generic residual intersection of polynomial ideals, especially in the context of syzygies; see [START_REF] Northcott | A homological investigation of a certain residual ideal[END_REF], [START_REF] Avramov | Poincaré Series of Modules over Local Rings of Small Embedding Codepth or Small Linking number[END_REF], [START_REF] Buchsbaum | Generic Free Resolutions and a family of Generically Perfect Ideals[END_REF], [START_REF] Bruns | The Resolution of the Generic Residual Intersection of a Complete Intersection[END_REF], [START_REF] Johnson | On equations defining Veronese Rings[END_REF]. Some important classes of ideals in this category are the Northcott ideals, the Herzog ideals; see Definition 3.4 in [START_REF] Avramov | Poincaré Series of Modules over Local Rings of Small Embedding Codepth or Small Linking number[END_REF] and the deviation two Gorenstein ideals defined in [START_REF] Huneke | On the Structure of Gorenstein Ideals of Deviation Two[END_REF]. Northcott ideals were resolved by Northcott in [START_REF] Northcott | A homological investigation of a certain residual ideal[END_REF]. Herzog gave a resolution of a special case of the Herzog ideals in [START_REF] Herzog | Certain Complexes Associated to a Sequence and a Matrix[END_REF]. These results were extended in [START_REF] Buchsbaum | Generic Free Resolutions and a family of Generically Perfect Ideals[END_REF]. In a similar vein, Bruns-Kustin-Miller [START_REF] Bruns | The Resolution of the Generic Residual Intersection of a Complete Intersection[END_REF] resolved the ideal I 1 (XY ) + I min(m,n) (X), where X is a generic m × n matrix and Y is a generic n × 1 matrix. proved certain properties for the ideals of the form I 1 (XY ) + I 2 (X), where X is a generic symmetric matrix and Y is either generic or generic alternating. One of the recent articles is [START_REF] Ilten | Unobstructed Stanley-Reisner degenerations for dual quotient bundles on G(2, n)[END_REF] which shows connection of ideals of this form with the ideal of the dual of the quotient bundle on the Grassmannian G(2, n).

Ideals of the form I + J also appear naturally in the study of some natural class of curves; see [START_REF] Gimenez | Minimal graded free resolution for monomial curves defined by arithmetic sequences[END_REF]. While computing Betti numbers for such ideals, a useful technique is often the iterated Mapping Cone. This technique requires a good understanding of successive colon ideals between I and J, which is often difficult to compute. It is helpful if Gröbner bases for I and J are known.

In this paper our aim is to produce some suitable Gröbner bases for ideals of the form I 1 (XY ), when Y is a generic column matrix and X is one of the following:

(1) X is a generic square matrix;

(2) X is a generic symmetric matrix;

(3) X is a generic (n + 1) × n matrix.

We have also studied I 1 (XY ), when (4) X is an (m × mn) generic matrix and Y is an (mn × n) generic matrix.

Our method is constructive and it would exhibit that the first two cases behave similarly. Newly constructed Gröbner bases will be used to compute the Betti numbers of I 1 (XY ). We will see that computing Betti numbers for I 1 (XY ) in the first two cases is not difficult, while the last two cases are not so straightforward. We will use some results from [START_REF] Saha | Primary decomposition of certain Determinanatal ideals[END_REF] and [START_REF] Saha | Betti numbers of certain sum ideals[END_REF] which have some more deep consequences of the Gröbner basis computation carried out in this paper.

DEFINING THE PROBLEMS

Let K be a field and

{x ij ; 1 ≤ i ≤ n + 1, 1 ≤ j ≤ n}, {y j ; 1 ≤ j ≤ n} be indeterminates over K. Let R = K[x ij , y j | 1 ≤ i, j ≤ n], R = K[x ij , y j | 1 ≤ i ≤ n + 1, 1 ≤ j ≤ n] denote polynomial K-algebras. Let X = (x ij ) n×n ,
such that X is either generic or generic symmetric. Let X = (x ij ) (n+1)×n and Y = (y j ) n×1 be generic matrices. We define I = I 1 (XY ) and J = I 1 ( XY ).

Let g i = n j=1 x ij y j , for 1 ≤ i ≤ n. Then, I = g 1 , .
. . , g n . Let us choose the lexicographic monomial order on R given by [START_REF] Avramov | Poincaré Series of Modules over Local Rings of Small Embedding Codepth or Small Linking number[END_REF]

x 11 > x 22 > • • • > x nn ; (2) x ij , y j < x nn for every 1 ≤ i = j ≤ n.
It is an interesting observation that the set {g 1 , . . . , g n } is a Gröbner basis for I with respect to the above monomial order and the elements g 1 , . . . , g n form a regular sequence as well; see Lemma 4.3 and Theorem 6.1. However, this Gröbner basis is too small in size to be of much help in applications like computing primary decomposition of I 1 (XY ) or computing Betti numbers of ideals of the form I 1 (XY ) + J, carried out in [START_REF] Saha | Primary decomposition of certain Determinanatal ideals[END_REF] and [START_REF] Saha | Betti numbers of certain sum ideals[END_REF] respectively. This motivated us to look for a a different Gröbner basis for I; see Theorem 4.1. This construction gives rise to a bigger picture and naturally generalizes to a Gröbner basis for the ideal J = I 1 ( XY ). As an application, we compute the Betti numbers for the ideals I and J ; see section 6. in the polynomial ring R (respectively R);

NOTATION (i) C

k := {a = (a 1 , • • • , a k ) | 1 ≤ a 1 < • • • < a k ≤ n}; denotes the collection of all ordered k-tuples from {1, • • • , n}. In case of J = I 1 ( XY ), the set C k would denote the collection of all ordered k-tuples (a 1 , • • • , a k ) from {1, • • • , n + 1}. (ii) Given a = (a 1 , . . . , a k ) ∈ C k ; • X a = [a 1 , • • • , a k |1,
• X a,m := [a 1 , • • • , a k |1, • • • , k -1, m] if m ≥ k; • X a = m≥k [a 1 , • • • , a k |1, • • • , k-1, m]y m = m≥k X a,m y m ;
• S k := { X a : X a ∈ S k } and I k denotes the ideal generated by S k in the polynomial ring R (respectively R);

• G k = ∪ i≥k S i ; • G = ∪ k≥1 G k ; • X a r := [a 1 , a 2 , • • • , âr , a r+1 • • • , a k |1, 2, • • • , k -1], if k ≥ 2. (iii) Suppose that C k = a 1 < . . . < a ( n k )
, where < is the lexicographic ordering. Given m ≥ k, the map

σ m : X a 1 ,m , . . . , X a ( n k ) ,m → 1, • • • , n k
is defined by σ m (X a i ,m ) = i. This is a bijective map. The map σ k will be denoted by σ, which is the bijection from

S k to {1, • • • , n k } given by σ(X a i ) = σ k (X a i ,k ) = i.

GR ÖBNER BASIS FOR I

We first construct a Gröbner basis for the ideal I. A similar computation works for computing a Gröbner basis for the ideal J , which will be discussed in the next section. Our aim in this section is to prove Theorem 4.1. The set G k is a reduced Gröbner Basis for the ideal I k , with respect to the lexicographic monomial order induced by the following order on the variables:

y 1 > y 2 > • • • > y n > x ij
for all i, j, such that x ij > x i j if i < i or if i = i and j < j . In particular, G = G 1 is a reduced Gröbner Basis for the ideal I 1 = I.

We first write down the main steps involved in the proof.

Let X a , X b ∈ G k = ∪ i≥k S i . Then, either X a , X b ∈ S k or X a ∈ S k , X b ∈ S k , for k > k. Our aim is to show that S( X a , X b ) → G k 0 and use Buchberger's criterion. (A) By Lemma 4.2, we have S(X a , X b ) -→ S k 0. We write m a X a + m b X b = S(X a , X b ) = ( n k ) t=1 α t X at -→ S k 0, such that X a i = X a
and X a j = X b , for some i and j. Therefore, by Schreyer's theorem the tuples (α 1 , . . . , α i -m a , . . . , α j -m b , . . . , α r ) generate Syz(I k ). (B) Syz(I k ) is precisely known by [START_REF] Eagon | Ideals defined by matrices and a certain complex associated to them[END_REF].

(C) S( X a , X b ) -→ S k S( X a , X b ) - ( n k ) t=1 α t X at by Lemma 4.8, if X a , X b ∈ S k and by Lemma 4.10, if X a ∈ S k , X b ∈ S k , for k > k. (D) S( X a , X b ) - ( n k ) t=1 α t X at = s ∈ I k+1 , by Lemma 4.8, if X a , X b ∈ S k . (E) S( X a , X b ) - ( n k ) t=1 α t X at = s ∈ I k +1 , by Lemma 4.10, if X a ∈ S k , X b ∈ S k , for k > k. (F) s -→ G k 0, proved in Theorem 4.1 for both the cases.
We first prove a number of Lemmas to complete the proof through the steps mentioned above.

Lemma 4.2. The set S k forms a Gröbner basis of I k with respect to the chosen monomial order on R.

Proof. We use Buchberger's criterion for the proof. Let c, d

∈ S k . Suppose that S(X c , X d ) S k -→ r. Then, S(X c , X d ) -a i ∈C i h i X a i = r.
If X is generic (respectively generic symmetric), we know by [START_REF] Sturmfels | Gröbner bases and Stanley decompositions of determinantal rings[END_REF] (respectively by [START_REF] Conca | Gröbner basis of Ideals of Minors of a Symmetric Matrix[END_REF]) that the set of all k × k minors of the matrix X forms a Gröbner basis for the ideal I k (X), with respect to the chosen monomial order. Therefore, there exists

[a 1 , a 2 , • • • , a k | b 1 , b 2 , • • • , b k ], such that its leading term k i=1 x a i b i divides Lt(r).
We see that if b k = k, the minor belongs to the set S k and we are done.

Let us now consider the case b

k ≥ k + 1. Let X be generic symmetric. Then, a k = k and b k ≥ k + 1 imply that the minor belongs to the set S k . If a k , b k ≥ k + 1, then x a k b k | Lt(r) but x a k b k doesn't divide
any term of elements in S k . Let X be generic. Then, for any a k and under the condition

b k ≥ k + 1, then x a k b k | Lt(r) but x a k b k doesn't divide any term of elements in S k . Lemma 4.3. Let h 1 , h 2 • • • , h n ∈
R be such that with respect to a suitable monomial order on R, the leading terms of them are pairwise coprime. Then, h 1 , h 2 • • • , h n is a Gröbner basis of the ideal generated by h 1 , h 2 • • • , h n with respect to the same monomial order and they form a regular sequence in R.

Proof. . The proof is a routine application of the division algorithm.

Lemma 4.4. Let 1 ≤ k ≤ n. The height of the ideal I k is n -k + 1, in case of X.
Proof. . Let us consider the case for X. We know that ht(I k ) ≤ n -k + 1. It suffices to find a regular sequence of that length in the ideal

I k . We claim that {[1 • • • k|1 • • • k], [2 • • • k +1|1 • • • k], . . . , [n-k +1 • • • n|1 • • • k]} forms a regular sequence. The leading term of [a 1 , a 2 , • • • , a k | b 1 , b 2 , • • • , b k ]
with respect to the chosen monomial order is k i=1 x a i b i . Therefore, leading terms of the above minors are mutually coprime and we are done by Lemma 4.3.

Remark 4.5. We now assume that X = (x ij ) is a generic n × n matrix. The proof for the symmetric case is exactly the same.

Description of generators of Syz(I k ). By Lemma 4.4 we conclude that a minimal free resolution of the ideal I k is given by the Eagon-Northcott complex. Let us describe the first syzygies of the Eagon-Northcott resolution of

I k . Let a = (a 1 , . . . , a k+1 ) ∈ C k+1 . For 1 ≤ r ≤ k + 1, we define X a r = [a 1 , . . . , âr , . . . , a k+1 |1, . . . , k]. Hence X a r ∈ S k .
We define the map φ as follows.

{1, 2, • • • , k} × C k+1 φ -→ R ( n k ) (j, a) → α such that α(i) = (-1) r i +1 x (ar i , j) if i = σ(X a r i ) for some r i ; 0 otherwise.
The map σ is the bijection from S k to {1, 2, • • • , n k }, defined before. The image of φ gives a complete list of generators of Syz(I k ).

Example 4.6. We give an example, by taking k = 3 and n = 5. Let σ : S 5 -→ {1, • • • 5 3 } be defined by,

• [1, 2, 3 | 1, 2, 3] → 1 • [1, 2, 4 | 1, 2, 3] → 2 • [1, 2, 5 | 1, 2, 3] → 3 • [1, 3, 4 | 1, 2, 3] → 4 • [1, 3, 5 | 1, 2, 3] → 5 • [1, 4, 5 | 1, 2, 3] → 6 • [2, 3, 4 | 1, 2, 3] → 7 • [2, 3, 5 | 1, 2, 3] → 8 • [2, 4, 5 | 1, 2, 3] → 9 • [3, 4, 5 | 1, 2, 3] → 10 In our example, φ : {1, • • • 3} × C 4 -→ R ( 5
3 ) and φ(j, a) → α. Let j = 2 and a = (1, 3, 4, 5). Then,

X a 1 = [3, 4, 5 | 1, 2, 3], X a 2 = [1, 4, 5 | 1, 2, 3], X a 3 = [1, 3, 5 | 1, 2, 3], X a 4 = [1, 3, 4 | 1, 2, 3]. Therefore, σ(X a 1 ) = 10, σ(X a 2 ) = 6, σ(X a 3 ) = 5, σ(X a 4 ) = 4.
Similarly, α(4) = (-1) 4+1 x 52 = -x 52 , α(5) = (-1) 3+1 x 42 = x 42 , α(6) = (-1) 2+1 x 32 = -x 32 , α(10) = (-1) 1+1 x 12 = x 12 . Therefore, α = (0, 0, 0, -x 52 , x 42 , -x 32 , 0, 0, 0,

x 12 ). Lemma 4.7. Let 1 ≤ k ≤ n -1 and let S k = X a 1 , . . . , X a ( n k ) be such that a 1 < . . . < a ( n k ) with respect to the lexicographic ordering. Suppose that α = (α 1 , • • • , α ( n k ) ) ∈ Syz 1 (I k ), then ( n k ) i=1 α i X a i = 0 and ( n k ) i=1 α i X a i ∈ I k+1 . Proof. We have X a i = m≥k σ -1 m (i)y m . Therefore ( n k ) i=1 α i X a i = i α i ( m≥k σ -1 m (i)y m ) = m≥k ( i α i σ -1 m (i)y m ).
It is enough to show that i α i σ -1 m (i)y m ∈ I k+1 , for every m ≥ k. We have α ∈ Syz(I k ) = Im(φ) . Without loss of generality we may assume that α ∈ Im(φ). There exists (j,

a k+1 ) ∈ {1, 2, • • • k} × C k+1 such that φ(j, a k+1 ) = α. We will show that α i • σ -1 m (i) ∈ I k+1 for every m ≥ k and each i. We have i = σ(X a k+1 r i ) since α i = 0. But σ -1 m (i) = [a 1 , . . . , âr i , . . . , a k+1 |1, . . . , k -1, m]. We have [a 1 , . . . , a k+1 |j, 1, . . . , k -1, m] = 0 for j ≤ k -1 and [a 1 , . . . , a k+1 |k, 1, . . . , k-1, m] = (-1) k [a 1 , . . . , a k+1 |1, . . . , k, m] ∈ I k+1 .
Therefore,

( n k ) i=1 α i • σ -1 m (i) = ( n k ) i=1 (-1) r i +1 x (ar i , j) [a 1 , . . . , âr i , . . . , a k+1 |1, . . . , k -1, m] = [a 1 , . . . , a k+1 |j, 1, . . . , k -1, m] ∈ I k+1 ;
Hence,

( n k ) i=1 α i X a i = ( n k ) i=1 α i • σ -1 m (i) = (-1) k ( n k ) i=1 [a 1 , . . . , a k+1 |1, . . . , k, m]y m ∈ I k+1 . Lemma 4.8. Let X a i , X a j ∈ S k = X a 1 , . . . , X a ( n k ) , for i = j. Then,
there exist monomials h t in R and a polynomial r ∈ I k+1 such that

(i) S(X a i , X a j ) = ( n k ) t=1 h t X at , upon division by S k ; (ii) S( X a i , X a j ) = ( n k ) t=1 h t X at + r, upon division by S k .
Proof. (i) The expression follows from the observation that S k is a Gröbner basis for the ideal I k .

(ii) We first note that, Lt( X at ) = Lt(X at )y k , for every X at ∈ S k . Let S(X a i , X a j ) = cX a i -dX a j , where c = lcm(Lt(X a i ), Lt(X a j ))

X a i and d = lcm(Lt(X a i ), Lt(X a j )) X a j Hence, S( X a i , X a j ) = c • X a i -d • X a i = m≥k [c • X a i ,m -d • X a j ,m ] y m .
It follows immediately that Lt(S( X a i , X a j )) = y k Lt(S(X a i , X a j )).

The set S k is a Gröbner basis for the ideal I k . Therefore, we have Lt(X at ) | Lt(S(X a i , X a j )), for some t. Then, Lt( X at ) | Lt(S( X a i , X a j )) and we

have h t = Lt(S(X a i , X a j )) Lt(X at ) = Lt(S( X a i , X a j ))
Lt( X at ) . We can write

r 1 := S( X a i , X a j ) -h t X at = m≥k [c • X a i ,m -d • X a j ,m -h t X at,m ]y m = m>k [c • X a i ,m -d • X a j ,m -h t X at,m ]y m + [c • X a i -d • X a j -h t X at ]y k
Note that r 1 ∈ I k and Lt(r 1 ) = Lt(S( X a i , X a j )-h t X at ) = y k Lt(S(X a i , X a j )h t X at ). We proceed as before with the polynomial S(X a i , X a j ) -h t X at ∈ I k and continue the process to obtain the desired expression involving the polynomial r.

We now show that the polynomial r is in the ideal I k+1 . Let us write

H j = h j +d, H i = h i -c and H t = h t for t = i, j. It follows from S(X a i , X a j ) = ( n k ) t=1 h t X at , that ( n k ) t=1 H t X at = 0. Therefore, H = (H 1 , . . . , H ( n k ) ) ∈ Syz(I k )
and by Lemma 4.7 we have

( n k ) t=1 H t X at ∈ I k+1 . Hence, r = S( X a i , X a j ) -t =i,j h t X at ∈ I k+1 . Lemma 4.9. (i) Let k > k and a = (a 1 , . . . , a k ) ∈ C k . Suppose that X a = bt∈C k β bt X bt is the Laplace expansion of X a . Then bt∈C k β bt X bt,i = [a 1 , . . . , a k |1, . . . , k -1, i, k + 1, . . . , k ]. (ii) Let k > k; a = (a 1 , . . . , a k ) ∈ C k , b = (b 1 , . . . , b k ) ∈ C k . Suppose that X a = p∈C k α p X p and S(X a , X b ) = cX a -dX b = p∈C k β p X p . Then c t≥k [a 1 , • • • , a k |1, • • • , k-1, t, k+1, • • • , k ]y t -d X b - p∈C k β p X p ∈ I k+1 .
Proof. (i) See [START_REF] Janjić | A proof of generalized Laplace's Expansion Theorem[END_REF].

(ii) We have S(X a , X b ) = cX a -dX b = p∈C k β p X p . By rearranging terms we get p∈C k (cα p -β p )X p -dX b = 0 and by separating out the term

(cα b -β b )X b we get p =b (cα p -β p )X p + (cα b -β b -d)X b = 0. Therefore, p =b (cα p -β p ) X p +(cα b -β b -d) X b ∈ I k+1 , by Lemma 4.7. Hence t≥k p =b (cα p -β p )X p,t y t + (cα b -β b -d) t≥k X b,t y t ∈ I k+1 . Now t≥k p∈C k α p X p,t = t≥k [a 1 , • • • , a k |1, • • • , k -1, t, k + 1, • • • , k ] by (i). Hence, c t≥k [a 1 , • • • , a k |1, • • • , k-1, t, k+1, • • • , k ]y t -d X b - p∈C k β p X p ∈ I k+1 . Lemma 4.10. Let k > k; a = (a 1 , . . . , a k ) ∈ C k , b = (b 1 , . . . , b k ) ∈ C k . Suppose that S k = X a 1 , . . . , X a ( n k )
, such that a 1 < . . . < a ( n k ) with respect to the lexicographic ordering. Then, there exist monomials h t ∈ R and a polynomial r ∈ I k+1 such that

(i) S(X a , X b ) = ( n k ) t=1 h t X at , upon division by S k . (ii) S( X a , X b ) = ( n k ) t=1 (h t X at )y k + r, upon division by S k .
Proof. (i) The expression follows from the observation that S k is a Gröbner basis for the ideal

I k . (ii) Let S(X a , X b ) = cX a -dX b , where c = lcm(Lt(X a ), Lt(X b )) X a and d = lcm(Lt(X a ), Lt(X b )) X b . Then, S( X a , X b ) = cy k X a -dy k X b = cy k t≥k X a,t y t -dy k t≥k X b,t y t = y k y k (cX a -dX b ) + terms devoid of y k .
We therefore have Lt(S( X a , X b )) = y k y k Lt(S(X a , X b )), since y k is the largest variable appearing in the above expression. The set S k being a Gröbner basis for the ideal I k , we have Lt(X a t ) dividing Lt(S(X a i , X a j ))

for some t. Let h t = Lt(cX a -dX b )
Lt(X at ) , with t = 1, . . . , n k . Moreover, Lt( X at ) being equal to y k Lt(X at ), it divides Lt(S( X a , X b )). Let

r 1 := S( X a , X b ) - Lt(S( X a , X b )) Lt( X at ) X at = S( X a , X b ) -y k h t X at ∈ I k .
We have

r 1 = y k y k (cX a -dX b ) -y k h t X at + terms devoid of y k = y k y k (cX a -dX b ) -y k h t i≥k X at,i y i + terms devoid of y k = y k y k (cX a -dX b -h t X at ) + terms devoid of y k = y k y k (S(X a , X b ) -h t X at ) + terms devoid of y k . Hence, Lt(r 1 ) = Lt(S(X a , X b ) -h t X at ) = y k y k Lt(S(X a , X b ) -h t X at ).
We proceed as before with the polynomial S(X a , X b ) -h t X at ∈ I k and continue the process to obtain the desired expression involving the polynomial r.

We now show that the polynomial r is in the ideal I k+1 . Let us write r = S( X a , X b ) -

( n k ) t=1 (h t X at )y k = cy k l≥k X a,l y l -dy k l≥k X b,l y l - ( n k ) t=1 l≥k h t X at,l y l y k + T -T ;
where T = c l≥k [a 1 , . . . , a k | 1, . . . , k -1, l, k + 1, . . . , k ]y l y k . After a rearrangement of terms, we may write

r =    T - ( n k ) t=1 l≥k h t X at,l y l y k -dy k l≥k X b,l y l    +   cy k l≥k X a,l y l   -T. Let T = c l>k [a 1 , . . . , a k | 1, . . . , k -1, l, k + 1, . . . , k ]y l y k . Now we note, cX a -dX b - ( n k ) t=1 h t X at = 0. Hence T - ( n k ) t=1 l≥k h t X at,l y l y k - dy k l≥k X b,l y l becomes equal to T - ( n k ) t=1 l>k
h t X at,l y l y k -dy k l>k X b,l y l .

We also have cy k l≥k X a,l y l -T = cy k l>k X a,l y l -T , since the term for l = k in cy k l≥k X a,l y l gets cancelled with the term appearing in T for l = k. Hence we write

r =    T - ( n k ) t=1 l>k h t X at,l y l y k -dy k l>k X b,l y l    1 +   cy k l>k X a,l y l   2 -T = ( ) 1 + ( ) 2 -T .
Clearly, the expression ( ) 1 belongs to I k+1 , by Lemma 4.9. We note that no term of ( ) 1 contains y k . So also for T . Hence, the leading term of r is the leading term of ( ) 2 . By an application of similar argument as above we see that the expression ( ) 2 , after division by elements of S k , further reduces to

-   l>k s≥k c[a 1 , . . . , a k |1, . . . , k -1, s, k + 1, . . . , k -1, l]y l y s   = -   l>k s>k c[a 1 , . . . , a k |1, . . . , k -1, s, k + 1, . . . , k -1, l]y l y s   -   l>k c[a 1 , . . . , a k |1, . . . , k -1, k , k + 1, . . . , k -1, l]y l y k   . Moreover, l>k c[a 1 , . . . , a k |1, . . . , k -1, k , k + 1, . . . , k -1, l]y l y k + T = 0 and l>k s>k c[a 1 , . . . , a k |1, . . . , k -1, s, k + 1, . . . , k -1, l]y l y k = 0.
Therefore, after division by elements of S k , the expression ( ) 1 + ( ) 2 -T reduces to ( ) 1 , which is in I k+1 .

Proof of Theorem 4.1. We use induction on n -k to prove that G k is a Gröbner basis for the ideal I k . For n -k = 0; the set G k = S n contains only one element and hence trivially forms a Gröbner basis. We apply Buchberger's algorithm to prove our claim. Let X a , X b ∈ G k . The following cases may arise:

• X a , X b ∈ S k , for a, b ∈ C k ; • X a ∈ S k and X b ∈ S k where k > k; a ∈ C k and b ∈ C k .
We have proved in Lemmas 4.8 and 4.10 that upon division by S k , the Spolynomial S( X a , X b ) -→ r for some r ∈ I k+1 , in both the cases. By induction hypothesis, G k+1 is a Gröbner basis for I k+1 . Hence r reduces to 0 modulo G k+1 and hence modulo G k , since G k+1 ⊂ G k .

We now show that G k is a reduced Gröbner basis for

I k . Let X a ∈ S k and X b ∈ S k where k ≥ k; a ∈ C k and b ∈ C k . Then, X a = i≥k X a,i y i and X b = i≥k X b,i y i . If k > k, then y k |Lt( X a ) but does not divide Lt( X b ). Hence, Lt( X a ) does not divide Lt( X b ). If k = k, then Lt( X a ) = x (a 1 ,1) • • • x (a k ,k) y k and Lt( X b ) = x (b 1 ,1) • • • x (b k ,k) y k . Therefore, X a | X b implies that a = b.
This proves that the Gröbner basis is reduced.

GR ÖBNER BASIS FOR J

Theorem 5.1. Let us consider the lexicographic monomial order induced by

y 1 > y 2 > • • • > y n > x 11 > x 12 > • • • > x (n+1),(n-1) > x (n+1),n on R = K[x ij , y j | 1 ≤ i ≤ n + 1, 1 ≤ j ≤ n].
The set G k is a reduced Gröbner Basis for the ideal I k . In particular, G = G 1 is a reduced Gröbner Basis for the ideal I 1 = J .

Proof. The scheme of the proof is the same as that for I, with suitable changes made for X in the Lemmas. We only reiterate the last part of the proof where we carry out induction on n -k. For n -k = 0, the set G k = S n = {∆ 1 y n , . . . , ∆ n+1 y n }, where ∆ i = det( X i ). We first note that Lt(∆ i ) and Lt(∆ j ) are coprime. Therefore,

S(∆ i y n , ∆ j y n ) = Lt(∆ j ) • (∆ i y n ) -Lt(∆ i ) • (∆ j y n ) = Lt(∆ j )(Lt(∆ i )y n + y n p i ) -Lt(∆ i )(Lt(∆ j )y n -y n p j ) = (Lt(∆ j )y n )p i -(Lt(∆ i )y n )p j = (∆ j y n -p j y n )p i -(∆ i y n -p i y n )p j = ∆ j y n p i -∆ i y n p j -→ Gn 0.
The rest of the proof is essentially the same as that for Theorem 4.1.

6. BETTI NUMBERS OF I AND J Theorem 6.1. Suppose that X = (x ij ) n×n is either a generic or a generic symmetric n × n matrix and Y a generic n × 1 matrix given by Y = (y j ) n×1 . If X is generic, we write g i = n j=1 x ij y j and

I = I 1 (XY ) = g 1 , g 2 , • • • , g n . If X is generic symmetric, we write g 1 = n j=1 x 1j y j , g n = ( 1≤k≤n x kn y k ) and g i = ( 1≤k<i x ki y k ) + ( i≤k≤n x ik y k ) for 1 < i < n and I = I 1 (XY ) = g 1 , • • • , g n . The generators g 1 , . . . , g n of I = I 1 (XY ) in either case form a regular sequence in the polynomial K-algebra R = K[x ij , y j | 1 ≤ i, j ≤ n].
Moreover, {g 1 , . . . , g n } form a Gröbner basis for I in either case with respect to the lexicographic monomial order which satisfies (1) and ( 2) given below:

(1)

x 11 > x 22 > • • > x nn ; (2) x ij , y j < x nn for every 1 ≤ i = j ≤ n.
Proof. The monomial order chosen is lexicographic order induced by the ordering among the variables given by ( 1) and ( 2). It is clear from the expressions of g i that their leading terms are pairwise coprime. Therefore, the proof follows from Lemma 4.3.

Corollary 6.2. I is minimally resolved by the Koszul complex G and the i-th Betti number of I is n i .

Theorem 6.3. Suppose that X = (x ij ) (n+1)×n is a generic (n+1)×n matrix and Y a generic n × 1 matrix given by Y = (y j ) n×1 . Let

g i = n+1 j=1 x ij y j and J = I 1 ( XY ) = g 1 , • • • , g n+1 . The total Betti numbers of the ideal J are β 0 = 1, β 1 = n + 1, β n+1 = n, β k+1 = n k + n k-1 + n k+1 for 1 ≤ k < n.
We first discuss the scheme of the proof below. We will use the following observations to compute the total Betti numbers of J .

Step 1. The minimal graded free resolution of I = g 1 , • • • , g n is given by the Koszul Resolution.

Step 2. We prove that g 1 , • • • , g n : g n+1 = g 1 , • • • , g n , ∆ ; where ∆ = det(X). This proof requires the fact that g 1 , • • • , g n , ∆ is a prime ideal, which has been proved in Theorem 5.4 in [START_REF] Saha | Primary decomposition of certain Determinanatal ideals[END_REF]. Step 3. We prove that g

1 , • • • g n : ∆ = y 1 , y 2 , • • • , y n .
Step 4. We construct a graded free resolution of g 1 , • • • , g n , ∆ using mapping cone between resolutions of g 1 , • • • , g n and y 1 , • • • , y n . We extract a minimal free resolution from this resolution. Step 5. Finally, we construct a graded free resolution of g 1 , • • • , g n , g n+1 using mapping cone between free resolutions of g 1 , • • • , g n , ∆ and g 1 , • • • , g n . We extract a minimal free resolution from this resolution.

Remark 6.4. We need detailed information about the ideal g 1 , • • • , g n , ∆ , where ∆ = det(X). We need the fact that this ideal is a prime ideal, which has been proved in Theorem 5.4 in [START_REF] Saha | Primary decomposition of certain Determinanatal ideals[END_REF]. We also need a minimal free resolution for this ideal, which has been proved below in Lemma 6.10. We came to know much later that g 1 , • • • , g n , ∆ was defined in [START_REF] Northcott | A homological investigation of a certain residual ideal[END_REF]. It is known as the generic Northcott ideal and a minimal free resolution can be found in [START_REF] Northcott | A homological investigation of a certain residual ideal[END_REF]. However, we give a different proof here using our Gröbner basis computation, which also shows the linking of nested complete intersection ideals. Moreover, Northcott's resolution can perhaps be used to prove that g 1 , • • • , g n , ∆ is a prime ideal, although our proof in [START_REF] Saha | Primary decomposition of certain Determinanatal ideals[END_REF] is absolutely different and uses the result in [START_REF] Ferrero | Prime Ideals in Polynomial Rings in Several Indeterminates[END_REF]. Lemma 6.5. ∆y i = n j=1 A ji g j , where A ji is the cofactor of x ji in X. Proof. We have

∆y i = n j=1 A ji x ji y i = n j=1 A ji n k=1 x jk y k - n j=1 A ji k =i x jk y k = n j=1 A ji g j , since n j=1 A ji k =i x jk y k = k =i n j=1 A ji x jk y k = 0. Lemma 6.6. g 1 , • • • , g n , ∆ ⊆ g 1 , • • • , g n : g n+1 . Proof. We have g i ∈ g 1 , • • • , g n : g n+1 , for every 1 ≤ i ≤ n. Moreover, y i ∆ ∈ g 1 , • • • , g n , by Lemma 6.5. Hence, g n+1 ∆ ∈ g 1 , • • • , g n . Lemma 6.7. g 1 , • • • , g n : g n+1 = g 1 , • • • , g n , ∆ Proof. We have proved that g 1 , • • • , g n , ∆ ⊆ g 1 , • • • , g n : g n+1 in Lemma 6.6. We now prove that g 1 , • • • , g n : g n+1 ⊆ g 1 , • • • , g n , ∆ . Let z ∈ g 1 , • • • , g n : g n+1 . Then zg n+1 ∈ g 1 , • • • , g n ⊂ g 1 , • • • , g n , ∆ . It is easy to see that g n+1 / ∈ g 1 , • • • , g n , ∆ . Therefore, z ∈ g 1 , • • • , g n , ∆ , since g 1 , • • • , g n , ∆
is a prime ideal by Theorem 5.4 in [START_REF] Saha | Primary decomposition of certain Determinanatal ideals[END_REF].

Lemma 6.8. g 1 , • • • , g n : ∆ = y 1 , • • • , y n Proof. We have y i ∆ ∈ g 1 , • • • , g n by Lemma 6.5; which implies that y 1 , • • • , y n ⊂ g 1 , • • • , g n : ∆ . Let z ∈ g 1 , • • • , g n : ∆ . Then z∆ ∈ g 1 , • • • , g n ⊆ y 1 , • • • , y n . Therefore, z ∈ y 1 , • • • , y n , since ∆ / ∈ y 1 , • • • , y n and y 1 , • • • , y n is a prime ideal.
Mapping Cones. The resolution for y 1 , • • • , y n is given by the Koszul complex F . We now give a resolution of g 1 , • • • , g n , ∆ by the mapping cone technique. We know that g 1 , • • • , g n : ∆ = y 1 , • • • , y n , by Lemma 6.8. We first construct a connecting homomorphism φ : F -→ G . Let φ 0 denote the multiplication by ∆. In order to make the map φ 0 a degree zero map, we set the grading as F 0 ∼ = (R(-n)) 1 and G 0 = (R(0)) 1 . Since F and G are both Koszul resolutions, we set the grading as

G i ∼ = (R(-2i)) ( n i ) and F i ∼ = (R(-n -i)) ( n i
) . Now we see that, i = n implies that -2i = -n -i. Hence the image of φ i for i = n is contained in the maximal ideal. We have F i = G i , only for i = n. If we can show that the map φ n is not the zero map, then this will be the only free part of the resolution which we can cancel out for obtaining the minimal resolution. Lemma 6.9. The map φ n is not the zero map.

Proof. We refer to [START_REF] Gimenez | Minimal graded free resolution for monomial curves defined by arithmetic sequences[END_REF]. If φ n is the zero map, then φ 0 (R) ⊆ δ 1 (G 1 ), where δ . denotes the differential of G. The image of δ 1 is the ideal g 1 , • • • , g n , which does not contain φ 0 (1) = ∆. The map φ n is not the zero map. Therefore, the above discussion proves the following Lemma. Lemma 6.10. Hence a minimal graded free resolution of g

1 , • • • , g n , ∆ is given by M , such that M i ∼ = (R(-n -i + 1)) ( n i-1 ) ⊕ (R(-2i)) ( n i ) for 0 < i < n, M 0 ∼ = R(0) and M n ∼ = (R(-2n)) n .
(Proof of Theorem 6.3.) We now find the Betti numbers for the ideal g 1 , • • • , g n+1 by constructing the mapping cone between the resolutions M and the resolution G of g 1 , • • • , g n . The connecting map ψ 0 is multiplication by g n+1 . Hence to make it degree zero we set,

G 0 = (R(2)) 1 and G i ∼ = (R(2 -2i)) ( n i ) for i > 0.
Here we note that 2 -2i = -2i and -n -i + 1 = 2 -2i for 1 ≤ i ≤ n. Hence, for each 1 ≤ i ≤ n, the image of ψ i is contained in the maximal ideal. This shows that the resolution obtained by the mapping cone between M and G is minimal. Hence the total Betti numbers of J are:

β 0 = 1, β 1 = n + 1; β n+1 = n; β k+1 = n k + n k-1 + n k+1 for 1 ≤ k < n.
Corollary 6.11. The ring R/I is Cohen-Macaulay and the ring R/J is not Cohen-Macaulay.

Proof. The polynomial ring R is Cohen-Macaulay and g 1 , . . . , g n is a regular sequence therefore the ring R/I is Cohen-Macaulay.

We have seen that projdim R R/J = n + 1. Therefore, by the Auslander-Bauchsbaum formula depth R R/J = n(n + 1) + n -(n + 1) = n 2 + n -1. We have proved in Lemma 5.5 in [START_REF] Saha | Primary decomposition of certain Determinanatal ideals[END_REF] that y 1 , . . . , y n is a minimal prime over J . Therefore, dim R/J ≥ dim R/ y 1 , . . . , y n = n 2 + n; hence the ring R/J is not Cohen-Macaulay. Finally, we consider the case when X = (x ij ) m×mn is a generic matrix of size m × mn and Y = (y ij ) mn×n is generic matrix of size mn × n. We define

I = I 1 (XY ). Let g ij = mn t=1 x it y tj , with 1 ≤ i ≤ m, 1 ≤ i ≤ n. Then, I = {g ij | 1 ≤ i ≤ m, 1 ≤ i ≤ n} .
In this section we construct a Gröbner basis for the ideal I with respect to a suitable monomial order and use that to show that the generators g ij , with 1 ≤ i ≤ m, 1 ≤ i ≤ n form a regular sequence. We first set a few notations before we prove the main results.

• X = A 1 • • • A n , where A s =   x 1(m(s-1)+1) • • • x 1(ms) . . . . . . . . . x m(m(s-1)+1) • • • x m(ms)  
is the m × m matrix for every 1 ≤ s ≤ n.

• [X] s = A s A 1 • • • A s • • • A n , for every 1 ≤ s ≤ n. • [Y ] s =         
y (m(s-1)+1)s . . . y (ms)s y 1s . . .

y (mn)s         
, for every 1 ≤ s ≤ n.

We will use Theorem 4.1 for constructing a Gröbner basis for the ideal I. A very important reason behind considering this class of ideals is that we get some nice examples of transversal intersection of ideals. Two results that would be useful for our purpose are the following: Lemma 7.1. Let > be a monomial ordering on R. Let I and J be ideals in R, such that m(I) and m(J) denote unique minimal generating sets for their leading ideals Lt(I) and Lt(J) respectively. Then, I ∩ J = IJ if the set of variables occurring in the set m(I) is disjointed from the the set of variables occurring in the set m(J).

Proof. See Lemma 3.6 in [START_REF] Saha | Betti numbers of certain sum ideals[END_REF].

Lemma 7.2. Let I and J be graded ideals in a graded ring R, such that I ∩ J = I • J. Suppose that F and G are minimal free resolutions of I and J respectively. Then F ⊗ G is a minimal free resolution for the graded ideal I + J.

Proof. See Lemma 3.7 in [START_REF] Saha | Betti numbers of certain sum ideals[END_REF]. Proof. We have P t = t s=1 I 1 ([X] s [Y ] s ), and we observe that if p ∈ G s and q ∈ G t for 1 ≤ s < t ≤ n, then gcd(Lt(p), Lt(q)) = 1. Therefore the S-polynomial of p, q reduces to zero after applying division upon G t . Theorem 7.4. Let us denote P t = t s=1 I 1 ([X] s [Y ] s ), for 1 ≤ t ≤ n -1.

Then P t ∩ I 1 ([X] t+1 [Y ] t+1 ) = P t • I 1 ([X] t+1 [Y ] t+1
). Hence the elements g ij = mn t=1 x it y tj , 1 ≤ i ≤ m, 1 ≤ i ≤ n form a regular sequence and the Koszul complex resolves R/I as an R-module minimally.

Proof. If p ∈ G s and q ∈ G t , for 1 ≤ s < t ≤ n. Then gcd(Lt(p), Lt(q)) = 1, therefore by theorem 7.3 and lemma 7.1, we have

P t ∩I 1 ([X] t+1 [Y ] t+1 ) = P t • I 1 ([X] t+1 [Y ] t+1 ).
By Theorem 6.1 the generators of the ideal P 1 form a regular sequence and also the generators of the ideal I 1 ([X] s [Y ] s ) form a regular sequence for each 1 ≤ s ≤ n. Hence the Koszul complex resolve R/P 1 and R/I 1 ([X] s [Y ] s ) minimally. Now P t ∩ I 1 ([X] t+1 [Y ] t+1 ) = P t • I 1 ([X] t+1 [Y ] t+1 ). Hence, by application of lemma 7.1 we can conclude that the Koszul complex resolves R/I minimaly.

7 .

 7 I 1 (XY ), WHERE X IS m × mn GENERIC MATRIX AND Y IS mn × n GENERIC MATRIX

Theorem 7 . 3 .

 73 Let us choose the lexicographic monomial order on R induced byy 11 > y 21 > • • • > y (mn)1 > y (m+1)2 > y (m+2)2 > • • • > y (2m)2 > y 12 > • • • y (mn)2 > • • • > y (m(n-1)+1)n > y (m(n-1)+2)n > • • • > y ((mn)n > y 1n > • • • y (m(n-1))n > x 11 > x 12 > • • • > x m(mn) . Let G s be the reduced Gröbner Basis of the ideal I 1 ([X] s [Y ] s ) for 1 ≤ s ≤ n, obtained by Theorem 4.1. Then G t = ∪ ts=1 G s is a reduced Gröbner Basis for the idealP t = t s=1 I 1 ([X] s [Y ] s ) for 1 ≤ t ≤ n.In particular, G n is a reduced Gröbner Basis for the ideal P n = I = I 1 (XY ).

  2, . . . , k] denotes the k × k minor of the matrix X, with a 1 , . . . , a k as rows and 1, . . . , k as columns. Similarly, X a = [a 1 , • • • , a k |1, . . . , k] denotes the k × k minor of the matrix X, with a 1 , . . . , a k as rows and 1, . . . , k as columns.• S k := {X a : a ∈ C k } and I k denotes the ideal generated by S k

ACKNOWLEDGEMENTS

The second author is the corresponding author who has been supported by the research project EMR/2015/000776, sponsored by the SERB, Government of India. The third author thanks SERB for the post-doctoral fellowship under the said project. The third author thanks CSIR for the Senior Research Fellowship for Ph.D. The authors thank the anonymous referees for their valuable comments and for drawing their attention to the references [5] and [18], extremely pertinent to this work.