L Busé

G Chèze

On the total order of reducibility of a pencil of algebraic plane curves

Keywords: Pencil of algebraic curves, spectrum of a rational function, algebraic de Rham's cohomology, Newton's polygon

In this paper, the problem of bounding the number of reducible curves in a pencil of algebraic plane curves is addressed. Unlike most of the previous related works, each reducible curve of the pencil is here counted with its appropriate multiplicity. It is proved that this number of reducible curves, counted with multiplicity, is bounded by d 2 -1 where d is the degree of the pencil. Then, a sharper bound is given by taking into account the Newton's polygon of the pencil.

Introduction

Given a pencil of algebraic plane curves such that a general element is irreducible, the purpose of this paper is to give a sharp upper bound for the number of reducible curves in this pencil. This question has been widely studied in the literature, but never, as far as we know, by counting the reducible factors with their multiplicities.

Let r(X, Y ) = f (X, Y )/g(X, Y ) be a rational function in K(X, Y ), where K is an algebraically closed field. It is commonly said to be non-composite if it cannot be written r = u • h where h(X, Y ) ∈ K(X, Y ) and u ∈ K(T ) such that deg(u) ≥ 2 (recall that the degree of a rational function is the maximum of the degrees of its numerator and denominator after reduction). If d = max(deg(f ), deg(g)), we define

f ♯ (X, Y, Z) = Z d f X Z , Y Z , g ♯ (X, Y, Z) = Z d g X Z , Y Z
that are two homogeneous polynomials of the same degree

d in K[X, Y, Z]. The set σ(f, g) = {(µ : λ) ∈ P 1 K | µf ♯ + λg ♯ is reducible in K[X, Y, Z]} ⊂ P 1 K
is the spectrum of r and a classical theorem of Bertini and Krull implies that it is finite if r is non-composite. Actually, σ(f, g) is finite if and only if r is non-composite and if and only if the pencil of projective algebraic plane curves µf ♯ + λg ♯ = 0, (µ : λ) ∈ P1 K , has an irreducible general element (see for instance [Jou79, Chapitre 2, Théorème 3.4.6] and [Bod08, Theorem 2.2] for detailed proofs). Notice that the study of σ(f, g) is trivial if d = 1. Therefore, throughout this paper we will always assume that d ≥ 2.

Given (µ : λ) ∈ σ(f, g), a complete factorization of the polynomial µf ♯ + λg ♯ is of the form µf ♯ + λg ♯ = n(µ:λ) i=1 P e (µ:λ),i (µ:λ),i (⋆)

where each polynomial P (µ:λ),i is irreducible and homogeneous in K[X, Y, Z]. If σ(f, g) is finite the total order of reducibility 1 ρ(f, g) of r is then defined by ρ(f, g) =

(µ:λ)∈P 1 K n(µ : λ) -1 .
Observe that the above sum is finite because n(µ : λ) = 1 implies that (µ : λ) ∈ σ(f, g).

It is known that ρ(f, g) is bounded above by d 2 -1 where d stands for the degree of r. As far as we know, the first related result has been given by Poincaré [START_REF] Poincaré | Sur l'intégration algébrique des équations différentielles du premier ordre[END_REF]. He showed that |σ(f, g)| ≤ (2d -1) 2 + 2d + 2. This bound was improved only very recently by Ruppert [START_REF] Ruppert | Reduzibilität Ebener Kurven[END_REF] who proves that |σ(f, g)| is bounded by d 2 -1. This result was obtained as a byproduct of a very interesting technique developed by the author to decide the reducibility of an algebraic plane curve. Later on, Stein studied in [START_REF] Stein | The total reducibility order of a polynomial in two variables[END_REF] a less general question but gave a stronger result: he proves that if g = 1 then ρ(f, 1) ≤ d -1. Its approach, based on the study of the multiplicative group of all the divisors of the reducible curves in the pencil, is entirely different from that of Ruppert. Then, Stein's bound was improved in [START_REF] Kaliman | Two remarks on polynomials in two variables[END_REF] and after that several papers [START_REF] Lorenzini | Reducibility of polynomials in two variables[END_REF][START_REF] Vistoli | The number of reducible hypersurfaces in a pencil[END_REF][START_REF] Shreeram | Translates of polynomials[END_REF][START_REF] Bodin | Reducibility of rational functions in several variables[END_REF] developed techniques with similar flavors to deal with the general case ρ(f, g). All of them obtained the bound ρ(f, g) ≤ d 2 -1 but also provide some various extensions: In [START_REF] Lorenzini | Reducibility of polynomials in two variables[END_REF] the bound is proved in arbitrary characteristic, in [START_REF] Bodin | Reducibility of rational functions in several variables[END_REF] it is shown that a direct generalization of Stein's result yields the bound ρ(f, g) ≤ d 2 + d -1, in [START_REF] Vistoli | The number of reducible hypersurfaces in a pencil[END_REF] the result is generalized to a very general ground variety and finally, in [START_REF] Shreeram | Translates of polynomials[END_REF] the authors were interested in a total reducibility order over a field K that is not necessarily algebraically closed. Incidentally, point out the paper [START_REF] Vitório | Completely reducible hypersurfaces in a pencil[END_REF] that deals with completely reducible curves in a pencil, a topic which is closely related.

The aim of this paper is to study the total order of reducibility by counting the multiplicities. More precisely, for each (µ : λ) ∈ σ(f, g) define m(µ : λ) := n(µ:λ) i=1 e (µ:λ),i from the factorization (⋆). This number is the number of factors of µf ♯ + λg ♯ where the multiplicities of the factors are counted. In particular, it is clear that n(µ : λ) ≤ m(µ : λ). We define the total order of reducibility with multiplicities of the rational function r as the integer

m(f, g) = (µ:λ)∈P 1 K m(µ : λ) -1 .
Obviously, it always holds that 0 ≤ ρ(f, g) ≤ m(f, g). Moreover, notice that unlike ρ(f, g), m(f, g) takes into account those curves in the pencil that are geometrically irreducible but scheme-theoretically non-reduced. However, it is proved in [AHS03, General Mixed Primset Theorem, p 74] that the number of such curves is at most 4 in our context; we will come back to this point in Section 2.

The first main result of this paper is that the upper bound d 2 -1 for ρ(f, g) is also valid for m(f, g). This is the content of Section 2 where it is assumed that the characteristic of K is zero. Our method, which is inspired by [START_REF] Ruppert | Reduzibilität Ebener Kurven[END_REF], is elementary compared to the previously mentioned papers. Roughly speaking, we will transform the pencil of curves into a pencil of matrices and obtain in this way the claimed bound as a consequence of rank computations of some matrices that we will study in Section 1. In this way, the known inequality ρ(f, g) ≤ d 2 -1 is easily obtained. Moreover, we will actually not only bound m(f, g) by d 2 -1, but a bigger quantity that takes into account the multiple factors of the reducible elements in the pencil. Notice that we will also show that the same bound holds in the case where r = f /g is a rational function in an arbitrary number of variables via a classical use of Bertini's Theorem at the end of Section 2.

The second main result of this paper, given in Section 3, is a refined upper bound for m(f, g) which is obtained by considering the Newton's polygons of the polynomials f and g. This result also gives a bound for the total order of reducibility ρ(f, g) which is new and sharper. Notice that in this section the characteristic of K will be assumed to be 0 or > d(d -1) where d denotes the degree of r = f /g.

Notations

Throughout this paper, K stands for an algebraically closed field of characteristic p. Given a polynomial f , deg(f ) denotes its total degree and ∂ X f (resp. ∂ Y f ) denotes the partial derivative of f with respect to the variable X (resp. to Y ). Also, for any integer n the notation K[X, Y ] ≤n stands for the set of all the polynomials in K[X, Y ] with total degree less or equal to n; the notation K[X, Y, Z] n stands for the set of all homogeneous polynomials of degree n in K[X, Y, Z].

Ruppert's linear map

In the paper [START_REF] Ruppert | Reduzibilität Ebener Kurven[END_REF], Ruppert introduced an original technique to decide whether a plane algebraic curve is reducible. Its formulation relies on the computation of the first de Rham's cohomology group of the complementary of the plane curve by means of linear algebra methods. Later, Gao followed this approach to obtain an algorithm for the factorization of a bivariate polynomial [START_REF] Gao | Factoring multivariate polynomials via partial differential equations[END_REF].

From now on, we will always assume in this section that the characteristic of the algebraically closed field K is p = 0.

For ν a positive integer and

f (X, Y ) ∈ K[X, Y ] a polynomial of degree d ≤ ν, define the K-linear map G ν (f ) : K[X, Y ] ≤ν-1 × K[X, Y ] ≤ν-1 -→ K[X, Y ] ≤ν+d-2 (G, H) → f 2 ∂ Y G f -∂ X H f = f ∂ Y f G ∂ Y G - f ∂ X f H ∂ X H . Let f 1 , . . . , f r be the irreducible factors of f . If gcd(f, ∂ X f ) is a nonzero constant in K then it is proved in [Gao03] that ker G d (f
) is a K-vector space of dimension r and that the set

f f i ∂ X f i , f f i ∂ Y f i | i = 1, . . . , r (1.1)
is a basis of this kernel. This result provides an explicit description of the kernel of the linear map G d (f ) if the polynomial f (X, Y ) does not have any square factor. In order to investigate this kernel in the general case, that is to say for an arbitrary polynomial f ∈ K[X, Y ] and for an arbitrary integer ν ≥ deg(f ), we interpret it in terms of algebraic de Rham cohomology.

Let ν be a positive integer and 0

= f (X, Y ) ∈ K[X, Y ] be a polynomial of degree d ≤ ν. Assume that f = f e 1 1 • • • f er
r is a factorization of f where each polynomial f i is irreducible and denote by C the algebraic curve defined by the equation f = 0. The first algebraic de Rham cohomology

H 1 (A 2 K \ C) is the quotient of the closed 1-differential forms w ∈ Ω K[X,Y ] f /K of K[X, Y ] f over K by the exact 1-forms. By definition of G ν (f ), a couple (G, H) ∈ K[X, Y ] ≤ν-1 × K[X, Y ] ≤ν-1
belongs to the kernel of G ν (f ) if and only if the 1-form 1 f (GdX + HdY ) is closed. Therefore, the kernel of G ν (f ) is in correspondence with the closed 1-differential forms w ∈ Ω K[X,Y ] f /K that can be written w = 1 f (GdX + HdY ) for some polynomials G and H of degree less or equal to ν -1. As a consequence of Ruppert's results in [START_REF] Ruppert | Reduzibilität Ebener Kurven[END_REF] (see also [START_REF] Scheiblechner | On the complexity of counting irreducible components and computing betti numbers of algebraic variety[END_REF]Theorem 8.3]), these particular closed 1-forms are sufficient to give a representation of any element in H 1 (A 2 K \ C), that is to say that the canonical map

ker G ν (f ) → H 1 (A 2 K \ C)
is surjective. Actually, the closed 1-forms for some P ∈ K[X, Y ] and s ∈ N. However, we claim that the following equality holds

B ν = w = 1 f (GdX + HdY ), (G, H) ∈ K[X, Y ] ≤ν-1 × K[X, Y ] ≤ν-1 such that ∃P ∈ K[X, Y ] ≤ν with d P f = w . (1.2)
It is a consequence of the following technical results.

Lemma 1. Let p, q be polynomials in K[X, Y ] such that p divides qdp. Then each irreducible factor of p divides q.

Proof. Let p 1 , . . . , p r be distinct irreducible factors of p such that p = r i=1 p e i i . Then the equality dp p = r i=1 e i dp i p i together with our hypothesis imply that p e i i divides q r j=1 e j p p j dp j . We deduce that p e i i must divide q p p i dp i and therefore that p i divides q.

Lemma 2. Let f ∈ K[X, Y ] of degree d and G, H ∈ K[X, Y ] of degree ≤ ν-1 with ν ≥ d. If P ∈ K[X,
d P f s = f dP -sP df f s+1
we have f dP -sP df = f s (GdX + HdY ).

(1.3) Assume that s ≥ 2 and denote by f = r i=1 f e i i an irreducible factorization of f . Equation (1.3) implies that f divides P df and therefore, by Lemma 1, that f i divides P for all i = 1, . . . , r. Furthermore, since

df f = r i=1 e i df i f i we get f dP -sP df = f dP -sP f r i=1 e i df i f i = f (dP -s r i=1 e i P f i df i ).
But f s divides f dP -sP df by (1.3), so we deduce that

f s-1 | dP -s r i=1 e i P f i df i .
Define Q := gcd(f, P ) = r i=1 f µ i i with 1 ≤ µ i ≤ e i for all i = 1, . . . , r and set R := P/Q. We obtain that f s-1 divides

QdR + RdQ -s r i=1 e i P f i df i = QdR + r i=1 (µ i -se i )R Q f i df i since dQ Q = r i=1 µ i df i f i .
As s ≥ 2, µ ise i < 0 for all i and hence f µ i i divides R Q f i df i . It follows that f i divides Rdf i and therefore that f i divides R by Lemma 1. But then f µ i +1 i divides P which implies that µ i = e i for all i. Therefore, we conclude that if s ≥ 2 then necessarily f divides P : a contradiction with our hypotheses. So we must have 0 ≤ s ≤ 1.

Suppose that s = 0. Then

GdX + HdY = f dP = f ∂ X P dX + f ∂ Y P dY
and hence deg(P ) ≤ νd. Now, assume that s = 1. We have

f GdX + f HdY = f dP -P df = (f ∂ X P -P ∂ X f )dX + (f ∂ Y P -P ∂ Y f )dY.
Denote by δ the degree of P and by P δ , resp. f d , the homogeneous part of highest degree of P , resp. f . If

f d ∂ X P δ -P δ ∂ X f d = 0 or f d ∂ Y P δ -P δ ∂ Y f d = 0 then necessarily δ ≤ ν since deg(f G) ≤ ν + d -1 and deg(f H) ≤ ν + d -1.
Otherwise, we obtain that

d P δ f d = f d dP δ -P δ df d (f d ) 2 = 0
and hence that δ = d ≤ ν.

We are now ready to compute the dimension of the kernel of the K-linear map G ν (f ) for all ν ≥ d.

Proposition 3. Let f (X, Y ) ∈ K[X, Y ] of degree d such that f = f e 1 1 • • • f er r is a factorization of f where each polynomial f i is irreducible of degree d i .
Then, for all ν ≥ d we have

dim K ker G ν (f ) = r -1 + 2 + ν -d + r i=1 d i (e i -1) 2 .
Proof. From the above discussion on the interpretation of ker G ν (f ) in terms of 1-differential forms, we know that

dim K ker G ν (f ) = dim K H 1 (A 2 K \ C) + dim K B ν
where B ν is defined by (1.2). Since we also know that dim K H 1 (A 2 K \ C) = r, it remains to compute the dimension of B ν . For that purpose, observe that the condition

d P f = w in the definition of B ν is equivalent to the system of equations f ∂ X P -P ∂ X f -Gf = 0 f ∂ Y P -P ∂ Y f -Hf = 0 with the constraints deg(G) ≤ ν -1, deg(H) ≤ ν -1 and deg(P ) ≤ ν.
Denote by L ν the vector space of those triples (G, H, P ) solution of this system. The canonical projection (G, H, P ) → (G, H) sends L ν to B ν . Moreover, the kernel of this projection are the triples (0, 0, P ) satisfying the condition d P f = 0 which implies that P is equal to f up to multiplication by an element in K. Therefore, dim K B ν = dim K L ν -1 and we are left with the computation of the dimension of L ν .

The first equation defining L ν , that can be rewritten as f (∂ X P -G) = P ∂ X f , implies that P must be of the form

P = Q 1 f gcd(f, ∂ X f ) where Q 1 is a polynomial of degree less or equal to ν -d + deg gcd(f, ∂ X f ).
Moreover, any such polynomial P provides a couple (P, G) that is solution of the above equation -once P is fixed then so does for G. A similar reasoning with the second defining equation of L ν shows that its solutions are in correspondence with the polynomials P of the form

Q 2 f / gcd(f, ∂ Y f ) where Q 2 is any polynomial of degree less or equal to ν -d + deg gcd(f, ∂ Y f ).
Now, to obtain the common solutions of the two defining equations of L ν we have to solve the equation

Q 2 gcd(f, ∂ X f ) = Q 1 gcd(f, ∂ Y f ).
But again, with similar arguments and using the fact that

gcd (gcd(f, ∂ X f ), gcd(f, ∂ Y f )) = gcd(f, ∂ X f, ∂ Y f ) we get that Q 1 = Q gcd(f, ∂ X f ) gcd(f, ∂ X f, ∂ Y f ) , Q 2 = Q gcd(f, ∂ Y f ) gcd(f, ∂ X f, ∂ Y f ) where Q is any polynomial in K[X, Y ] of degree less or equal to ν -d + deg gcd(f, ∂ X f, ∂ Y f ) = ν -d + r i=1 d i (e i -1).
(1.4) Therefore, we deduce that the dimension of L ν is equal to the dimension of the K-vector space of polynomials in K[X, Y ] of degree less or equal to the quantity (1.4), that is to say

2 + ν -d + r i=1 d i (e i -1) 2
and the claimed formula is proved. Following Ruppert's approach in [START_REF] Ruppert | Reduzibilität Ebener Kurven[END_REF], we introduce a new K-linear map which is similar to G ν (f ) but with a source of smaller dimension. This property will be very important in the next section. To be more precise, for all positive integer ν consider the K-vector space

E ν = {(G, H) ∈ K[X, Y ] ≤ν-1 ×K[X, Y ] ≤ν-1 such that deg(XG+Y H) ≤ ν-1}.
It is of dimension ν 2 -1 and has the following property.

f 2 ∂ Y G f -∂ X H f has degree at most ν + d -3 Proof. Denote by G ν-1 , resp. H ν-1 , f d , the homogeneous component of G, resp. H, f of degree ν -1, resp. ν -1, d. We have f 2 d d XG ν-1 + Y H ν-1 f d = f d d(XG ν-1 + Y H ν-1 ) -(XG ν-1 + Y H ν-1 )df d = f d (G ν-1 + X∂ X G ν-1 + Y ∂ X H ν-1 )dX +f d (H ν-1 + X∂ Y G ν-1 + Y ∂ Y H ν-1 )dY -(XG ν-1 ∂ X f d + Y H ν-1 ∂ X f d )dX -(Y H ν-1 ∂ Y f d + XG ν-1 ∂ Y f d )dY.
So, using Euler's relation the coefficient of dX is

f d (νG ν-1 -Y ∂ Y G ν-1 + Y ∂ X H ν-1 ) -G ν-1 (df d -Y ∂ Y f d ) -Y H ν-1 ∂ X f d that is to say (ν -d)f d G ν-1 -Y f 2 d ∂ Y G ν-1 f d -∂ X H ν-1 f d . (1.5) Similarly, the coefficient of dY is (ν -d)f d H ν-1 -Xf 2 d ∂ Y G ν-1 f d -∂ X H ν-1 f d . (1.6) Now, since (G, H) ∈ E ν we have XG ν-1 + Y H ν-1 = 0.
Therefore the quantities (1.5) and (1.6) are both equal to zero. It follows that

0 = X × (1.5) + Y × (1.6) = (ν -d)f d (XG ν-1 + Y H ν-1 ) -2XY f 2 d ∂ Y G ν-1 f d -∂ X H ν-1 f d = -2XY f 2 d ∂ Y G ν-1 f d -∂ X H ν-1 f d and the lemma is proved. Let f (X, Y ) ∈ K[X, Y ] of degree d. For all integer ν ≥ d we define the K-linear map R ν (f ) : E ν -→ K[X, Y ] ≤ν+d-3 : (G, H) → f 2 ∂ Y G f -∂ X H f .
Point out that the operator R ν (-) is K-linear, that is to say that for all couples (f, g) ∈ K[X, Y ] ≤ν and all couple (u, v) ∈ K 2 , we have

R ν (uf + vg) = uR ν (f ) + vR ν (g).
Of course, a similar property holds for the operator G ν (f ).

Proposition 5. Let f (X, Y ) ∈ K[X, Y ] of degree d. Then dim K ker R d (f ) = dim K ker G d (f ) -1 and for all ν > d dim K ker R ν (f ) = dim K ker G ν-1 (f ). Proof. Denote by G ν-1 , resp. H ν-1 , f d , the homogeneous component of G, resp. H, f of degree ν -1, resp. ν -1, d.
First, notice that for all integer ν ≥ d and all couple (G, H) ∈ ker G ν (f ) we have

d XG ν-1 + Y H ν-1 f d = (ν -d) G ν-1 dX + H ν-1 dY f d . (1.7)
Indeed, this follows from the computation we did in the proof of Lemma 4, more precisely the coefficients (1.5) and (1.6). Now, let

f = f e 1 1 • • • f er r be a factorization of f where each polynomial f i is irreducible of degree d i . By definition of both maps G d (f ) and R d (f ), it is obvious to notice that any element in the kernel of R d (f ) is also in the kernel of G d (f ). Moreover, it is easy to check that f f 1 ∂ X f 1 , f f 1 ∂ Y f 1 ∈ ker G d (f ) but does not belong to the kernel of R d (f ) because X f f 1 ∂ X f 1 + Y f f 1 ∂ Y f 1 = f f 1 (X∂ X f 1 + Y ∂ Y f 1 ) = d 1 f + f f 1 f1 (1.8)
where deg( f1 ) < d 1 (by Euler's relation). Nevertheless, for all couple (G, H) ∈ ker G d (f ), Equation (1.7) shows that there exists α ∈ K such that

XG d-1 + Y H d-1 = αf d . It follows that (G, H) - α d 1 f f 1 ∂ X f 1 , f f 1 ∂ Y f 1 ∈ ker R d (f ) and therefore dim K ker R d (f ) = dim K ker G d (f ) -1.
To finish the proof, fix an integer ν > d. It is clear from the definitions that ker

G ν-1 (f ) ⊆ ker R ν (f ) ⊆ ker G ν (f ). Pick a couple (G, H) ∈ ker R ν (f ). It satisfies XG ν-1 + Y H ν-1 = 0.
Therefore, using (1.7) we deduce that

G ν-1 dX + H ν-1 dY f d = 0 that is to say that G ν-1 = H ν-1 = 0. It follows that (G, H) ∈ ker G ν-1 (f ). Corollary 6. Let f (X, Y ) ∈ K[X, Y ] of degree d such that f = f e 1 1 • • • f er r is a factorization of f where each polynomial f i is irreducible of degree d i . Then dim K ker R d (f ) = r -2 + 2 + r i=1 d i (e i -1) 2 . In particular, f (X, Y ) is irreducible if and only if dim K ker R d (f ) = 0.
Remark 7. If f is a square-free polynomial, it is not hard to check that the set

-d i f f 1 ∂ X f 1 + d 1 f f i ∂ X f i , -d i f f 1 ∂ Y f 1 + d 1 f f i ∂ Y f i , i = 2, . . . , r (1.9)
form a basis of the kernel of R d (f ). Indeed, Equation (1.8) implies that the elements of (1.9) belongs to E d . Furthermore, as already mentioned, the set (1.1) form a basis of the kernel of G d (f ) when f is square-free. It is then straightforward to check that the elements of (1.9) are linearly independent over K and then, using Corollary 6, to deduce that (1.9) form a basis of the kernel of R d (f ).

Since we will often deal with homogeneous polynomials in the rest of this paper, we need to extend Corollary 6 to the case of a homogeneous polynomial. To proceed, it is first necessary to define Ruppert's matrix in this setting

. If f (X, Y, Z) ∈ K[X, Y, Z] is a homogenous polynomial of degree d, we define R(f ) : E -→ K[X, Y, Z] 2d-3 : (G, H) → 1 Z f 2 ∂ Y G f -∂ X H f
where

E = {(G, H) ∈ K[X, Y, Z] d-1 × K[X, Y, Z] d-1 such that Z|XG + Y H}.
Observe that the division by Z in this definition is justified by Lemma 4.

Here is the main result of this section.

Theorem 8. Let f (X, Y, Z) ∈ K[X, Y, Z] homogeneous of degree d and sup- pose that f = f e 1 1 • • • f er r where each polynomial f i (X, Y, Z) is irreducible and homogeneous of degree d i . Then dim K ker R(f ) = r -2 + 2 + r i=1 d i (e i -1) 2 . In particular, f (X, Y, Z) is irreducible if and only if dim K ker R(f ) = 0. Proof. Denote f (X, Y ) = f (X, Y, 1) ∈ K[X, Y ] and consider the map R d ( f ).
We claim that the kernels of R(f ) and R d ( f ) are isomorphic K-vector spaces.

Indeed, let ( G(X, Y ), H(X, Y )) ∈ ker R d ( f ) and set G(X, Y, Z) = Z d-1 G X Z , Y Z , H(X, Y, Z) = Z d-1 H X Z , Y Z .
Multiplying by Z 2d-2 the equality

f X Z , Y Z ∂ Y G X Z , Y Z -G X Z , Y Z ∂ Y f X Z , Y Z -f X Z , Y Z ∂ X H X Z , Y Z + H X Z , Y Z ∂ X f X Z , Y Z = 0 we get f ∂ Y G -G∂ Y f -f ∂ X H + H∂ X f = ZR(f )(G, H) = 0. Moreover, since deg(X G + Y H) ≤ d -1 we deduce that Z divides XG + Y H and conclude that (G, H) belongs to the kernel of R(f ). Similarly, if (G, H) ∈ ker R(f ) then ( G, H) = (G(X, Y, 1), H(X, Y, 1)) ∈ ker R d ( f ). Therefore, we have proved that dim K ker R(f ) = dim K ker R d ( f ).
From here, if deg( f ) = d then the claimed equality follows from Corollary 6. Now, if deg( f ) < d then, by Proposition 5, dim

K ker R d ( f ) = dim K ker G d-1 ( f ). As deg( f ) < d, we can suppose that f r (X, Y, Z) = Z er , d r = 1, and then f (X, Y ) = f e 1 1 (X, Y, 1) • • • f e r-1
r-1 (X, Y, 1). Thus deg( f ) = de r and f has r -1 factors. Therefore, Proposition 3 applied to f yields the equality

dim K ker G d-1 ( f ) = (r -1) -1 + 2 + (d -1) -(d -e r ) + r-1 i=1 d i (e i -1
) 2 that gives the expected formula.

An upper bound for the total order of reducibility

In this section, given a non-composite rational function r = f /g ∈ K(X, Y ) we establish an upper bound for its total order of reducibility counting multiplicities m(f, g) (recall that if r is composite then σ(f, g) is not a finite set). It turns out that this upper bound is the same as the known upper bound for the usual total order of reducibility ρ(f, g) [START_REF] Lorenzini | Reducibility of polynomials in two variables[END_REF][START_REF] Vistoli | The number of reducible hypersurfaces in a pencil[END_REF]. Notice that we will actually prove a stronger result by considering a quantity which is bigger than m(f, g). To proceed, we first need some notations.

Throughout this section, we will assume that the algebraically closed field K has characteristic p = 0.

Given a non-composite rational function r = f /g ∈ K(X, Y ) of degree d, define the two homogeneous polynomials of degree

d in K[X, Y, Z] f ♯ (X, Y, Z) = Z d f X Z , Y Z , g ♯ (X, Y, Z) = Z d g X Z , Y Z . If (µ : λ) ∈ σ(f, g) and µf ♯ (X, Y, Z) + λg ♯ (X, Y, Z) = n(µ:λ) i=1 P e (µ:λ),i (µ:λ),i (2.1)
where each polynomial P (µ:λ),i is irreducible and homogeneous in K[X, Y, Z], then ρ(f, g) =

(µ:λ)∈P 1 K n(µ : λ) -1 and m(f, g) = (µ:λ)∈P 1 K (m(µ : λ) -1) = (µ:λ)∈P 1 K     n(µ:λ) i=1 e (µ:λ),i   -1   .
The number of multiple factors of µf ♯ (X, Y, Z) + λg ♯ (X, Y, Z), counted with multiplicity, is n(µ:λ) i=1 e (µ:λ),i -1 .

In the sequel we will actually balance each multiplicity in this sum with the degree of its corresponding factor, that is to say, we will rather consider the number

ω(µ : λ) = n(µ:λ) i=1 deg(P (µ:λ),i ) e (µ:λ),i -1 ≥ n(µ:λ) i=1 e (µ:λ),i -1 .
Consequently, we define

ω(f, g) = (µ:λ)∈P 1 K ω(µ : λ).
Before going further in the notation, let us make a digression on the interesting quantity ω(f, g) that first appears in the works of Darboux [Dar78] and Poincaré [START_REF] Poincaré | Sur l'intégration algébrique des équations différentielles du premier ordre[END_REF] on the qualitative study of first order differential equations. In particular, they knew the following result:

Lemma 9. Let r = f /g ∈ K(X, Y ) a non-composite reduced rational func- tion of degree d. Then, ω(f, g) ≤ 2d -2.
Proof. See [Jou79, Chapitre 2, Corollaire 3.5.6] for a detailed proof of this result valid with an arbitrary number of variables.

It is also interesting to emphasize how Lemma 9 implies that the cardinal of the set

γ(f, g) := (µ : λ) ∈ P 1 K such that µf ♯ + λg ♯ = P e (µ:λ) (µ:λ)
with e (µ:λ) ≥ 2 and

P (µ:λ) ∈ K[X, Y, Z] irreducible ⊂ P 1 K
that is to say of the set of geometrically irreducible but reduced fibers2 , is less or equal to 3. Indeed, Lemma 9 yields

(µ:λ)∈γ(f,g) deg(P (µ:λ) )(e (µ:λ) -1) ≤ 2d -2.
But obviously, deg(P (µ:λ) ) ≤ d 2 for all (µ : λ) ∈ γ(f, g) and, denoting by |γ(f, g)| the cardinal of γ(f, g), it follows that

d |γ(f, g)| = (µ:λ)∈γ(f,g) e (µ:λ) deg(P (µ:λ) ) ≤ 2d -2 + (µ:λ)∈γ(f,g) deg(P (µ:λ) ) ≤ 2d -2 + d 2 |γ(f, g)|.
Therefore, since d is a positive integer we have |γ(f, g)| ≤ 3. Mention that one can also be interested in fibers that are non reduced and geometrically irreducible on the affine space A 2 K , say with variables X, Y , that is to say fibers of the pencil of curves µf ♯ + λg ♯ of the form Z e∞ P e where P is an irreducible and homogeneous polynomial and e deg(P ) + e ∞ = d. Since there is at most one point (µ : λ) ∈ P 1 K such that Z divides µf ♯ + λg ♯ , we deduce from the inequality |γ(f, g)| ≤ 3 that the number of such fibers is at most 4. This property actually appears in [AHS03, General Mixed Primset Theorem, p 74].

Closing this parenthesis on the quantity ω(f, g), we finish with the notation by defining from (2.1) the quantity

θ(µ, λ) = ω(µ : λ) + 1 2 - n(µ:λ) i=1 (e (µ:λ),i -1)
which is positive since

θ(µ, λ) ≥ ω(µ : λ) + 1 2 -ω(µ : λ) = ω(µ : λ) 2 .
Finally, we set θ(f, g) =

(µ:λ)∈P 1 K θ(µ : λ).
It is important to notice that we defined θ(µ, λ) in order to have the equality

m(µ : λ) -1 + ω(µ : λ) + θ(µ : λ) = dim ker R(µf ♯ + λg ♯ ) (2.2)
according to Theorem 8.

Theorem 10. Let r = f /g ∈ K(X, Y ) a non-composite reduced rational function and set d = deg(r) = max(deg(f ), deg(g)). We have

0 ≤ ρ(f, g) ≤ m(f, g) + ω(f, g) + θ(f, g) ≤ d 2 -1.
Proof. For all (µ : λ) ∈ P 1 K , consider the linear map

R(µf ♯ + vg ♯ ) = µR(f ♯ ) + vR(g ♯ )
and its matrix M(µf

♯ + vg ♯ ) = µM(f ♯ ) + vM(g ♯ ),
where arbitrary bases for the K-vector spaces E and K[X, Y, Z] 2d-3 have been chosen. They form a pencil of matrices that has d 2 -1 columns and more rows. We define the polynomial Spect(U, V ) ∈ K[U, V ] as the greatest common divisor of all the (d 2 -1)-minors of the matrix

UM(f ♯ ) + V M(g ♯ ).
(2.3)

It is a homogeneous polynomial of degree ≤ d 2 -1, since each entry of (2.3) is a linear form in K[U, V ].
First, notice that Spect(U, V ) is nonzero. Indeed, since r = f /g is reduced and non-composite, the spectrum σ(f, g) is finite and hence there exists (µ : λ) / ∈ σ(f, g). By Theorem 8, it follows that ker M(µf ♯ + λg ♯ ) = {0} and therefore that at least one of the (d 2 -1)-minors of (2.3) is nonzero since it has to be nonzero after the specializations of U to µ and V to λ. Now, let (µ : λ) ∈ σ(f, g). By Theorem 8 dim ker M(µf ♯ + λg ♯ ) = m(µ : λ) -1 + ω(µ : λ) + θ(µ : λ) > 0.

(2.4) Therefore, (µ : λ) is a root of Spect(U, V ). Moreover, by a well-known property of characteristic polynomials, (µ : λ) is a root of Spect(U, V ) of multiplicity at least

m(µ : λ) -1 + ω(µ : λ) + θ(µ : λ).
Summing all these multiplicities over all the elements in the spectrum σ(f, g), we obtain the quantity m(f, g) + ω(f, g) + θ(f, g). It is bounded above by d 2 -1 because Spect(U, V ) is a polynomial of degree less or equal to d 2 -1.

Observe that the term m(f, g) + ω(f, g) + θ(f, g) depends quadratically on the degrees and on the multiplicities of the irreducible components of the reducible curves in the pencil µf ♯ + λg ♯ . This has to be compared with the bound d 2 -1 which depends quadratically on the total degree d of the pencil.

As mentioned earlier, the inequality ρ(f, g) ≤ d 2 -1 has been proved in [START_REF] Lorenzini | Reducibility of polynomials in two variables[END_REF][START_REF] Vistoli | The number of reducible hypersurfaces in a pencil[END_REF]. This bound is known to be reached only for d = 1, 2, 3 and several authors raised the question of the optimality of this bound for an arbitrary degree d (see for instance [AHS03, Question 1, p. 79] or [Vis93, top of p. 254]). Coming back to the total order of reducibility counting multiplicities, we do not know whether the bound d 2 -1 given in Theorem 10 is optimal. Of course, it is optimal for d = 1, 2, 3 since this is the case for the bound ρ(f, g) ≤ d 2 -1. Nevertheless, as a consequence of Theorem 10 we obtain the

Corollary 11. Let r = f /g ∈ K(X, Y ) a non-composite reduced rational function of degree d. If ρ(f, g) = d 2 -1 then ω(f, g) = 0.
In other words, if there exists a pencil of curves with total order of reducibility equal to d 2 -1 then it must have all its reducible members schemetheoretically reduced.

In the same spirit, given a polynomial f ∈ K[X, Y ] of degree d, one may ask if there exists a sharper bound for the spectrum m(f ) := m(f, 1) than d 2 -1. Indeed, as a consequence of a result of Stein [START_REF] Stein | The total reducibility order of a polynomial in two variables[END_REF] (see also [START_REF] Lorenzini | Reducibility of polynomials in two variables[END_REF] and [START_REF] Shreeram | Translates of polynomials[END_REF]), such a phenomenon appears when multiplicities of the irreducible factors are not considered; one has ρ(f ) ≤ d -1 (and this bound is reached). As pointed out to us by Dino Lorenzini, it turns out that the later inequality combined with Lemma 9 implies that m(f ) ≤ 3d -3.

The technique we used for proving Theorem 10 allows to show that

m(f ) + ω(f ) + θ(f ) ≤ d(d -1)/2 (2.5)
providing f is a non-composite polynomial. It follows from the fact that R d (1) has rank d(d -1)/2, this rank being easy to compute since the linear map R d (1) sends a couple (G, H) to the difference ∂ Y G -∂ X H. We do not know if a bound linear in the degree d holds for the quantity m(f

) + ω(f ) + θ(f ).
Although beyond the scope of this paper, we would like to mention that our approach can be directly applied for a collection of polynomials (f 1 , . . . , f r ) rather than a couple of polynomials (f, g). The problem is then to investigate the variety S of points (λ 1 , . . . , λ r ) such that the polynomial

λ 1 f ♯ 1 + • • • + λ r f ♯
r is reducible, assuming that this latter is generically irreducible. As an immediate consequence of our approach, the degree of S is less or equal to d 2 -1. Notice that the study of S has already been considered in [START_REF] Bodin | Irreducibility of hypersurfaces[END_REF] in arbitrary characteristic.

Finally, before closing this section we establish a result similar to Theorem 10 in the multivariate case. This kind of result is based on a classical use of Bertini's Theorem under the following form.

Lemma 12. Let f = |e|≤d c e 1 ,...,en X e 1 1 . . . X en n ∈ K[X 1 , . . . , X n ] set |e| = e 1 + • • • + e n and L := K(U 1 , . . . , U n , V 1 , . . . , V n , W 1 , . . . , W n )
where U 1 , . . . , U n , V 1 , . . . , V n , W 1 , . . . , W n are algebraically independent indeterminates.

Then, the bivariate polynomial

f (X, Y ) = f (U 1 X + V 1 Y + W 1 , . . . , U n X + V n Y + W n ) ∈ L[X, Y ] is irreducible in L[X, Y ] if and only if f is irreducible in K[X 1 , . . . , X n ].
Proof. See [START_REF] Kaltofen | Effective Noether irreducibility forms and applications[END_REF]lemma 7]. See also [START_REF] Jouanolou | Théorèmes de Bertini et applications[END_REF] for a complete treatment of Bertini's Theorem.

In the following theorem, the quantities m(f, g), ω(f, g) and θ(f, g) that we have defined for a rational function r = f /g in two variables are straightforwardly extended to a rational function in several variables, denoting by X 0 the homogenizing variable.

Theorem 13. Let r = f /g ∈ K(X 1 , . . . , X n ) a non-composite reduced ratio- nal function of degree d. We have m(f, g) + ω(f, g) + θ(f, g) ≤ d 2 -1. Proof. Given (µ : λ) ∈ P 1 K , Lemma 12 implies that µf ♯ + λg ♯ = n(µ:λ) i=1 P e (µ:λ),i (µ:λ),i
with P (µ:λ),i homogeneous and irreducible in K[X 0 , X 1 , . . . , X n ], if and only if

µ f ♯ + λg ♯ = n(µ:λ) i=1 P e (µ:λ),i (µ:λ),i with P(µ:λ),i homogeneous irreducible in L[X, Y, Z]. Therefore, m(f, g) = m( f , g), ω(f, g) = ω( f , g) and θ(f, g) = θ( f , g).
The claimed result then follows from Theorem 10 applied to the rational function r = f /g ∈ K(X, Y ).

Exploiting Newton's polygon

In the previous section we considered rational functions f /g with a certain fixed degree. In this section, we will refine this characterization by considering the Newton's polygons of f and g. In this way, we will give an upper bound for the total order of reducibility counting multiplicities m(f, g) that improves the one of Theorem 10 in many cases. In particular, an example for which this bound is almost reached for an arbitrary degree is presented.

To obtain this upper bound, we will follow a more basic approach than in Section 2. Indeed, instead of using Theorem 8 we will exhibit explicit elements in the kernel of a suitable Ruppert's linear map and show that they are linearly independent. This has the advantage to allow us working in non-zero characteristic, but has the disadvantage to provide a bound for the quantity m(f, g) and not m(f, g) + ω(f, g) + θ(f, g) as in Theorem 8.

Before going further into details, mention that a bound for the total order of reducibility ρ(f, g) related to the Newton's polygons of f and g is contained in the result of Vistoli [Vis93, Theorem 2.2] since this amounts to homogenize the corresponding pencil of curves over a certain toric variety which is built from the Newton's polygons of f and g. The bound provided in [START_REF] Vistoli | The number of reducible hypersurfaces in a pencil[END_REF] is then expressed in terms of invariants of this variety and of the pencil of curves that are not easy to make explicit.

Recall that p stands for the characteristic of the algebraically closed field K. We begin with some notations and preliminary materials.

Given a polynomial f (X, Y ) ∈ K[X, Y ], its support is the set S f of integer points (i, j) such that the monomial X i Y j appears in f with a non zero coefficient. The convex hull, in the real space R 2 , of S f is denoted by N(f ) and called the Newton's polygon of f . It is contained in the first quadrant of the plane R 2 . Recall that the Minkowski sum A + B of two sets A and B ∈ R 2 is the set of all elements a + b with a ∈ A and b ∈ B. We have the following classical result due to Ostrowski: let f, f 1 , . . . , f r be polynomials in

K[X, Y ] such that f = f 1 . . . f r , then N(f ) = N(f 1 ) + • • • + N(f r ). (3.1)
Now, we introduce another polygon.

Definition 14. N + (f ) is the smallest convex set that contains N(f ) and the origin, and that is bordered by edges having non-positive slopes (horizontal and vertical edges are hence allowed).

Remark: An equivalent definition of N + (f ) is the following: For any integer point (i, j) ∈ N × N we define its boxed Newton Polygon B + (i, j) to be all integer points in the rectangle with opposite corners (0, 0), (i, j). Then N + (f ) is the convex hull of all B + (i, j) with (i, j) in the support of f . As N(f ), N + (f ) is also contained in the first quadrant of the plane R 2 . For example, N + (XY ) is the square with vertices (0, 0), (0, 1), (1, 1) and (1, 0) and N + (X + Y + X 2 Y 2 ) is the polygon with vertices (0, 0), (1, 0), (2, 2) and (0, 2). Moreover, it will be useful in the sequel to notice that if f, g ∈ K[X, Y ] are such that g divides f then clearly N(f /g) ⊂ N + (f ).

The notion of total degree of a polynomial f ∈ K[X, Y ] can be refined in many ways in the sparse context. For instance

, if f (X, Y ) = i,j f i,j X i Y j in K[X, Y ],
given a couple (a, b) ∈ Z 2 the (a, b)-weighted degree, or simply weighted degree, of f is defined by

d a,b (f ) = max (i,j)∈N 2 {ai + bj | f i,j = 0}.
Thus, the total degree of a polynomial f is nothing but d 1,1 (f ) and the degree of f with respect to the variable X, resp. Y , corresponds to deg

1,0 (f ), resp. deg 0,1 (f ). If E is an edge of a given convex set N , denote by a E X + b E Y = c E one of its integer equation. Then, it is clear that d a E ,b E (m) = d a E ,b E (n) if m, n ∈ E, and that d a E ,b E (m) = d a E ,b E (n) if m ∈ E, n ∈ E.
In what follows we will use this remark for particular edges that we will call good edges.

Definition 15. Suppose given a convex set N in the first quadrant of the plane. An edge E of N is called a good edge if the two following conditions hold:

• there exists (a E , b E ) ∈ N 2 \ (0, 0) and c E ∈ N such that a E X + b E Y = c E is an equation of E, • if n ∈ N , n ∈ E and m ∈ E then 0 ≤ d a E ,b E (n) < d a E ,b E (m).
Remarks: A good edge is a vertical edge or an edge with a non positive slope such that the convex set is below or to the left of this edge. A good edge does not always exist. Consider for example the triangle formed by (1, 0), (2, 2) and (0, 1).

We are now ready to state the main result of this section.

Theorem 16. Let N be a convex set in R 2 . Denote by p its number of integral points and by p X , resp. p Y , the number of points in N lying on the X-axis, resp. Y -axis. If N possesses a good edge E, then p E stands for the number of integral points in N lying on E; otherwise set p E = 0.

Suppose given a non-composite reduced rational function r = f /g ∈ K(X, Y ) of degree d, assume that N ⊆ N (1 + X + Y ) d and that the characteristic p of K is such that p = 0 or p > d(d -1).

• If N(f ) ⊂ N and N(g) ⊂ N then ρ(f, g) ≤ 2p -p X -p Y -p E + κ.
(3.2)

• If N + (f ) ⊂ N and N + (g) ⊂ N then m(f, g) ≤ 2p -p X -p Y -p E + κ. (3.3) • If N(f ) ⊂ N , N(g) ⊂ N and (-g(0, 0) : f (0, 0)) ∈ σ(f, g) then m(f, g) ≤ 2p -p X -p Y -p E + κ. (3.4)
where κ = max(e ∞ -1, 0) with e ∞ the multiplicity (possibly 0) of the line at infinity {Z = 0} in the pencil of curves µf ♯ + λg ♯ .

Before proceeding with the proof of this theorem, we comment it and illustrate it through three examples. First, consider the dense case which corresponds to the situation studied in Section 2. Here, we have

N = N(f ) = N + (f ) = N(g) = N + (g) = N((1 + X + Y ) d )
and therefore p = (d + 2)(d + 1)/2, p X = p Y = p E = d + 1, the good edge E being the diagonal joining the vertices (0, d) and (d, 0). Moreover, since we are in the dense case, a linear change of coordinates leaves invariant N(f ), N(g), ρ(f, g) and m(f, g). Thus we can assume that κ = 0, that is to say that the line at infinity is not a factor of any member of pencil of curves µf ♯ + λg ♯ . It follows that we obtain the expected bounds ρ(f, g) ≤ d 2 -1 and m(f, g) ≤ d 2 -1.

Our next example, taken from [Lor93, Remark 5], is to show that the bound (3.3) is almost reached for an arbitrary degree d. Indeed, set

f (X, Y ) = X(X + 1) • • • (X + d -2)Y + X, g(X, Y ) = 1.
It is not hard to check that r = f /g is non-composite (see [START_REF] Lorenzini | Reducibility of polynomials in two variables[END_REF]Remark 5]) and that m(f, g) ≥ 2d-2. Now, defining N = N + (f +g) which is a rectangle with vertices (0, 0), (d -1, 0), (d -1, 1) and (0, 1), we have p = 2d, p X = d, p Y = 2. Furthermore, we choose the horizontal good edge corresponding to a ε = 0, b ε = 1, c ε = 1 and obtain p E = d. Since κ = d -1, the bound given in (3.3) is equal to 2d -1 and we obtain 2d -2 ≤ m(f, g) ≤ 2d -1. Finally, our last example is to justify why we chose to state (3.4) despite the technical hypothesis requiring that the projective point (-g(0, 0) : f (0, 0)) does not belong to the spectrum of f /g. Consider the example

f (X, Y ) = a 0 + a 1 XY + a 2 X 2 Y 2 + a 3 X 3 Y 2 + a 4 X 2 Y 3 g(X, Y ) = b 0 + b 1 XY + b 2 X 2 Y 2 + b 3 X 3 Y 2 + b 4 X 2 Y 3
where the coefficients a i 's and b j 's are all assumed to be nonzero and such that the above mentioned hypothesis is satisfied. We have N(g) = N(f ) and it is clear that N(f ) N + (f ). Taking N = N + (f ) and defining the good edge E as, for instance, the top horizontal edge of N + (f ), we get p = 15, p x = 4, p y = 4, p E = 3, d = 5. Therefore, (3.3) yields m(f, g) ≤ 2pp xp yp E = 19 < d 2 -1 = 24.

(3.5) Now, choosing N = N(f ) there is only one choice for the good edge E and we obtain p = 5, p x = 1, p y = 1, p E = 2, d = 5. Consequently, (3.4) gives m(f, g) ≤ 10, to be compared with (3.5). The following picture shows the different polytopes involved in this example. 0 -6 Y X s s s s s q q q q q q q q q q q q q q q q q p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p @ @

We now turn to the proof of Theorem 16. We begin with the following preliminary definition and result.

Definition 17. Let f (X, Y ) ∈ K[X, Y ], let (a E , b E ) ∈ Z 2 and let f = f e 1
1 • • • f er r be a factorization of f where each polynomial f i is irreducible. For all i = 2, . . . , r, we set

G (1) i = -d a E ,b E (f i ) f f 1 ∂ X f 1 + d a E ,b E (f 1 ) f f i ∂ X f i H (1) i = -d a E ,b E (f i ) f f 1 ∂ Y f 1 + d a E ,b E (f 1 ) f f i ∂ Y f i
and for all i = 1, . . . , r and k = 2, . . . , e i we set

G (k) i = f f k i ∂ X f i , H (k) i = f f k i ∂ Y f i . Proposition 18. Let f (X, Y ) ∈ K[X, Y ] be a polynomial of degree d, let f = f e 1 1 • • • f er
r be a factorization of f where each polynomial f i is irreducible and assume that the characteristic p of K is such that p = 0 or p > d.

(i) For all i = 2, . . . , r and all (a

E , b E ) ∈ Z 2 , N XG (1) i ⊂ N(f ) and N Y H (1) i ⊂ N(f ).
(ii) For all i = 1, . . . r and all k = 2, . . . , e i , N XG

(k) i ⊂ N + (f ) and N Y H (k) i ⊂ N + (f ).
Furthermore, if f (0, 0) = 0 then for all i = 1, . . . r and all k = 2, . . . , e i , N XG

(k) i ⊂ N(f ) and N Y H (k) i ⊂ N(f ). (iii) If E is a good edge of N(f ) with equation a E X + b E Y = c E , then for all i = 1, . . . r and all k = 1, . . . , e i , d a E ,b E a E XG (k) i + b E Y H (k) i ≤ d a E ,b E (f ) -1. (iv) The ( r i=1 e i ) -1 elements G (k) i , H (k) i
, i = 1, . . . , r, k = 1, . . . , e i , (i, k) = (1, 1) are K-linearly independent.

Proof. We begin with the proof of (i) and (ii).

By Ostrowski's formula, see

(3.1), N X f f k i ∂ X f i = N f f k i +N(X∂ X f i ), and since N(X∂ X f i ) ⊂ N(f i ) we get N X f f k i ∂ X f i ⊂ N f f k i + N(f i ) = N f f k-1 i ⊂ N + (f ). (3.6) If k = 1, Equation (3.6), shows that N X f f i ∂ X f i ⊂ N(f ) for all i = 1, . . . , r and hence that N(XG (1)
i ) ⊂ N(f ) for all i = 2, . . . , r. If k > 1 then, by (3.1) we have

N(f ) = N f f k-1 i + N(f k-1 i ). So, if f (0, 0) = 0 then f k-1 i (0, 0) = 0 and hence (0, 0) ∈ N(f k-1 i ). It follows that N f f k-1 i ⊂ N(f ), that proves that N XG (k) i
⊂ N(f ) for all i = 1, . . . r and all k = 2, . . . , e i .

We can proceed similarly with the polynomials h (k) i and conclude this way the proof of (i) and (ii).

We turn to the proof of (iii). If k > 1, then by Definition 15 we clearly have

d a E ,b E (a E XG (k) i + b E Y H (k) i ) ≤ d a E ,b E (f ) -1. If k = 1, denote by f top the homogeneous part of f with maximum weighted degree d a E ,b E (f ). Then, Euler's relation a E X∂ X f top + b E Y ∂ Y f top = d a E ,b E (f )f top allows to conclude.
It remains to prove (iv). For all i = 1, . . . , r and k = 1, . . . , e i , set

g (k) i = f f k i ∂ X f i , h (k)
i = f f k i ∂ Y f i . (3.7)
We will prove that the r i=1 e i elements (g

(k) i , h (k)
i ) defined by (3.7) are Klinearly independent and then the desired result will follow directly. So suppose that there exists a collection of λ

i,k ∈ K such that r i=1 e i k=1 λ i,k g (k) i = 0, r i=1 e i k=1 λ i,k h (k) i = 0
and choose an integer j ∈ {1, . . . , r}. We have

r i=1 e i k=1 λ i,k g (k) i g (e j ) j = 0 and since f divides g (k) i g (e j ) j
for all i = j, we deduce that

e j k=1 λ j,k g (k) j g (e j ) j = 0 mod f. Equivalently, there exists a polynomial T ∈ K[X, Y ] such that f T = λ j,1 f f j f f e j j (∂ X f j ) 2 + • • • + λ j,e j f f e j j f f e j j (∂ X f j ) 2
that is to say, such that

f e j j T = (∂ X f j ) 2 f f e j j (λ j,1 f e j -1 j + • • • + λ j,e j ).
Therefore, either f j divides ∂ X f j or either f e j j divides (λ j,1 f e j -1 j

+ • • • + λ j,e j ). A similar reasoning by replacing g (k) i with h (k) i shows that either f j divides ∂ Y f j or either f e j j divides (λ j,1 f e j -1 j + • • • + λ j,e j ). Now, since deg f j ≤ d < p and f is not a constant polynomial, (∂ X f j , ∂ Y f j ) = (
0, 0) and hence f j cannot divide one of its partial derivative. It follows that necessarily f e j j divides λ j,1 f e j -1 j

+ • • • + λ j,e j . But since deg(f e j j ) > deg(λ j,1 f e j -1 j + • • • + λ j,e j ) we must have λ j,1 f e j -1 j + • • • + λ j,e j = 0. Furthermore deg f e j j > deg f e j -1 j > • • • > deg f j ,
so that λ j,k = 0 for all j, k. This proves that the (g

(k) i , h (k) i ), hence the (G (k) i , H (k)
i ), are linearly independent over K. Observe that the technical hypothesis f (0, 0) = 0 in (ii) is necessary. Indeed, if f = X 3 (Y 2 +X +1), f 1 = X and e 1 = 3 then XG e 1 1 = X(Y 2 +X +1) and its Newton's polygon is not included in the Newton's polygon of f . Proof of Theorem 16. We will proceed similarly to what we did to prove Theorem 10. Given a polynomial h(X, Y ) ∈ K[X, Y ], we define the K-linear map

SR(h) : E N -→ K[X, Y ] (G, H) -→ h 2 ∂ Y G h -∂ X H h
where

E N = {(G, H) ∈ K[X, Y ]×K[X, Y ] such that N(XG) ⊂ N , N(Y H) ⊂ N , d a E ,b E (a E XG + b E Y H) ≤ d a E ,b E (h) -1}.
Notice that the last condition in the above definition has to be forgotten if there does not exist a good edge E. Also, observe that all the elements introduced in Definition 17 belong to the kernel of SR(f ), keeping the notation of loc. cit.

Since we assumed that N ⊆ N (1 + X + Y ) d , E N is a subvector space of the K-vector space K[X, Y ] ≤d-1 ×K[X, Y ] ≤d-1 , so that SR(h) is a restriction of the K-linear map G d (h) introduced in Section 1, to E N . Let us compute the dimension of this latter vector space. Pick (G, H) ∈ E N . We have N(XG) ⊂ N , hence XG has at most pp x nonzero coefficients and so does G, because XG and G have the same number of nonzero coefficients. Similarly, we get that H has pp y nonzero coefficients. The condition

d a E ,b E (a E XG + b E Y H) ≤ d a E ,b E (f ) -1
means that the weighted homogeneous part of highest degree of G and H are related. That is to say, we can write the homogeneous part of weighted degree d a E ,b E (f )-1 of H in terms of the homogeneous part of weighted degree d a E ,b E (f ) -1 of G. Consequently, we obtain dim K E N = 2pp xp yp E .

(3.8)

Now, for all (µ : λ) ∈ P 1 K , consider the linear map SR(µf + λg) = µSR(f ) + λSR(g) and, choosing bases for E N and K[X, Y ] ≤2d-2 , the corresponding matrix M(µf + λg) = µM(f ) + λM(g).

They form a pencil of matrices that has dim K E N columns and more rows. Then, define the polynomial Spect(U, V ) ∈ K[U, V ] as the greatest common divisor of all the minors of size dim K E N of the matrix

UM(f ) + V M(g).
It is a homogeneous polynomial of degree lower or equal to dim K E N .

The polynomial Spect(U, V ) is nonzero for the same reason as the one given in Theorem 10, since the linear maps SR(-) are restrictions of the linear maps G d (-). The fact that this property remains valid if p > d(d -1) is a consequence of [Gao03, Lemma 2.4] where Gao studied the property of the linear map G d (-) for square-free polynomials in positive characteristic. Now, let (µ : λ) ∈ σ(f, g). Then dim ker M(µf + λg) > 0 and (µ : λ) is root of Spect(U, V ) of multiplicity, say η. Since η ≥ dim ker M(µf + λg), Proposition 18 gives some lower bounds for η that allow to conclude the proof of this theorem as follows.

• If N(f ), N(g) ⊂ N , then Proposition 18,(i),(iii) and (iv) implies that η ≥ n(µ : λ) -1 if deg(µf + λg) = d or η + κ ≥ n(µ : λ) -1 otherwise. Summing over all the elements in σ(f, g) we deduce the bound (3.2).

• If N(f ) + , N(g) + ⊂ N , then Proposition 18,(i), (ii) -first part, (iii) and (iv) implies that η ≥ m(µ : λ) -1 if deg(µf + λg) = d or η + κ ≥ m(µ : λ) -1 otherwise. Summing over all the elements in σ(f, g) we deduce the bound (3.3).

• If N(f ), N(g) ⊂ N and (-g(0, 0) : f (0, 0)) / ∈ σ(f, g), then Proposition 18,(i), (ii) -second part, (iii) and (iv) implies that η ≥ m(µ : λ) -1 if deg(µf + λg) = d or η + κ ≥ m(µ : λ) -1 otherwise. Summing over all the elements in σ(f, g) we deduce the bound (3.4).

Notice that we used the fact that the polynomial Spect(U, V ) is of degree lower or equal to dim K E N .

To finish, point out that we can not state a result similar to Theorem 16 in terms of the Newton's polygon of f and g ∈ K[X 1 , . . . , X n ] following the above strategy because we are not able to preserve the sparsity of the polynomials through Bertini's Theorem.

Lemma 4 .

 4 Let f ∈ K[X, Y ] of degree d. For all positive integer ν and all couple (G, H) ∈ E ν , the polynomial

This terminology is taken from[START_REF] Stein | The total reducibility order of a polynomial in two variables[END_REF].

Notice that these fibers appear in the work of Poincaré[START_REF] Poincaré | Sur l'intégration algébrique des équations différentielles du premier ordre[END_REF] as the critical remarkable values of fifth type.