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User:D.Lazard/Bézout theorem

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Introduction

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Bézout's theorem states: if polynomials in variables have a finite number of common zeros, including the zeros at infinity, then the number of these zeros is the product of the degrees of the polynomials, if the zeros, including those at infinity are counted with their multiplicities. Despite its apparently simple statement, this theorem needed almost a century for being completely proved. The main difficulty was to give an accurate definition of multiplicities, and this requires some machiney of commutative algebra. Most proofs of this theorem proceed by recurrence on the number of polynomials, and use the concept of degree of a polynomial ideal. These proofs obtain the theorem as a corollary of if a homogeneous polynomial of degree is not a zero divisor modulo a homogeneous ideal of degree then the degree of the ideal is

If some hypotheses are relaxed, such as counting multiplicities or working with homogeneous polynomials, one gets only inequalities, commonly called Bézout inequalities. For example, if polynomials in variables have a finite number of common zeros, then the number of these zeros, counted with their multiplicities is at most the product of the largest degrees of the polynomials. Surprisingly, this Bézout inequality is not a corollary of Bézout's theorem, and seems to have not been proved before 1983 (cite MW).

All these results require the definition of the degree of a polynomial ideal. Many definitions have been given, either in terms of algebraic geometry or in terms of commutative algebra. Most, but not all, deal with homogeneous ideals, and it is rather difficult to compare them. One of the objectives of this article is to give a general definition in terms of elementary commutative algebra and to prove that all other definitions are special cases. In fact, we give two different definitions that are equal for equi-dimensional ideals, but not in general.

Basically, the degree of an algebraic variety is the number of points of its intersection with a generic linear variety of a convenient dimension. This definition is not algebraic but can easily be translated into an algebraic one. However the resulting definition is not intrinsic, involving auxiliary generic polynomials, which makes proofs unnecessarily complicated. Therefore, we use the definition through Hilbert series, which provides a simpler presentation of the theory. In the case of homogeneous ideals, this approach is not new, although is seems unknown by many specialists of algebraic geometry, and we do not know any published presentation of it. Personally, we have learnt it from an early version by Carlo Traverso alone of [Robbiano-Traverso]. In the first part of our article, we extend this approach to a unified presentation for the homogeneous and the non-homogeneous cases.

In the case of non-equidimensional ideals, this definition of the degree does not depend on the components of lower dimension and the embedded components. For taking the isolated components into account, Masser and Wüstholz have introduced another notion of degree, which is the sum of the degrees of the isolated components of any dimemsion. They have proved that, for an ideal generated by polynomials of degrees the degree of the intersection of the isolated components of height is at most So, if the height of the ideal is the Masser–Wüstholz degree is at most

The main new result of this article is that bounds not only the degree of the intersection of the isolated components of height but also the sum of the degrees of all isolated primary components of height at most Moreover, although Masser–Wüstholz proof involves analytic geometry, our proof is purely algebraic.

Gradation and Hilbert series

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In this article, we work with the polynomial ring in indeterminates over a fiels $K$. This ring is a graded by the degree, and this gradation extends to homogeneous ideals and quotients by such ideals. In all these graded modules, the homogeneous part of degree $d$ is a finite-dimensional $K$-vector space for every integer $d$.

Definition —  If

is a graded object, where $A_d$ is a finite-dimensional $K$-vector space for evry $d$, then the Hilbert series of $A$ is the formal power series

The main property of Hilbert series is to be additive under exact sequence,

Proposition —  If

is an exact sequence of homomorphisms of graded modules, then

Proof. Results immediately from the similar formula for dimension of vector spaces.

Proposition —  If $A$ is a graded algebra, and $x$ a homogeneous element of degree $d$ that is not a zero divisor in $A$, then

Proof. Results immediately from the exact sequence

where $A^{[d]}$ is $A$ with its gradation shifted by $d$, which multiply its Hilbert series by $t^d$.

Corollary —  The Hilbert series of is

Proof. Proof by recurrence on $n$, using the preceding proposition with The recurrence starts with the trivial result

Corollary —  If is a regular sequence of homogeneous elements in of respective degrees (this means that each is not a zero divisor modulo the preceding s) then

Theorem —  The Hilbert series of any graded module $M$ (without elements of negative degree) can be summed as

where is a polynomial not divisible by , and is a nonnegative integer not larger than .

Proof. If is a free graded module whose basis elements have degrees , then the form of , and the addivity of Hilbert series under direct sums show that . In the general case, this results from Hilbert's syzygy theorem, which asserts that every graded module has a free resolution of length at most . The additivity of Hilbert series under exact sequences inplies thus that every Hilbert series is an alternating sum of Hilbert series of free modules.

Definition or theorem —  If is a homogeneous ideal in , and where is a polynomial not divisible by , then is the dimension of , and is the degree of

In this article, we take the preceding statement as a definition. However, we will prove, below, that this definition is equivalent with the usual ones. In particular, the dimension of is the Krull dimension of .

General settings

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  • n: number of variables
  • k, ground field, supposed to be infinite for some results
  • polynomials of degrees such that (this can always be achieved by permuting the polynomials)
  • Degree of an ideal, denoted as defined in Hilbert series and Hilbert polynomial
  • Strong degree of an ideal, denoted the sum of the degrees of the isolated primary components

Lemmas

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Lemma 1 (Regularity lemma) — By replacing by for sufficiently generic scalars one can suppose that, for every i, and every associated prime p of , if then for every (must be restated for including the homogeneous case)

Proof

Standard, see, for example, [Kaplanski, Commutative rings]. Idea: if some

Degree of an affine ideal

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When considering non-homogeneous polynomials, that is the filtration by the degree that is considered, rather than the gradation.

Deg vs deg

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By definition of Deg, one has Deg I = deg I for every primary ideal I.

Lemma — Let I be a strictly equi-dimensional ideal of dimension d, that is an ideal whose all associated primes have dimension d. Let J be an ideal of dimension at most d that has no associated prime in common with I. Then, if dim J = d, then

and if dim J < d, then

Proof: The exact sequence

implies that

As all quotient rings have dimension at most d, the Hilbert series can be written and one has P(1) = 0 if the dimension is smaller than d. Otherwise P(1) is the degree of the ideal.

One has dim (I + J) < d. In fact, every minimal prime p of I + J must contain a minimal prime p1 of I and a minimal prime p2 of J. If the dimension of p would be d, this would imply that both p1 and p2 have dimension d, and therefore that they are equal to p, which is excluded by the hypotheses.

The result follows immediately by removing the denominators in the above equation between Hulbert series, and substituting t for 1 in the resulting equality of polynomials.

Proposition — If I is an ideal of dimension d, then deg I is the sum of the degrees of the (isolated) primary components of I of dimension d (by definition, deg q = Deg q for a primary ideal q)

Proof: Let be the primary components of dimension d, and be the intersection of all other primary components. The dimension of is thus smaller than d. The result is obtained by applying recursively the above lemma to and for        

Corollary — For every ideal I, one has deg I ≤ Deg I, and the equality is true if and only if all minimal primes have the same dimension.

Corollary — Let be the intersection of the isolated primary components of an ideakl Then and

Strong Bézout inequality

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The strong Bézout inequality bounds the MW-degree of an ideal in terms of the degrees of its generators. It is

Theorem (Strong Bézout inequality) — Let be nonzero polynomials of respective degrees which are sorted in order that If the height of the ideal is then

Moreover for every the sum of the degrees of the isolated primary components of of height at most is at most

Technical lemmas

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Lemma 5.1 — If and are two ideals, and is a polynomial, then and have the same minimal primes.

Moreover, if there is no inclusion between the associated primes of and (that is, if no associated prime of one ideal contains the other ideal), then and a minimal primary decomposition of is the union of minimal primary resolutions of and

Proof: One has

As the inteersection and the product of two ideals have the same radical, and have the same radical and thus the same minimal primes.

The hypothesis of the last assertion implies that all above inclusions are equalities, and the result follows thus immediatly.

Thus, it remains to prove that if two ideals and satisfy the hypothesis of the last assertion, then In fact, no associated prime of either ideal can contain So, there is an element that does not belong to either associated prime. If then If it follows that The hypothesis implies thus that the inverse image in of a primary decomposition of this localized ideal is a primary decomposition of both and which are thus equal.

Lemma 5.2 —  Let be a minimal prime of an ideal There is an element that belongs to all other associated primes of If is the multiplicative set of the powers of then the -primary component of is

Proof: As is a minimal prime, each other associated prime contains an element that is not in The product of these elements is the desired element The last assertion results of the classical property of stability of primary decompositions under localization.

Lemma 5.3 —  Let be two ideals that have a common minimal prime Let and be their respective -primary components. Then and

Proof: The first assertion results from Lemma 5.2, since localization and intersection preserve inclusion.

By definition of Hilbert series, the inclusion implies that each coefficient of the Hilbert series is not smaller than the corresponding coefficient of Thus for Writing these series as a rational fractions with denominator one see that the same inequality applies to the numerators, and thus to their limits when which are the degrees of the ideals.

Lemma 5.4 —  Let be an ideal, and be a polynomial. If is a minimal prime of such that then is a minimal prime of Moreover, if and are the -primary components of and respectively, then

Proof: By hypothesis, Any minimal prime of contains and thus some minimal prime of Therefore, cannot be strictly included in that is, is a minimal prime of Then, the inequality on the degrees follows directly from Lemma 5.3.

Recursion

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Input: of degrees satisfying the regularity condition, that is, for and every associated prime p of if , then for all

If the hypothesis can be achieved by adding to a sufficiently generic linear combination of In the homogeneous case, the coefficients of in the linear combination must be a homogeneous polynomial of degrees

Let for We have first to study the minimal primes of By Krull's height theorem, such the height of such a minimal prime is at most

Lemma 6.1 — If if a minimal prime of of height then is a minimal prime of and for

Proof: Let us choose recursively, for a minimal prime of that is contained in As the height of is less than these minimal primes cannot be all distinct. Thus, let be the lowest index such This implies that and, by regularity hypothesis, for Therefore Finally, since the height of is at most and it is at least , as is a strictly increasing sequence of primes.

Lemma 6.2 — Let be a minimal prime of whose height is less than It is also a minimal prime of Let and be the corresponding primary components of and Then

Proof: The first assertion results immediately from lemma 6.1, and the second one is a special case of lemma 5.3.

Corollary 6.3 — If then

Now, we have to study, for the isolated primary components of whose height is (the case being trivial). So, let such that has a minimal prime of height Let be a polynomial that does not belong to any minimal prime of height of but belongs to all other associated primes of Let the multiplicative set generated by

The ideal is the intersection of the isolated primary components of height of We will prove the following lemma by recursion on

Lemma 6.4 —  For every the sequence is a regular sequence in and the ideal is unmixed of height (that is, all its primary components have height

Moreover, is the intersection of some isolated primary components of height of whose associated prime does not contain

Proof: The cases and are trivial. So we suppose that the assertions are true for some and we prove them for

The definition of as the inverse image of a localization implies that is the intersection of some primary components of and that the minimal primes of are also minimal primes of None of these minimal primes can contain In fact, if such a prime would contains it would contain by the regularity condition, and thus it would be a minimal prime of of height this is impossible since has been chosen for having a non-empty intersection with such a prime. As is unmixed, we can deduce that is not a zero divisor modulo and, using the recurrence hypothesis, that is a regular sequence in

This shows that the primary components of are primary components of that they have height and that their associated primes do not contain

Lemma 6.5 —  Let the intersection of all primary components of height of and be the intersection of the isolated primary components of height of whose associated prime do not contain Then

Proof: Using previous notations, and the last assertion of lemma 6.4, we have and the two ideals have the same height. Thus

In the preceding lemma, we have proved that is not a zero divisor modulo So,

It results from the proof ofth epreceding lemma that and that these ideals have the same height. So

The result follows immediately, by combining these inequalities.

End of the proof

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Given an ideal let us denote by the sum of the degrees of its isolated primary components of height at most The strong Bézout inequality results immediately, by recurrence on from the following result.

Proposition —  Using notations of preceding section, one has for every and every

If and then

Otherwise, that is, if or

(By Krull's height theorem, one has always )

Proof: The second inequality is obtained by summing the inequalities of lemma 6.2 over the minimal primes of whose height is The first inequality is obtained by adding to the inequality of lemma 6.5 the inequalities of lemma 6.2, relaxed to This gives the result since no minimal prime involved in lemma 6.2 is a minimal primes of the ideal of lemma 6.5.