User:D.Lazard/Bézout theorem

Introduction

Bézout's theorem states: if $n$ polynomials in $n$ 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 $f$ of degree $d$ is not a zero divisor modulo a homogeneous ideal $I$ of degree $D,$ then the degree of the ideal $I+\langle f\rangle$ is $dD.$

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 $m$ polynomials in $n$ 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 $n$ 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 $d_{2}\geq \dots \geq d_{k}\geq d_{1},$ the degree of the intersection of the isolated components of height $h$ is at most $d_{1}d_{2}\dots d_{h}.$ So, if the height of the ideal is $h,$ the Masser–Wüstholz degree is at most $hd_{1}d_{2}\dots d_{h}.$

The main new result of this article is that $d_{1}d_{2}\dots d_{h}$ bounds not only the degree of the intersection of the isolated components of height $h,$ but also the sum of the degrees of all isolated primary components of height at most $h.$ Moreover, although Masser–Wüstholz proof involves analytic geometry, our proof is purely algebraic.

Gradation and Hilbert series

In this article, we work with the polynomial ring $R_{n}=K[X_{1},\dots ,X_{n}]$ in $n$ 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

$A=\bigoplus _{d\geq 0}A_{d}$

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

$HS_{A}(t)=\sum _{d=0}^{\infty }t^{d}\dim _{K}(A_{d}).$

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

Proposition —  If

$0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0$

is an exact sequence of homomorphisms of graded modules, then

$HS_{B}(t)=HS_{A}(t)+HS_{C}(t)$

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 $HS_{A/\langle f\rangle }(t)=(1-t^{d})HS_{A}(t).$

Proof. Results immediately from the exact sequence

$0\longrightarrow A^{[d]}\longrightarrow A\longrightarrow A/\langle f\rangle \longrightarrow 0,$

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

Corollary —  The Hilbert series of $R_{n}=K[x_{1},\dots ,x_{n}]$ is

$HS_{R_{n}}(t)={\frac {1}{(1-t)^{n}}}.$

Proof. Proof by recurrence on $n$, using the preceding proposition with $f=x_{n}.$ The recurrence starts with the trivial result $HS_{R_{0}}(t)=HS_{K}(t)=1.$

Corollary —  If $f_{1},\dots ,f_{k}$ is a regular sequence of homogeneous elements in $R_{n},$ of respective degrees $d_{1},\dots ,d_{k}$ (this means that each $f_{i}$ is not a zero divisor modulo the preceding $f_{j}$s) then

$HS_{R_{n}/\langle f_{1},\dots ,f_{k}\rangle }(t)={\frac {\prod (1-t^{d_{i}})}{1-t^{k}}}.$

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

$HS_{M}(t)={\frac {P(t)}{(1-t)^{d}}},$

where $P(t)$ is a polynomial not divisible by $1-t$, and $d$ is a nonnegative integer not larger than $n$.

Proof. If $M$ is a free graded module whose basis elements have degrees $e_{1},\dots ,e_{h}$, then the form of $HS_{R_{n}}(t)$, and the addivity of Hilbert series under direct sums show that $HS_{M}(t)={\frac {t^{e_{1}}+\dots +t^{e_{h}}}{(1-t)^{n}}}$. In the general case, this results from Hilbert's syzygy theorem, which asserts that every graded $R_{n}$ module has a free resolution of length at most $n$. The additivity of Hilbert series under exact sequences inplies thus that every Hilbert series is an alternating sum of Hilbert series of free modules. $\blacksquare$

Definition or theorem —  If $I$ is a homogeneous ideal in $R_{n}$, and $HS_{M}(t)={\frac {P(t)}{(1-t)^{d}}},$ where $P(t)$ is a polynomial not divisible by $1-t$, then $n$ is the dimension of $I$, and $P(1)$ is the degree of $I.$

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 $I$ is the Krull dimension of $R_{n}/I$.

General settings

• n: number of variables
• k, ground field, supposed to be infinite for some results
• $R=k[x_{1},\dots ,x_{n}]$
• $f_{1},\dots ,f_{k}:$ polynomials of degrees $d_{1},\dots ,d_{k},$ such that $d_{1}\leq d_{k}\leq \dots \leq d_{2}$ (this can always be achieved by permuting the polynomials)
• Degree of an ideal, denoted $\deg ,$ as defined in Hilbert series and Hilbert polynomial
• Strong degree of an ideal, denoted $\operatorname {Deg} ,$ the sum of the degrees of the isolated primary components

Lemmas

Lemma 1 (Regularity lemma) — By replacing $f_{i}$ by $f_{i}+c_{i,i+1}f_{i+1}+\dots +c_{i,k}f_{k}$ for sufficiently generic scalars $c_{i,j},$ one can suppose that, for every i, and every associated prime p of $\langle f_{1},\dots ,f_{i}\rangle$, if $f_{i+1}\in p$ then $f_{j}\in p$ for every $j>i.$ (must be restated for including the homogeneous case)

Proof

Standard, see, for example, [Kaplanski, Commutative rings]. Idea: if some $f_{j}\not \in p\dots$

Degree of an affine ideal

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

Deg vs deg

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

$\deg(I\cap J)=\deg I+\deg J,$

and if dim J < d, then

$\deg(I\cap J)=\deg I.$

Proof: The exact sequence

$0\;\rightarrow \;R/(I\cap J)\;\rightarrow \;R/I\oplus R/J\;\rightarrow \;R/(I+J)\;\rightarrow \;0$

implies that

$HS_{R/I}(t)+HS_{R/I}(t)=HS_{I\cap J}(t)+HS_{I+J}(t).$

As all quotient rings have dimension at most d, the Hilbert series can be written ${\frac {P(t)}{(1-t)^{d}}},$ 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 $(1-t)^{d}$ 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 $q_{1},\dots ,q_{k}$ be the primary components of dimension d, and $q_{k+1}$ be the intersection of all other primary components. The dimension of $q_{k+1}$ is thus smaller than d. The result is obtained by applying recursively the above lemma to $q_{1}\cap q_{2}\cap \dots \cap q_{i}$ and $q_{i+1}$ for $i=1,\dots ,k.$        

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 $J$ be the intersection of the isolated primary components of an ideakl $I.$ Then $\deg J=\deg I,$ and $\operatorname {Deg} I=\operatorname {Deg} J.$

Strong Bézout inequality

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 $f_{1},\dots ,f_{k}$ be nonzero polynomials of respective degrees $d_{1},\dots ,d_{k},$ which are sorted in order that $\deg f_{2}\geq \deg f_{3}\geq \dots \geq \deg f_{k}\geq \deg f_{1}.$ If the height of the ideal $I=\langle f_{1},\dots ,f_{k}\rangle$ is $h,$ then

$\operatorname {Deg} I\leq d_{1}\dots d_{h}.$

Moreover for every $r\leq h,$ the sum of the degrees of the isolated primary components of $I$ of height at most $r$ is at most $d_{1}\dots d_{r}.$

Technical lemmas

Lemma 5.1 — If $I$ and $J$ are two ideals, and $f$ is a polynomial, then $(I\cap J)+\langle f\rangle$ and $(I+\langle f\rangle )\cap (J+\langle f\rangle )$ have the same minimal primes.

Moreover, if there is no inclusion between the associated primes of $I+\langle f\rangle$ and $J+\langle f\rangle ,$ (that is, if no associated prime of one ideal contains the other ideal), then $(I\cap J)+\langle f\rangle =(I+\langle f\rangle )\cap (J+\langle f\rangle ),$ and a minimal primary decomposition of $(I\cap J)+\langle f\rangle$ is the union of minimal primary resolutions of $I+\langle f\rangle$ and $J+\langle f\rangle .$

Proof: One has

$(I+\langle f\rangle )\cdot (J+\langle f\rangle )\subseteq (I\cap J)+\langle f\rangle \subseteq (I+\langle f\rangle )\cap (J+\langle f\rangle ).$

As the inteersection and the product of two ideals have the same radical, $(I\cap J)+\langle f\rangle$ and $(I+\langle f\rangle )\cap (J+\langle f\rangle )$ 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 $H$ and $K$ satisfy the hypothesis of the last assertion, then $H\cap K=H.K.$ In fact, no associated prime of either ideal can contain $H+K.$ So, there is an element $z\in H+K$ that does not belong to either associated prime. If $x\in H\cap K,$ then $zx\in H.K.$ If $S=\{1,z,z^{2},\dots ,$ it follows that $S^{-1}(H\cap K)=S^{-1}(H.K).$ The hypothesis implies thus that the inverse image in $R$ of a primary decomposition of this localized ideal is a primary decomposition of both $H\cap K$ and $H.K,$ which are thus equal.

Lemma 5.2 —  Let $p$ be a minimal prime of an ideal $I.$ There is an element $t\in R\setminus p$ that belongs to all other associated primes of $I.$ If $S$ is the multiplicative set of the powers of $t,$ then the $p$-primary component of $I$ is $R\cap S^{-1}I.$

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

Lemma 5.3 —  Let $I\subset J$ be two ideals that have a common minimal prime $p.$ Let $q$ and $q'$ be their respective $p$-primary components. Then $q\subseteq q',$ and $\deg q\geq \deg q'.$

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

By definition of Hilbert series, the inclusion $q\subseteq q'$ implies that each coefficient of the Hilbert series $HS_{R/q}(t)$ is not smaller than the corresponding coefficient of $HS_{R/q'}(t).$ Thus $HS_{R/q}(t)\geq HS_{R/q'}(t)$ for $t<1.$ Writing these series as a rational fractions with denominator $(1-t)^{\delta },$ one see that the same inequality applies to the numerators, and thus to their limits when $t\to 1,$ which are the degrees of the ideals.

Lemma 5.4 —  Let $I$ be an ideal, and $f$ be a polynomial. If $p$ is a minimal prime of $f$ such that $f\in p,$ then $p$ is a minimal prime of $I+\langle f\rangle .$ Moreover, if $q$ and $q'$ are the $p$-primary components of $I$ and $I+\langle f\rangle ,$ respectively, then $\deg q\geq \deg q'.$

Proof: By hypothesis, $I+\langle f\rangle \subseteq p.$ Any minimal prime $m$ of $I+\langle f\rangle$ contains $I,$ and thus some minimal prime of $I.$ Therefore, $m$ cannot be strictly included in $p;$ that is, $p$ is a minimal prime of $I+\langle f\rangle .$ Then, the inequality on the degrees follows directly from Lemma 5.3.

Recursion

Input: $f_{1},\dots ,f_{k}$ of degrees $d_{1},\dots ,d_{k}$ satisfying the regularity condition, that is, for $i<k,$ and every associated prime p of $\langle f_{1},\dots ,f_{i}\rangle ,$ if $f_{i+1}\in p$, then $f_{j}\in p$ for all $j\geq i.$

If $d_{2}\geq \dots \geq d_{k}\geq d_{1},$ the hypothesis can be achieved by adding to $f_{i}$ a sufficiently generic linear combination of $p_{i+1},\dots ,p_{k}.$ In the homogeneous case, the coefficients of $p_{j}$ in the linear combination must be a homogeneous polynomial of degrees $d_{i}-d_{j}.$

Let $I_{r}=\langle f_{1},\dots ,f_{r},$ for $r=1,\dots ,k.$ We have first to study the minimal primes of $I_{r}.$ By Krull's height theorem, such the height of such a minimal prime is at most $r.$

Lemma 6.1 — If $p_{r}$ if a minimal prime of $I_{r}$ of height $h<r,$ then $p_{r}$ is a minimal prime of $I_{h},$ and $f_{i}\in p_{r}$ for $i>h.$

Proof: Let us choose recursively, for $i=r,r-1,\dots ,1,$ a minimal prime $p_{i}$ of $I_{i}$ that is contained in $p_{i+1}.$ As the height of $p_{r}$ is less than $r,$ these minimal primes cannot be all distinct. Thus, let $i$ be the lowest index such $p_{i}=p_{i+1}.$ This implies that $f_{i+1}\in p_{i},$ and, by regularity hypothesis, $f_{j}\in p_{i}$ for $j>i.$ Therefore $p_{r}=p_{i}.$ Finally, $i=h,$ since the height of $p_{i}$ is at most $i,$ and it is at least $i$, as $p_{1},\dots ,p_{i}$ is a strictly increasing sequence of primes. $\blacksquare$

Lemma 6.2 — Let $p$ be a minimal prime of $I_{r}$ whose height is less than $r.$ It is also a minimal prime of $I_{r-1}.$ Let $q_{r}$ and $q_{r-1}$ be the corresponding primary components of $I_{r}$ and $I_{r-1}.$ Then $\deg q_{r}\leq q_{r-1}.$

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

Corollary 6.3 — If $h:=\operatorname {height} (I_{k})<k,$ then $\operatorname {Deg} I_{k}\leq \operatorname {Deg} I_{h}.$

Now, we have to study, for $r>1$ the isolated primary components of $I_{r}$ whose height is $r.$ (the case $r=1$ being trivial). So, let $r>1$ such that $I_{r}=\langle f_{1},\dots ,f_{r}\rangle$ has a minimal prime of height $r.$ Let $s$ be a polynomial that does not belong to any minimal prime of height $r$ of $I_{r},$ but belongs to all other associated primes of $I_{r}.$ Let $S_{r}=\{1,s,s^{2},\dots ,s^{i},\dots \}$ the multiplicative set generated by $s.$

The ideal $J_{r}=R\cap S_{r}^{-1}I_{r}$ is the intersection of the isolated primary components of height $r$ of $I_{r}.$ We will prove the following lemma by recursion on $j.$

Lemma 6.4 —  For every $j<r,$ the sequence $(f_{1},\dots ,f_{j})$ is a regular sequence in $S_{r}^{-1}R,$ and the ideal $J_{j,r}=R\cap S_{r}^{-1}I_{j}$ is unmixed of height $j$ (that is, all its primary components have height $j).$

Moreover, $J_{j,r}$ is the intersection of some isolated primary components of height $j$ of $I_{j}$ whose associated prime does not contain $f_{j+1}.$

Proof: The cases $j=0$ and $j=1$ are trivial. So we suppose that the assertions are true for some $j<r,$ and we prove them for $j+1.$

The definition of $J_{j,r}$ as the inverse image of a localization implies that $J_{j,r}$ is the intersection of some primary components of $I_{j},$ and that the minimal primes of $J_{j,r}$ are also minimal primes of $I_{j}.$ None of these minimal primes can contain $f_{j+1}.$ In fact, if such a prime would contains $f_{j+1},$ it would contain $f_{j+1},\dots ,f_{r}$ by the regularity condition, and thus it would be a minimal prime of $I_{r}$ of height $<r;$ this is impossible since $S$ has been chosen for having a non-empty intersection with such a prime. As $J_{j,r}$ is unmixed, we can deduce that $f_{j+1}$ is not a zero divisor modulo $J_{j,r},$ and, using the recurrence hypothesis, that $f_{1},\dots ,f_{j+1}$ is a regular sequence in $S^{-1}R.$

This shows that the primary components of $J_{j,r}$ are primary components of $I_{j},$ that they have height $j$ and that their associated primes do not contain $f_{j+1}.$ $\blacksquare$

Lemma 6.5 —  Let $J_{r}$ the intersection of all primary components of height $r$ of $I_{r},$ and $J'_{r-1}$ be the intersection of the isolated primary components of height $r-1$ of $I_{r-1}$ whose associated prime do not contain $f_{r}.$ Then $\deg J_{r}\leq \deg f_{r}\cdot \deg J'_{r-1}.$

Proof: Using previous notations, and the last assertion of lemma 6.4, we have $J'{r-1}\subseteq J_{r-1,r},$ and the two ideals have the same height. Thus

$\deg J_{r-1,r}\leq \deg J'_{r-1}.$

In the preceding lemma, we have proved that $f_{r}$ is not a zero divisor modulo $\deg J_{r-1,r}.$ So,

$\deg(J'_{r-1}+\langle f_{r}\rangle )\leq \deg f_{r}\cdot \deg J_{r-1,r}.$

It results from the proof ofth epreceding lemma that $J'_{r-1}+\langle f_{r}\rangle \subseteq J_{r},$ and that these ideals have the same height. So

$\deg J_{r}\leq \deg(J'_{r-1}+\langle f_{r}\rangle ).$

The result follows immediately, by combining these inequalities. $\blacksquare$

End of the proof

Given an ideal $I,$ let us denote by $\deg _{h}I$ the sum of the degrees of its isolated primary components of height at most $h.$ The strong Bézout inequality results immediately, by recurrence on $r,$ from the following result.

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

If $\operatorname {height} I_{r}=r,$ and $h\geq r$ then

$\deg _{h}I_{r}\leq d_{r}\deg _{h}I_{r-1}.$

Otherwise, that is, if $\operatorname {height} I_{r}<r,$ or $h<r$

$\deg _{h}I_{r}\leq \deg _{h}I_{r-1}.$

(By Krull's height theorem, one has always $\operatorname {height} I_{r}\leq r.$)

Proof: The second inequality is obtained by summing the inequalities of lemma 6.2 over the minimal primes of $I_{r}$ whose height is $\leq h.$ The first inequality is obtained by adding to the inequality of lemma 6.5 the inequalities of lemma 6.2, relaxed to $\deg q_{r}\leq \deg f_{r}\cdot \deg q_{r-1}.$ This gives the result since no minimal prime involved in lemma 6.2 is a minimal primes of the ideal $J'_{r-1}$ of lemma 6.5. $\blacksquare$