# Why is this theorem attributed to Serre?

Page $117$ of Atiyah, MacDonald's Introduction to Commutative Algebra text has the following theorem. Let $P(M,t)$ denote the Poincare- series of $M$.

• $\textbf{Theorem.}$ $\bigl(\mathsf{Hilbert-Serre}\bigr)$. $P(M,t)$ is a rational function in $t$ of the form $f(t)/\prod_{i=1}^{s} (1-t^{k_i})$, where $f(t) \in \mathbf{Z}[t]$.

This theorem appears in the section of the book called Hilbert-Functions (page 116), so one understands that it could have possibly been discovered by Hilbert.

• But why is the above theorem attributed to Serre? References about when Serre was credited to the above theorem would be helpful.
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Probably, it is the theorem which is attributed to Serre. – Mariano Suárez-Alvarez Jun 18 '12 at 17:01
I don't remember having proved this theorem. – Denis Serre Jun 19 '12 at 6:42
Small humour and the internet are pretty inmiscible sometimes! – Mariano Suárez-Alvarez Jun 19 '12 at 7:47
About humour: Mariano understood the humour in my post (I am Jean-Pierre's nephew, and very proud to have such a reknowned mathematician in my family), but apparently Ralph didn't. Whence Mariano's comment about whether humour is soluble in internet. – Denis Serre Jun 19 '12 at 8:04
@Denis Serre: It's only because you are posting the same kind of comment as soon as the name "Serre" occurs. Compare for the one to Igor's answer in mathoverflow.net/questions/97976/approachable-french-masters/…. – Ralph Jun 19 '12 at 8:17

I don't know what Atiyah-Macdonald were thinking, but I can tell you a theorem which is attributed to Serre (correctly, I think), and is relevant to this question.

Let $M$ be a finitely-generated graded $k[x_0, x_1, \ldots, x_n]$ module. Let $H^0(M)$, $H^1(M)$, ..., $H^n(M)$ be the local cohomology modules of $M$ with respect to the maximal ideal $\langle x_0,\ldots, x_n \rangle$. These are graded modules which satisfy the following properties:

Theorem: For all integers $d$, the function $$\dim M_d - \sum_{r=0}^n (-1)^r \dim H^r(M)_d$$ is polynomial in $d$.

Theorem (Serre vanishing) For $d$ sufficiently large, $H^r(M)_d=0$.

So Serre vanishing separates Hilbert's theorem into two parts: A certain function is a polynomial for all $d$, and that function is equal to the Hilbert function for large $d$.

I'm presenting this using the language of commutative algebra, which I don't think is the language Serre used. In sheaf cohomology language, let $\mathcal{M}$ be the sheaf on $\mathbb{P}^{n-1}$ corresponding to $M$ and let $\mathcal{H}^r(M) = \bigoplus_{d=-\infty}^{\infty} H^r(\mathbb{P}^{n-1}, \mathcal{M} \otimes \mathcal{O}(-d))$. Then the relation between sheaf cohomology and local cohomology is that $$\mathcal{H}^r(M) \cong H^{r+1}(M)$$ for $r \geq 1$ and there is a short exact sequence $$0 \to H^0(M) \to M \to \mathcal{H}^0(M) \to H^1(M) \to 0.$$ In this language, Serre vanishing says that, for $d$ large, $\mathcal{H}^r(M)_d=0$ for $r>0$ and $M_d \cong \mathcal{H}^0(M)_d$; this is how the result is usually stated. The first theorem in this language is that $\dim \sum_{r=0}^{n-1} (-1)^r \mathcal{H}^r(M)_d$ is a polynomial in $d$.

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Minor quibble: according to your notation, it should be $\mathbb{P}^n$, since you start at $x_0$. – Ryan Reich Jun 20 '12 at 6:30
@David Speyer: Thanks David. – S.C. Jun 20 '12 at 11:07