Why is this theorem attributed to Serre? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T20:30:04Z http://mathoverflow.net/feeds/question/99916 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99916/why-is-this-theorem-attributed-to-serre Why is this theorem attributed to Serre? Chandrasekhar 2012-06-18T16:43:20Z 2012-06-22T08:07:33Z <p>Page $117$ of Atiyah, MacDonald's <em><a href="http://books.google.co.in/books?id=HOASFid4x18C&amp;printsec=frontcover&amp;source=gbs_ge_summary_r&amp;cad=0#v=onepage&amp;q&amp;f=false" rel="nofollow">Introduction to Commutative Algebra</a></em> text has the following theorem. Let $P(M,t)$ denote the <em><a href="http://en.wikipedia.org/wiki/Hilbert%25E2%2580%2593Poincar%25C3%25A9_series" rel="nofollow">Poincare- series</a></em> of $M$.</p> <ul> <li>$\textbf{Theorem.}$ $\bigl(\mathsf{Hilbert-Serre}\bigr)$. $P(M,t)$ is a rational function in $t$ of the form $f(t)/\prod_{i=1}^{s} (1-t^{k_i})$, where $f(t) \in \mathbf{Z}[t]$.</li> </ul> <p>This theorem appears in the section of the book called <em>Hilbert-Functions</em> (page 116), so one understands that it could have possibly been discovered by <em>Hilbert</em>. </p> <ul> <li>But why is the above theorem attributed to <a href="http://en.wikipedia.org/wiki/Jean-Pierre_Serre" rel="nofollow">Serre</a>? References about when Serre was credited to the above theorem would be helpful. </li> </ul> http://mathoverflow.net/questions/99916/why-is-this-theorem-attributed-to-serre/100023#100023 Answer by David Speyer for Why is this theorem attributed to Serre? David Speyer 2012-06-19T16:22:25Z 2012-06-19T16:22:25Z <p>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.</p> <p>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 <a href="http://homepages.math.uic.edu/~bshipley/huneke.pdf" rel="nofollow">local cohomology modules</a> of $M$ with respect to the maximal ideal $\langle x_0,\ldots, x_n \rangle$. These are graded modules which satisfy the following properties:</p> <p><b>Theorem:</b> For <i>all</i> integers $d$, the function $$\dim M_d - \sum_{r=0}^n (-1)^r \dim H^r(M)_d$$ is polynomial in $d$.</p> <p><b>Theorem (Serre vanishing)</b> For $d$ sufficiently large, $H^r(M)_d=0$.</p> <p>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 $d$.</p> <p>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 <code>$$\mathcal{H}^r(M) \cong H^{r+1}(M)$$</code> for $r \geq 1$ and there is a short exact sequence <code>$$0 \to H^0(M) \to M \to \mathcal{H}^0(M) \to H^1(M) \to 0.$$</code> 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 <code>$\dim \sum_{r=0}^{n-1} (-1)^r \mathcal{H}^r(M)_d$</code> is a polynomial in $d$.</p>