Existence of Hilbert's polynomial

Question

I heard that Hilbert's syzygy theorem can be used to show the existence of Hilbert polynomials. How does the construction works? Namely, why do every coherent $O$-module $\mathscr F$ the $$P_{\mathscr F}(n)=\chi (\mathscr F (n))$$ is a polynomial with respect to $n\in\mathbb Z$?

Answer 1

You know the cohomology groups of $\mathscr{O}_{\mathbf{P}^n}(m)$, so compute $\chi(\mathscr{O}_{\mathbf{P}^n}(m))$ and show that the result is a polynomial in $m$. Then you probably need to recall or prove the fact that if $$0 \to \mathscr{F}_1 \to \cdots \to \mathscr{F}_n \to 0$$ is exact then the alternating sum of the Euler characteristics of the $\mathscr{F}_i$ is zero. [Split the problem up into short exact sequences and use the long exact sequence on cohomology to reduce to a statement about exact sequences of vector spaces.]