0
$\begingroup$

Let $X$ be a projective scheme over algebraically closed field $k$, $L$ is invertible sheaf on $X$ and $\mathcal F \in \operatorname{Coh}(X)$ , we define Hilbert polynomial $P_{\mathcal F}(n)=\chi(\mathcal F \otimes L^n).$ Is there any way to prove that $P$ is polynomial not using Riemann-Roch and Chern classes. I think that there should be some argument similar to mdeland's argument(Why is the Euler characteristic of powers of a line bundle a polynomial in the power?), which works for locally free $\mathcal F$.

$\endgroup$
  • 2
    $\begingroup$ Why should something be proved without using very simple and standard techniques? $\endgroup$ – Alex Degtyarev Jan 14 '15 at 20:01
  • $\begingroup$ Do you know how to compute the Hilbert polynomial of a projective variety, say as outlined in Hartshorne I.7? $\endgroup$ – Ben Lim Jan 14 '15 at 20:07
  • $\begingroup$ @BenLim Yes, I'm able to prove it in the case of very ample $L$. $\endgroup$ – David Jan 14 '15 at 21:33
7
$\begingroup$

Prove that $\chi (\mathcal F \otimes L_1^{n} \otimes L_2^m)$ is a polynomial function of $n$ and $m$ of degree $d$ when $L_1$ and $L_2$ are very ample. You can do this by exactly the same induction argument. (If you have a function of two variables whose successive differences in both variables are degree $d-1$ polynomials, then it is a degree $d$ polynomial.)

Then write your divisor as a difference of two ample divisors and set $m=-n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.