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 $\mathcal F$.
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2$\begingroup$ Why should something be proved without using very simple and standard techniques? $\endgroup$– Alex DegtyarevCommented Jan 14, 2015 at 20:01
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$\begingroup$ Do you know how to compute the Hilbert polynomial of a projective variety, say as outlined in Hartshorne I.7? $\endgroup$– David Benjamin LimCommented Jan 14, 2015 at 20:07
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$\begingroup$ @BenLim Yes, I'm able to prove it in the case of very ample $L$. $\endgroup$– DavidCommented Jan 14, 2015 at 21:33
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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 $m=-n$.
answered Jan 15, 2015 at 0:47