Mumford's book Abelian Varieties asserts that for a line bundle L on a projective variety (if necessary, you can assume it is as nice as possible), the Euler characteristic $\chi(L^k)$ of tensor powers of $L$ is a polynomial in $k$. If L is very ample, this is just the Hilbert polynomial, and this can be proven by an induction argument twisting the short exact sequence $0 \to \mathcal O(-1) \to \mathcal O \to K \to 0$. More generally, if $L$ (or $L^\*$) is an ideal sheaf, the same argument should work. Why does the result still hold for arbitrary $L$?


Edit: I'd be particularly interested in an elementary proof that does not involve proving an entire Riemann-Roch theorem--Mumford is using this result to *prove* Riemann-Roch for abelian varieties!



Why is the Euler characteristic of powers of a line bundle a polynomial in the power?
OK, here is another way to see it more in line with what you had in mind I think. Write your $L$ as $\mathcal O(D)$ for some divisor $D$ on $X$. Set $J\_1$ to be the ideal sheaf defined by $\mathcal O(-D) \cap \mathcal O\_X$ and $J\_2$ to be the ideal sheaf defined by $\mathcal O(D) \cap \mathcal O\_X$ (intersections taken inside of $K\_X$). Let $Y\_i$ be the closed subschemes of $X$ defined by these ideal sheaves (they have dimension smaller than that of $X$). Then we have the exact sequences


$$0 \to J\_1(kD) \to \mathcal O(kD) \to \mathcal O\_{Y\_1}(kD) \to 0$$


$$0 \to J\_2((k-1)D) \to \mathcal O((k-1)D) \to \mathcal O\_{Y\_2}((k-1)D) \to 0$$


The two left hand terms are equal by construction. Then by the induction hypothesis, and chasing the Euler characteristics, $\chi(kD) - \chi((k-1)D)$ is a numerical polynomial. This implies that that $\chi(kD)$ itself is a numerical polynomial (Section 1.7 of Harshorne's Algebraic Geometry).


(Here I swept something under the rug, because the subschemes $Y\_i$ may not be as nice as $X$ was. But they are at least proper, and we should show that the result we want is that for a proper variety $W$, $\chi(kD)$ is polynomial for a divisor $D$. Then reduce this to the case where $W$ is reduced by looking at the inclusion of $W\_\mathrm{red}$ into $W$. Then further reduce to the case where $W$ is integral.)


