TITLE: Why is the Euler characteristic of powers of a line bundle a polynomial in the power?
QUESTION [8 upvotes]: 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!

REPLY [6 votes]: See EGA III$_1$ 2.5.3 and EGA IV$_2$ 5.3. The elegant generalization there which incorporates an auxiliary coherent sheaf opens the door to using Grothendieck's unscrewing lemma (EGA III$_1$ 3.1.2) to vary that and get the result in considerably more generality: any line bundle on any proper scheme over a field (or artin local ring, using length in place of dimension).  The method is via Chow's Lemma to reduce to projective case and ultimately use slicing methods in the style of DeLand's answer to get back to the very ample case.