# Hilbert Polynomial using non ample line bundles

In his book on abelian varieties, in the Appendix to section 6, Mumford says that if $X$ is any complete variety, $F$ is a coherent sheaf and $L$ an invertible sheaf, then the function defined as $$P(F,L,n)=\chi(F\otimes L^n)$$ is a polynomial in $n$. This is an exercise in Hartshorne, Chapter 3, 5.2, for a very ample line bundle.

I can see why the above is true for $L$ a very ample line bundle. In fact the argument is by induction on the dimension of the support of $F$. We write a short exact sequence

$$0\to R(n)\to F(n-1)\to F(n)\to F\vert_H(n)\to 0$$ Note that $R$ is supported on a proper subvariety of $X$ since at the generic point $F(-1)=F$. Taking Euler characteristic, we get the result.

The above proof does not work in the ample case since an ample line bundle may not have any global sections (what is an example of this??).

How does one prove the result in the general case?

• One way is to done in (if memory serves me) the Riemann-Roch SGA (I don't off hand remember the number)where one use the fact that the augmentation ideal of the K-group is nilpotent, the polynomiality is a formal consequence of this. – Torsten Ekedahl Sep 21 '11 at 10:52
• A proof using Grothendieck-Riemann-Roch was pointed out to me some time ago. If $X$ is smooth projective, then we immediately reduce to the locally free sheaves case using a resolution by locally free sheaves. $\chi(E)=P(c_1,...,c_n)$ and $c_i(E\otimes L^n)$ can be written as a polynomial in the Chern classes of $E$ and the Chern classes of $L$. – Rex Sep 21 '11 at 11:18
• Sorry I didn't make myself clear. The proof I am referring to does not use GRR. – Torsten Ekedahl Sep 21 '11 at 15:14

It is just a consequence of the Riemann-Roch theorem. But another way to see it, is using the fact that if $L$ is any line bundle then $L=A\otimes B^{-1}$ where $A$ and $B$ are very ample line bundles. Thus, if $Z$ is the zero set of a general section of $B$ you have a sequence
$0\to F\otimes A^n\otimes B^{-(i+1)}\to F\otimes A^n\otimes B^{-i}\to F\otimes A^n\otimes B^{-i}\otimes \mathcal O_Z\to 0$
Now just sum over $i$ and proceed by induction on the dimension.
• Ah! That's clever. But this gives that $P(F\otimes A^n,B,i)$ is a polynomial in $i$, and so it looks like $a_0(n)i^d+a_1(n)i^{d-1}+\cdots a_d(n)$. Here the coefficients depend on $n$ since we are taking $F\otimes A^n$. From this I don't see how to conclude that $P(F,L,n)$ is a polynomial. – Rex Sep 21 '11 at 11:02
• As I said, you need to sum over $i$ from $1$ to $n-1$. I guess, it is clear that the sum of your polynomial (which depends on $i$) is a polynomial in $n$. – mrw Sep 21 '11 at 11:07
• But why will the coefficients be independent of $n$. That is the problem. – Rex Sep 21 '11 at 11:14
• Consider any function $f:\mathbb{Z}\to \mathbb{Z}$. I can always write $f(n)=a_0(n)\cdot 1$, where 1 is the constant polynomial which only the value 1, and $a_0(n)=f(n)$. Does that mean that $f$ is a polynomial? No. In the answer you give $P(F\otimes A^n,B,i)$ is definitely a polynomial in $i$ whose coefficients depend on $n$. This is like the previous example. I hope that clarifies my point. – Rex Sep 21 '11 at 11:23