# Hartshorne Exercise III.4.7 (cohomology of closed subschemes in $\mathbb{P}^2$)

I have some questions about the following exercise in Hartshorne (III.4.7):

Let $f \in k[x_0,x_1,x_2]$ be a homogeneous polynomial of degree $d \geq 1$ and $f \neq 0$ and let $X$ be the closed subscheme of $\mathbb{P}^2_k$ defined by $f$. Then $\dim H^0(X,\mathcal{O}_X) = 1, \dim H^1(X,\mathcal{O}_X) = (d-1)(d-2)/2$. This is done using Cech cohomology.

1 - Hartshorne makes the assumption $f(1,0,0) \neq 0$. Is this necessary?

This implies that $f$ is monic in $x_0$ and yields a very nice description of the Cech complex (if necessary, I'll add this), which makes the computation possible. But what about the general case?

It's not hard to see that $f$ is mapped by a graded isomorphism of $k[x_0,x_1,x_2]$ to a polynomial, which does not vanish in $(1,0,0)$, if and only if $f$ does not vanish on $k^3$. Thus if $k$ is infinite, you're done. But what happens when $k$ is finite? For example

$f = xy \prod_{\alpha \in k} (x - \alpha y)$

is a nontrivial homogeneous polynomial of degree $|k|+2$ and vanishes on $k^2$ (and thus on $k^3$).

2 - Is the finite case important for some applications (for example in arithmetic geometry)?

3 - Is it surprising that the cohomology only depends on $d$?

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It's pretty much the default hypothesis in Hartshorne that $k$ is an algebraically closed field. If you are worried about the finite field case, consider the same variety over $k^{alg}$ and show that the dimensions of cohomology don't change. –  Robin Chapman May 31 '10 at 9:08
I suspected that this hypothesis was in the first chapter of Hartshorne. –  Martin Brandenburg May 31 '10 at 9:10
Ah, the dimensions don't change because of the flat base change theorem? –  Martin Brandenburg May 31 '10 at 9:11
Yeah. Also it's not surprising since h^1 is the genus which you could have computed using adjunction. –  Frank May 31 '10 at 10:17
What do you mean by adjunction here? –  Martin Brandenburg May 31 '10 at 12:39

You can compute the cohomology via the Koszul resolution. If $i:X \to {\mathbb P}^2_k$ is the embedding then the triple $0 \to O_{{\mathbb P}^{2}}(-d) \stackrel{f}\to O_{{\mathbb P}^2} \to i_*O_X \to 0$ is exact. So, you can compute $H^t(X,O_X) = H^t({\mathbb P}^2_k,i_*O_X)$ using the long exact sequence associated with this triple.