Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $d$?

share|improve this question
    
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

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.