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 $d$?



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


