Solutions to Hartshorne's Algebraic Geometry/Cech Cohomology

Exercise 4.7

Problem Statement:

Let $f$ be an equation cutting out a degree d curve $X$ in ${\mathbb{P}}_k^2$. Suppose that $X$ doesn't contain the point $(1,0,0)$. Use \v{C}ech cohomology to calculate the dimensions of $H^0,H^1$ of $X$.

Solution:

The degree d curve $X$ is the vanishing locus of $f$, so we have a short exact sequence:

$0\to 𝒪(-d)\to 𝒪\to {𝒪}_X\to 0$

where $𝒪$ without further decoration denotes the structure sheaf of ${\mathbb{P}}_k^2$. Specifically, the map on the left is multiplication by our polynomial f, which is a degree d map $𝒪\to 𝒪$, but a degree 0 map $𝒪(-d)\to 𝒪$. This is an injective map, and we are quotienting out precisely by its image, so it's equivalent to the usual short exact sequence associated to a closed subscheme.

Then apply the H functor to get a long exact sequence:

$0\to \mathrm{\Gamma }(𝒪(-d))\to \mathrm{\Gamma }(𝒪)\to \mathrm{\Gamma }({𝒪}_X)\to H^1(𝒪(-d))\to H^1(𝒪)\to H^1({𝒪}_X)\to H^2(𝒪(-d))\to H^2(𝒪)\to H^2({𝒪}_X)\to 0$

Which vanishes in higher degrees by dimensional vanishing.

Now to figure out what these things are:

$H^i(𝒪(e))=0$ for $0<i<r$ in projective space ${\mathbb{P}}_A^r$.

This gives us that $H^1(𝒪(-d))=H^1(𝒪)=0$. Furthermore, assuming degrees must be positive $\mathrm{\Gamma }(𝒪(-d))=0$.

$H^2({𝒪}_X)$ actually vanishes again by dimensional vanishing. $\mathrm{\Gamma }(𝒪)=k$, either by general knowledge (constants are the only globally defined homogeneous polynomials with degree zero on any of the standard open affines) or by the fact that $h^0(𝒪(e))=(\genfrac{}{}{0ex}{}{r+e}{r})$ in general; when e = 0, this gives dimension 1 over k. ($h^i:=\text{dim}{H}^i$).

Our last trick we shall use is Serre duality (here just for projective space):

$\mathrm{\Gamma }(𝒪(-e))\cong (H^r(e-r-1){)}^{\vee }$, where ${\cdot }^{\vee }$ represents the dual.

Since the dimension of a vector space (these H's are vector spaces in this context because of III 5.2 in Hartshorne, pg 228) is the same as its dual, $\text{dim}{H}^2(𝒪)=\text{dim}(H^2(𝒪){)}^{\vee }$. Moreover, $(H^2(𝒪){)}^{\vee }\cong \mathrm{\Gamma }(𝒪(-r-1))$, which has dimension $(\genfrac{}{}{0ex}{}{-1}{r})=0$, so it's 0. Hence $H^2(𝒪)=0$.

Moreover, $\text{dim}{H}^2(𝒪(-d))=\text{dim}(H^2(𝒪(-d)){)}^{\vee }$, and by the same trick (Serre duality), $(H^2(𝒪(-d)){)}^{\vee }\cong \mathrm{\Gamma }(𝒪(d-r-1))$, which has well-known dimension (e.g., Vakil 14.1.c) of $(\genfrac{}{}{0ex}{}{r+d-r-1}{r})=(\genfrac{}{}{0ex}{}{d-1}{2})=(\genfrac{}{}{0ex}{}{d-1}{d-3})=\frac{1}{2}(d-1)(d-2)$.

Combining all of the above results, we get two short exact sequences:

$0\to k\to \mathrm{\Gamma }({𝒪}_X)\to 0$

$0\to H^1({𝒪}_X)\to \mathrm{\Gamma }(𝒪(d-r-1))\to 0$

So we have $h^0({𝒪}_X)=1$ and $h^1({𝒪}_X)=\frac{1}{2}(d-1)(d-2)$.