Solutions to Hartshorne's Algebraic Geometry/Cech Cohomology

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:


 * $$0 \rightarrow \mathcal{O}(-d) \rightarrow \mathcal{O} \rightarrow \mathcal{O}_X \rightarrow 0$$

where $$\mathcal{O}$$ 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 $$\mathcal{O} \rightarrow \mathcal{O}$$, but a degree 0 map $$\mathcal{O}(-d) \rightarrow \mathcal{O}$$. 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:


 * $$0 \rightarrow \Gamma(\mathcal{O}(-d)) \rightarrow \Gamma(\mathcal{O}) \rightarrow \Gamma(\mathcal{O}_X) \rightarrow H^1(\mathcal{O}(-d)) \rightarrow H^1(\mathcal{O}) \rightarrow H^1(\mathcal{O}_X) \rightarrow H^2(\mathcal{O}(-d)) \rightarrow H^2(\mathcal{O}) \rightarrow H^2(\mathcal{O}_X) \rightarrow 0$$

Which vanishes in higher degrees by dimensional vanishing.

Now to figure out what these things are:

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

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

$$H^2(\mathcal{O}_X)$$ actually vanishes again by dimensional vanishing. $$\Gamma(\mathcal{O}) = k$$, 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 $$h^0(\mathcal{O}(e)) = \dbinom{r + e}{r}$$ 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):

$$\Gamma(\mathcal{O}(-e)) \cong (H^r(e - r - 1))^{\vee}$$, 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, $$\text{dim}H^2(\mathcal{O}) = \text{dim}(H^2(\mathcal{O}))^{\vee}$$. Moreover, $$(H^2(\mathcal{O}))^\vee \cong \Gamma(\mathcal{O}(-r-1))$$, which has dimension $$\dbinom{-1}{r} = 0$$, so it's 0. Hence $$H^2(\mathcal{O}) = 0$$.

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

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


 * $$0 \rightarrow k \rightarrow \Gamma(\mathcal{O}_X) \rightarrow 0$$


 * $$0 \rightarrow H^1(\mathcal{O}_X) \rightarrow \Gamma(\mathcal{O}(d-r-1)) \rightarrow 0$$

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