# 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:

$0\rightarrow {\mathcal {O}}(-d)\rightarrow {\mathcal {O}}\rightarrow {\mathcal {O}}_{X}\rightarrow 0$ 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 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 $h^{1}({\mathcal {O}}_{X})={\dfrac {1}{2}}(d-1)(d-2)$ .