# Solutions to Hartshorne's Algebraic Geometry/Cech Cohomology

## Exercise 4.7

### Problem Statement:

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

### Solution:

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

${\displaystyle 0\rightarrow {\mathcal {O}}(-d)\rightarrow {\mathcal {O}}\rightarrow {\mathcal {O}}_{X}\rightarrow 0}$

where ${\displaystyle {\mathcal {O}}}$ without further decoration denotes the structure sheaf of ${\displaystyle \mathbb {P} _{k}^{2}}$. Specifically, the map on the left is multiplication by our polynomial f, which is a degree d map ${\displaystyle {\mathcal {O}}\rightarrow {\mathcal {O}}}$, but a degree 0 map ${\displaystyle {\mathcal {O}}(-d)\rightarrow {\mathcal {O}}}$. 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:

${\displaystyle 0\rightarrow \Gamma ({\mathcal {O}}(-d))\rightarrow \Gamma ({\mathcal {O}})\rightarrow \Gamma ({\mathcal {O}}_{X})\rightarrow H^{1}({\mathcal {O}}(-d))\rightarrow H^{1}({\mathcal {O}})\rightarrow H^{1}({\mathcal {O}}_{X})\rightarrow H^{2}({\mathcal {O}}(-d))\rightarrow H^{2}({\mathcal {O}})\rightarrow H^{2}({\mathcal {O}}_{X})\rightarrow 0}$

Which vanishes in higher degrees by dimensional vanishing.

Now to figure out what these things are:

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

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

${\displaystyle H^{2}({\mathcal {O}}_{X})}$ actually vanishes again by dimensional vanishing. ${\displaystyle \Gamma ({\mathcal {O}})=k}$, 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 ${\displaystyle h^{0}({\mathcal {O}}(e))={\dbinom {r+e}{r}}}$ in general; when e = 0, this gives dimension 1 over k. (${\displaystyle h^{i}:={\text{dim}}H^{i}}$).

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

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

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

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

${\displaystyle 0\rightarrow k\rightarrow \Gamma ({\mathcal {O}}_{X})\rightarrow 0}$
${\displaystyle 0\rightarrow H^{1}({\mathcal {O}}_{X})\rightarrow \Gamma ({\mathcal {O}}(d-r-1))\rightarrow 0}$

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