# Friday, December 03

:::{.theorem title="?"}
For $X\da \PP^n\slice k$ and $D\in \Div(X)$, define 
\[
\deg D \da \sum n_i \deg Y_i && \text{where } D = \sum n_i [Y_i]
.\]
Let $H \da \ts{x_0 = 0}$ by a hyperplane, then

- $D\sim \deg(D) H$
- $f\in k\units \implies \deg(f) = 0$,
- $\deg: \Cl(X)\to \ZZ$ is an isomorphism.

:::

:::{.proof title="?"}
Missed, see Hartshorne.
:::

:::{.proposition title="?"}
For $Z \subseteq X$ proper and closed with $U \da X\sm Z$, if $\codim Z \eq 2$, then $\Cl(X) \cong \Cl(U)$.
If $Z$ is irreducible and $\codim Z = 1$, there is an exact sequence $\ZZ \mapsvia{f} \Cl(X) \to \Cl(U) \to 0$ where $f(1) = [Z]$.
:::

:::{.remark}
Note that this $\ZZ\to \Cl(X)$ isn't injective in general: take $X\da \AA^n$ so $\Cl(X) = 1$.
:::

:::{.proof title="?"}
Define a map
\[
\phi: \Div(X) &\to \Div(U) \\
Y &\mapsto
\begin{cases}
Y \intersect U &  Y \intersect U\neq \emptyset
\\
0 & Y \intersect U = \emptyset.
\end{cases}
\]
Then $\phi$ is $\ZZ\dash$linear, and $k\units(X) \cong k\units(U)$, and descends to a map $\Cl(X) \to \Cl(U)$.
Moreover $\ker \phi$ is generated by prime divisors contained in $Z$, so if $\codim Z\geq 2$ this is empty and we have an isomorphism.
Otherwise if $\codim Z = 1$ with $Z$ irreducible, then the only prime divisors in $Z$ is $Z$ itself, so $Z$ generates $\ker \phi$.
:::

:::{.example title="?"}
Let $Z \subseteq \PP^2$ be an irreducible degree $d$ curve and let $U = \PP^2\slice k\smz$.
Then $\Cl(U) = C_d$ is cyclic of order $d$.
We have
\[
\ZZ &\to \Cl(\PP^2\slice k) \cong \ZZ \to \Cl(U) \to 0 \\
1 &\mapsto [Z] \cong \deg d
.\]
:::

## Divisors on Curves

:::{.definition title="Curve"}
Let $k=\bar k$, then a **curve** $X\in \Sch\slice k$ is an integral (so reduced) separated of finite type of dimension 1.
We say $X$ is **complete** if $X\to \spec k$ is proper, and **smooth** if $X$ is regular (equivalently regular in codimension 1).
:::

:::{.example title="?"}
$X\da V(f) \subseteq \PP^2\slice k$ where $f$ is irreducible.
This is complete since it is closed in $\PP^2\slice k$, which is proper.
:::

:::{.proposition title="?"}
If $f:X\to Y$ is a morphism of curves and $X$ is complete and nonsingular, then $\im f$ is either a point or all of $Y$.
If $\im = Y$, then $f$ is finite.
:::

:::{.proof title="?"}
$X$ proper implies $f(X)$ is closed in $Y$, and $X$ irreducible implies $f(X)$ is irreducible.
Since $Y$ is irreducible, this forces $f(X) = \pt$ or $Y$.
Let $V \subseteq Y$ be an affine open and $U \da f\inv(V)$, the claim is that $U$ is affine and the pullback $f^*: \OO_Y(V) \to \OO_X(U)$ is a module-finite extension.
We have a map on function fields $f^*:k(X) \to k(Y)$, and since $\dim X, \dim Y =1$, these are fields of transcendence degree 1 over $k$.
Therefore $f^*$ is a finite extension of fields (use Noether normalization), and $\OO_Y(V) \subseteq k(Y)$.
We can write $\OO_Y(V) = \intersect _{p\in V} \OO_{Y, p}$.
The following gives module-finiteness:

:::{.claim}
The integral closure of $\OO_Y(V)$ in $k(X)$ is $\OO_X(U)$.
:::

To see that $U$ is affine: exercise!
:::