# Tuesday, December 07

:::{.proposition title="?"}
For $f\in K(C)\units$ for $C$ a smooth complete curve, $\deg(\div f) = 0$.
:::

:::{.proof title="?"}
If $f$ is constant this is trivial, so assume not.
Define 
\[
U_1 &\da \ts{p\in C \st v_p(F) \geq 0 } \\
U_2 &\da \ts{p\in C \st v_p(F) \leq 0 }
.\]
Then $U_1 \intersect U_2$ is precisely the set of closed points of $C$.
Suppose $f$ is regular on $U_1$, so $1/f$ is regular on $U_2$.
Define a map $\hat{f}: C\to \PP^1$ by writing $\PP^1 = \AA_1 \union \AA_1$ and defining $\ro{f}{U_i}: U_i\to \AA^1$ to map into the $i$th factor.
Note that $\ro{\hat f}{U_2} = 1/f$.
Then 
\[
\div(f) = \sum_{p\in C} v_p(f) [p] = f^*\qty{ [0] - [\infty]} \da f^*(D)
.\]

Then $\deg(\div(f)) = \deg f \cdot \deg (D) = 0$ since $\deg(D) = 0$, noting that $\deg f$ is the degree of the corresponding field extension.
:::

:::{.corollary title="?"}
For a smooth complete curve, the degree map descends to a well-defined map:
\[
\deg: \Cl(C) &\to \ZZ \\
\sum n_p p &\mapsto \sum n_p
.\]
:::

:::{.definition title="$\Cl_0$"}
Define
\[
\Cl_0(C) \da \ker\qty{\Cl(C) \mapsvia{\deg} \ZZ}
.\]
:::

:::{.definition title="Elliptic curve"}
An **elliptic curve** $E$ is a smooth complete genus 1 curve over $k$ with a distinguished closed point $0\in E$ called the origin.
Complex analytically, $E = \CC/ \Lambda$ where $\Lambda \subset \CC$ is an integral lattice with $\Lambda \tensor_\ZZ \RR \cong \CC$, so the basis vectors are independent.

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-12-07_11-49.pdf_tex} };
\end{tikzpicture}

:::

:::{.example title="?"}
The cubic in $\PP^3\slice k$ defined by 
\[
C \da V(zy^2 = x^3 + axz^2 + bz^3), \quad 0 \da [0: 1 : 0]
.\]
:::

:::{.definition title="Weierstrass $\wp$ function"}
Define a complex analytic function
\[
\wp: \CC &\to \PP^1 \\
z &\mapsto {1\over z^2} + \sum_{\lambda\in \Lambda\smz} {1\over (z-\lambda)^2} - {1\over \lambda^2} = z^{-2} + \bigo(z^2)
.\]
:::

:::{.remark}
Note that 

- $\wp(z + \lambda) = \wp(z)$ for all $\lambda \in \Lambda$
- $\wp: E\to \PP^1$, so $\wp\in \CC(E)\units$
- $\wp'(z) = \sum_{\lambda\in \Lambda} {-2 \over (z - \lambda)^3}\in \CC(E)\units = -{2\over z^3} + \bigo(z)$.
- $\wp'(z)^2 = {4\over z^6} + {c_1\over z^2} + \bigo(1)$
- $\wp(z)^3 = z^{-6} + c_2z^{-2} + \bigo(1)$

So there is a relation
\[
F: \quad \wp'(z)^2 = 4\wp(z)^3 + G_2(\Lambda)\wp(z) + G_3( \Lambda) + \bigo(z)
.\]

Note that this cancels the poles at the lattice points, making it a bounded holomorphic function and thus constant.
Since it's $\bigo(z)$, this forces it to be zero.
So define a map
\[
E &\to \PP^2\slice \CC \\
z &\mapsto [\wp(z): \wp'(z): 1]
,\]
and note that $0\mapsto [0:1:0]$ so this factors through $V(zy^2 = 4x^3 + G_2( \Lambda)xz^2 + G_3(\Lambda) z^3 )$ biholomorphically, using that $\deg \wp, \wp' = 2,3$ to get injectivity.
This makes $E$ an algebraic variety.
:::

:::{.remark}
An aside: suppose $f: C_1\to C_2$ is a degree 1 holomorphic map of compact complex curves.
Then $f'=0$ at only finitely many points, so $f$ is invertible on an open set and $f\inv$ extends continuously to $C_2$.
Now use the Riemann removable singularity theorem: extending a holomorphic function continuously over a puncture implies that the new function is holomorphic.
:::

:::{.remark}
Why is this algebraic structure unique?
Use an overpowered theorem: Serre's GAGA, i.e. there is a unique variety structure on a compact complex manifold over $\CC$.
In our case, it suffices to show $\wp(z-c), \wp'(z-c)\in \CC(\wp, \wp')$.
In fact, the rational functions are given by $K(X) = \ff\qty{\CC[\wp, \wp']/\gens{\text{a cubic}}}$.
:::

:::{.remark}
Write $E$ for the vanishing locus of the cubic $F$ above, and consider a map
\[
\CC/ \Lambda \iso E \subseteq \PP^2\slice \CC
.\]
Consider a line $L \subseteq \PP^2\slice \CC$, then $L \intersect C = p+q+r = \div(L)$ generically.
We claim $p+q+r \equiv 0 \mod \Lambda$.

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-12-07_12-11.pdf_tex} };
\end{tikzpicture}

To prove this:
write $L_0 \da V(z)$, so $\div(L_0) = 3[0]$, and consider $L_0 \intersect V(F)$.
We have $\div(L/L_0) = [p] + [q] +[r] - 3[0]$.


:::{.claim}
If $f\in K(E)\units$ and $\div f = \sum n_p [p]$ then $\sum n_p p \equiv 0 \mod \Lambda$ after taking these as honest points in $\CC$.
:::

To prove this, do some kind of contour integral over the fundamental domain and use lattice periodicity of $f$.
This yields $p+q+r-3\cdot 0 \in \Lambda$, so given any two points we can solve for the third.
:::

:::{.remark}
This can be used as a reduction algorithm:
\[
[p_1] + [p_2] + 2[p_3] - 4[p_4] 
&= [p_1 + p_2] + [0] + 2[p_3] - 4[p_4] \in \Div^0(E) \\
&= 2[p_3 - p_4] - 2[p_4] - 2[0] \\
\cdots &= [p] - [0]
\implies 
,\]
so there is an isomorphism
\[
E &\to \Cl^0(E) \\
p &\mapsto [p] - [0]
.\]
:::