# Curves and Divisors: Ramification and Degree (Monday, December 06)

:::{.remark}
Recall that we defined a curve as a 1-dimensional integral separated scheme of finite type over an algebraically closed field.
Here nonsingular corresponds to regular, and complete corresponds to proper.
We were proving the following:
:::

:::{.proposition title="?"}
If $f:X\to Y$ is a morphism of curves with $X$ complete and nonsingular, then

- $f(X) = \pt$ or all of $Y$
- If $f(X) = Y$, then $f$ is finite and $f^*: K(Y) \to K(X)$ is a finite extension of fields.

:::

:::{.remark}
If $\spec B \injects Y$ is an affine open, then defining $A$ as the integral closure of $B$ in $K(X)$ we get $\spec A\injects X$ and $\spec A = f\inv(\spec B)$.
This relies on $X$ being complete and nonsingular -- prove this as an exercise.
:::

:::{.definition title="Degree of a surjective morphism of curves"}
Let $f:X\to Y$ as before and suppose $f$ is surjective.
The **degree** of $f$ is defined as 
\[
\deg f \da [K(X): K(Y)]
.\]
:::

:::{.remark}
Define a pullback of divisors
\[
f^*: \Div(Y) &\to \Div(X)
,\]
defined on closed points (and extended $\ZZ\dash$linearly) as follows: let $q\in Y$ be a closed point, then since $Y$ regular in codimension 1 there exists a generator $t\in K(Y)$ such that $t\in \OO_{Y, q}$ and $\mfm_q = \gens{t}$.
We'll call $t$ a **local parameter** at $q$.
Take an open $U\ni q$ where $t$ is regular and $V(t) = \ts{q}$.

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Write $f^* t\in \OO_X(f\inv(U))$, then 
\[
\div_{f\inv(U)} f^*(t) \da f^*[q] = \sum_{f(p) = q} v_p(f^* t)[p]
.\]
:::

:::{.example title="?"}
Consider $C = \ts{y^2 = x^3+ax+b}$ and $X\da V(C)$, and take the projection
\[
X &\to \AA^1\slice k\\
(x, y) &\mapsto x
.\]

Assume the discriminant $\Delta(a, b)\neq 0$ so $X$ is nonsingular and the roots of $f(x) \da x^3 + ax + b$ are distinct:

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Consider $(a, b) = (0, -1)$ so $y^2 = x^3-1$ and work over $\CC$.
Check that $(x, y) = (0, \pm i)$ are solutions, and we can write
\[
f^* [0] = \sum_{f(p) = 0} v_p(f^* x) = 1\cdot [(0, i)] + 1\cdot [(0, -i)]
,\]
since $\mfm_p = \gens{ x, y\mp i} = \gens{x}$.
Similarly,
\[
f^*[1] = v_{(1, 0)}\gens{x-1} = 2[(1, 0)]
,\]
so the function $x$ is not a local coordinate at 1, but $y$ is.
Consider $\qty{ k[x, y] \over \gens{y^2-x^3-+1} }\localize{\gens{x-1, y}}$; then $\mfm = \gens{x-1, y}$ and we can factor $y^2 = x^3-1 = (x-1)(x^2 + x +1)$, and we can invert to write $x-1 = {y^2\over x^2+x+1} \in \mfm^2\sm \mfm^3$.
:::

:::{.remark}
The punchline: even though the size of the set-theoretic fibers changed, in both cases we pulled back degree 1 divisors and got degree 2 divisors, and this is evidently a 2-to-1 cover.
Note that this example wasn't complete, but we can take the projective closure by homogenizing to get $V(y^2z = ^3 + axz^2 + bz^3)$, and we can extend our map $\pi:X\to \AA^1$ to $\tilde \pi: \tilde X\to \PP^1$ by mapping the new point to $\infty$.
:::

:::{.definition title="Ramification, branching"}
Let $f:X\to Y$ be a morphism of smooth complete curves with $f(X) = Y$.
A **ramification point** of $f$ is a point $p\in X$ where $e_p(f) \da v_p(f^* t) >1$ for $t$ a local parameter at $q=f(p)$.
Such a point $q$ is said to be a **branch point**.
The **ramification divisor** of $f$ is defined as 
\[
R_f \da \sum_{p\in X} \qty{e_p(f) - 1}[p]
.\]
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:::{.remark}
This is a finite sum: show that for all but finitely many points (i.e. a Zariski open), the pullback of a local parameter will again be a local parameter on the cover, potentially after subtracting a constant to shift the image to 0.
More precisely, for any $f\in K(X)$, $f-f(p)$ will be a local parameter at $p$ for a Zariski open.
:::

:::{.proposition title="?"}
Let $f:X\to Y$ be a nonconstant morphism of smooth complete curves, then
\[
\deg f^* D = \deg f \cdot \deg D
,\]
where $\deg\qty{\sum n_p [p]} = \sum n_p$.
:::

:::{.proof title="?"}
The 30s version: write $\spec V \subset Y$ with $A$ defined as the integral closure of $B$ in $K(X)$.
Then $B\to A$ is module-finite of dimension $\deg f = [K(X) : K(Y)]$.
Taking there is an induced map on the local ring
\[
B\localize{q} &\to \bigoplus _{f(p) = q} A\localize{p} \\
t &\mapsto \bigoplus f^* t
.\]
Then \[
\dim\qty{\bigoplus A\localize{p} / t \bigoplus A\localize{p} / B\localize{q} / tB\localize{q}} = 
\dim\qty{ \bigoplus _{f(p) = q} k[t] / t^{e_p(f)} / k} = \deg f
.\] 

Note that this uses CRT: $A/tA \cong \bigoplus _{f(p) = q} A\localize{p} / tA\localize{p}$.
:::