# Thursday, January 19 (Divisors)

:::{.remark}
Recall that $\exp$ is surjective as a map of sheaves.
On open contractible subsets $U \subseteq \CC$, for any $g\in \OO_{\hol}\units(U)$ there is an $f\da \log(g)$, but $z\mapsto \log(z)\not\in \OO_{\hol}(\CC\units)$.
Thus surjections of sheaves need not induce surjections on global sections, the failure is measured by sheaf cohomology.
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:::{.definition title="Divisor class group"}
Define the **principal Weil divisors** as 
\[
\Prin\mathrm{W}Div = \ts{\div(f) \st f\in K(X)}
,\]
divisors of nonzero rational functions.
Here $\div(f) = \sum n_Y [Y]$ where $n_Y$ is the order of vanishing/poles along $Y$.
We then define the **(Weil) divisor class group** as
\[
\mathrm{W}\Cl(X) \da \WDiv(X) / \Prin\Div(X)
.\]

:::

:::{.example title="?"}
On $\PP^1$, $\div\qty{s^2-t^2\over st} = [1] + [-1] - [0] - [\infty]$, regarding $\infty = s/t$.
:::

:::{.example title="?"}
$\Cl(\AA^1) = 0$ since $\sum n_p [p] = \div f$ where $f = \prod (x-p)^{n_p}$.
:::

:::{.example title="?"}
There is an isomorphism
\[
\deg: \mathrm{W}\Cl(\PP^1) &\iso \ZZ \\
\sum n_p [p] &\mapsto \sum n_p
.\]
E.g. considering $\div\qty{}$ $[1] + [-1] - [0] = [\infty]$ in $\Cl(\PP^1)$.
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:::{.example title="?"}
Consider $\Cl(\Spec \ZZ)$: principal divisors are primes, so $\WDiv(\spec \ZZ) = \ts{\sum_{p\text{ prime}} n_p [p] }$.
Rational functions on $\spec \ZZ$ are identified with $\QQ$, and if $r = \prod p_i^{n_i}\in \QQ$ then $\div(r) = \sum n_i [p_i]$, so $\Cl(\spec \ZZ) = 0$ since every $\sum_{p \text{ prime}} n_p [p]$ is the divisor of some $r\in \ZZ\subseteq \QQ$.
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:::{.example title="?"}
For $X = \spec R$ for $R\da \ZZ\adjoin{\sqrt{-5}}$, we have $\WDiv(X) = \sum_{p\neq 0 \text{ prime ideals}} n_p [p]$, and the rational functions on $X$ are $\QQ(\sqrt{-5})$.
Since $R\in \DD$, there is unique factorization of (fractional) ideals, so writing $(r) = \prod p_i^{n_i}$ we have $\div r = \sum n_i [p_i]$.
However, $R$ is not a UFD, considering $2\cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$.

> Consider $r = 1\cdot [p]$ for $p = (2, 1 + \sqrt{-5})$? Ask.

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:::{.definition title="Cartier divisors"}
A **Cartier divisor** is a collection of rational functions $f_i$ on $U_i$ such that $\div(f_i) = \div(f_j)$ on $U_{ij}$.
These form a group $\CDiv(X)$, and there is a corresponding class group $\mathrm{Ca}\Cl(X) \da \CDiv(X) / \Prin\CDiv(X)$.
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:::{.example title="?"}
Write $\PP^1 = U_0 \union U_1$ and consider

- $f_0 = s$ on $U_0 = \spec \CC[s]$
- $f_1 = 1$ on $U_1 = \spec \CC[t]$

Note $\div f_0 = [0]$ on $U_0$ and $\div f_1 = 0$ on $U_1$, but $\ro{f_0}{U_{01}}$ has no poles or zeros and thus $\div \ro{f_0}{U_{01}}= 0 = \div \ro{f_1}{U_{01}}$.
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:::{.fact}
If $X$ is smooth then $\WDiv(X) = \CDiv(X)$.
Note all Weil divisors are Cartier: consider $X = V(xy-z^2) \subseteq \AA^3$, which is a circular cone.
Note that $V(z)$ is a union of two lines along the edge of the cone.
Consider $D = V_X(z, y)$, an irreducible codimension 1 subvariety, so $D\in \WDiv(X)$.
This is locally principal away from the origin, since one can slice by the plane $z=0$.
Suppose $f(x,y,z)$ cuts out $D$ at $0$, then write $f = c_0 + c_1x + c_2y + c_3z + \cdots$.
Since $f(0) = 0$ we have $c_0 = 0$, and the remaining terms always cut out *two* lines.
On the other hand, $2D$ is Cartier and principal, since the tangent plane along the cone $V_X(y) = V_X(y, z^2)$ cuts out a doubled line.
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:::{.remark}
Recall that line bundles are $\mcl\in\oxmods$ locally isomorphic to $\OO_{U_i}$.
Given $D\in\CDiv(X)$ with Cartier data $\ts{(f_i, U_i)}$ with $f_i$ rational on $U_i$ and $\div(f_i) = \div(f_j)$ on overlaps.
Define $\OO_X(D)\in\Pic(X)$ to be the sheaf whose sections over $U$ are $\ts{ (s_i) \in \OO_X(U \intersect U_i) s_i f_i = s_j f_j }$, so the sections are related by $s_j = {f_i\over f_j} s_i$ on $U_{ij}$.
Write $t_{ij} = {f_j\over f_i}\in \OO_X(U_i \intersect U_j)$ for the transition functions.
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:::{.example title="?"}
Write $\PP^1 = U_0 \union U_1$ and $D = \ts{(s, U_0), (1, U_1)} \in \CDiv(\PP^1)$, and consider $\OO_{\PP^1}(D)$.
This is given by $\ts{p\in k[s], q\in k[t] \st p = sq}$.
Writing $t=s\inv$, we have $p(s) = sq(s\inv)$, so if $q=1$ then $p=s$ and if $q(t) = t$ then $p=1$.
One can check $\OO_{\PP^1}(D) = \CC\gens{s, 1} \oplus \CC\gens{1, t}$.
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:::{.exercise title="?"}
Show that if $D = \ts{(s^k, U_0), (1, U_1)}$ then $\OO_{\PP^1}(D) = \OO_{\PP^1}(k)$, whose global sections are homogeneous degree $k$ polynomials on $\PP^1$.
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