# Monday, November 29

:::{.remark}
Standing assumption: $X\in \Sch$ is

- Integral: covered by $\spec R$ for $R$ integral domains
- Noetherian: covered by Noetherian rings.
- Separated: $\Delta$ is closed.
- Regular in codimension 1: $\dim \OO_{X, x} = 1\implies \OO_{X, x}$ is regular and thus a DVR.
:::

:::{.definition title="Prime and Weil divisors"}
A **prime divisor** $Y \subseteq X$ is a closed integral subscheme of codimension 1, and a **Weil divisor** is a formal $\ZZ\dash$linear combination $\sum_{1\leq i \leq k} n_i Y_i$.
The divisor is **effective** if $n_i \geq 0$ for all $i$.
:::

:::{.example title="?"}
If $\eta\in Y$ is the generic point, $\OO_{X, \eta} \da \OO_{X, Y}$ is a local ring of dimension 1, and thus a DVR.
This yields a valuation:
\[
v_Y: \ff(\OO_{X, Y})\units = k\units \to \ZZ
,\]
where $k$ is the residue field of the generic point of $X$, also called the *rational functions on $X$*.
:::

:::{.definition title="Principal divisors"}
If $f\in k\units$ then there is an associated divisor:
\[
\div(f) \da \sum_{Y \text{prime divisors}} v_Y(f) [Y]
.\]
Any divisor of a rational function is **principal**.
:::

:::{.example title="?"}
For $X\da \spec \ZZ$, the generic point is $\gens{0}$ and the prime divisors are prime ideals of height 1, so here just prime ideals $\gens{p}$.
So the integral closed codimension subschemes correspond to primes $p\in \ZZ$, and there are valuations $v_p: \QQ\units \to \ZZ$.
Write $k = \ff(\OO_{X, \gens{0}}) = \QQ$, then e.g. $\div(4/7) = 2[2] - 1[7]$.
:::

:::{.example title="?"}
Set $X\da \AA^2\slice \CC$ and $f(x, y) = x/y$.
What is $v_{[V(x)]}(f)$?
Then $\div(f) = [V(x)] - [V(y)]$, and
\[
v_{ [V(x)] }: \CC[x,y]\localize{\gens{x}} \to ?
.\]
So the answer is 1.
:::

:::{.proposition title="?"}
$\div(f)$ is well-defined, i.e. $v_f(Y) = 0$ for all but finitely many $Y$.
:::

:::{.proof title="?"}
Let $f\in k\units = \ff(A)$ for $\spec A \subseteq X$ an affine open.
Write $f=a/b$ for some $b\in A\smz$, noting that $A$ is a domain since we assumed $X$ integral.
Passing to $D(b)$, we can assume $f$ is a regular function on some affine open $U\subseteq X$.
Since $X\sm U$ is a proper closed subset and $X$ is Noetherian, it contains only finitely many prime divisors -- each irreducible component has $\codim_X \geq 1$, and conversely any prime divisor must be an irreducible component, and Noetherian spaces have finitely many irreducible components.
So it suffices to show $\div(f)$ is well-defined for $f\in A$ when $X=\spec A$.
Just use that $V(f) \subseteq \spec A$ is a proper closed subset, the same argument shows $V(f)$ contains finitely many prime divisors.
Since $f\in A$, we have $f\in \OO_{X, Y}$ and thus $v_Y(f) \geq 0$.
Moreover if $v_Y(f) = 0$ then $Y \subseteq V(f)$ -- use that $f\in pA\localize{p}$ and $p A\localize{p} \intersect A = p$ to get $f\in p$.
:::

:::{.definition title="Divisor class groups"}
The **divisor class group** of $X$ is defined as 
\[
\Cl(X) = \Div(X)/\Prin\Div(X)
,\]
where $\Prin\Div(X) = \ts{\div f \st f\in k\units}$.
Since $v_Y(fg) = v_Y(f) = v_Y(g)$, $\Prin\Div(X)$ forms a subgroup of $\Div(X)$.
:::

:::{.example title="?"}
Consider $X\da \spec \ZZ$, then
\[
\Div(X) = \bigoplus_{p \text{ prime}} \ZZ[p]
.\]

Then $\Prin\Div(\spec \ZZ) = \Div(X)$ by sending $\sum n_p [p] \to \div (\prod_p p^{n_p})$, so $\Cl(\spec \ZZ) = 0$.
:::

:::{.example title="?"}
For $K \in \Number\Field$ and $\OO_K$ its ring of integers, we can consider $\Cl(\OO_K)$.
For example, $\Cl(\spec \ZZ[\sqrt{-5}] ) = C_2 = \gens{2, 1+\sqrt{-5}}$, using the Dedekind domains admit unique factorization into prime ideals.
:::

:::{.proposition title="?"}
Let $A$ be a Noetherian domain, then $A$ is a UFD iff $\Cl(\spec A) = 1$ is trivial.
:::

:::{.proof title="?"}
Use the lemma that $A$ is a UFD $\iff$ every height 1 prime ideal is principal.
Note that $\gens{2, 1 + \sqrt{-5}}$ is height 1 but not principal!

$\implies$:
Let $Y \subseteq X = \spec A$ be a prime divisor, so $Y = V(p)$ for $p$ a height 1 prime ideal, so we can write $V(p) = V(f)$ for some $f$.
Then $\div(f) = [Y]$, and any prime divisor is principal, and now just use that $[Y]$ generate $\Div(X)$.

$\impliedby$:
Suppose $\Cl(X) = 0$ and let $Y \subseteq X$ be a prime divisor with $\div(f) = Y$ for some $f\in k\units$.
We want to show $V(f) = Y$.
If $\div(f) = [Y]$, then for all $Y' \subseteq X$ prime divisors we have $v_{Y'}(f) \geq 0$.
By ring theory, $f\in A$.
If $V(f) = 1$ then $f\in pA\localize{p}$ for $p=V(y)$, so $f\in pA\localize{p} \intersect A$ and thus $f\in p$.
The claim is that $p = \gens{f}$ -- suppose $g\in p$, then $v_{Y'}(g) \geq 0$ and $v_Y(g) \geq 1$ implies $v_{Y'}(g/f) \geq 0$ for all $Y'$.
So $g/f\in A$, making $g\in \gens{f}$.
:::

:::{.remark}
Use that valuations are non-negative on prime divisors and that the valuations are either 0 or 1.
:::