# Wednesday, December 01

:::{.remark}
Recall:

- $X\in\Sch$ is Noetherian, integral, separated, regular in codimension 1, 

- $Y \subseteq X$ a prime divisor is an integral closed codimension 1 subscheme,

- $\Div(X) = \ZZ\adjoin{ \ts{\text{Prime divisors}} }$,

- $\Prin\Div(X) = \ts{\div(f) \st f\in k\units, \text{ the rational functions on } X}$ where $\div(f) \da \sum_{Y} v_Y(f) [Y]$,

- $\Cl(X) \da \Div(X) / \Prin\Div(X)$.

We proved that if $X=\spec A$, 
\[
\Cl(X) = 1 \iff A \text{ is a UFD}
.\]
This shows that any height 1 prime ideal contained in $A$ is principal.
The key commutative algebra fact was \[
\Intersect_{\height(p) = 1} A\localize{p} = A
.\].
:::

:::{.remark}
A quick review of why $\Cl(X) = 1 \implies$ every $p\in \Spec A$ with $\height(p) = 1$ is principal.
Let $Y = V(p)$, then $[Y]\in \Cl(X) = 1$ means that $[Y] = \div(\phi)$ for some $\phi \in k\units$.
Since $v_{Y'} \geq 0$ for all $Y'$ (since they're just zero for $Y\neq Y'$) implies that $f\in A$, and the claim is that $\gens{\phi} = p$.
Taking $f\in p$, then $\div(f) = [Y] + \eps$ where $\eps\geq 0$ is effective.
Then $\div(f/\phi)\geq 0$ is effective, i.e. $f/\phi \in A\localize{p'}$ for all $p'$.
But then $f/\phi\in \Intersect_{\height(p) = 1} A\localize{p} = A$, so $f\in \gens{\phi}$.
So $p = \gens{\phi}$.
:::

:::{.proof title="of the commutative algebra fact"}
To show \( \Intersect _{\height(p)=1} A\localize{p} = A \),
let $a/b\in k\units$, then $\dim A/\gens{b} = \dim A - 1$.
Why?
Let $\bar p_0 \subset \cdots \subset A/\gens{b}$, then $q\inv(\bar p_0) = p \ni b$.
:::

:::{.remark}
The geometric analog: if $X\da \spec A$ and $V(p)$ is a variety, intersecting with a hyperplane yields a codimension 1 locus (in nice cases).
:::

:::{.example title="Affine space"}
For $k\in \Field$ not necessarily algebraically closed,
\[
\Cl(\AA^n\slice k) = 1
.\]

Proof: $k$ is a UFD, so $\kxn$ is a UFD, so apply the proposition.
:::

:::{.example title="Number fields"}
For $K\in \Number\Field$ and $\OO_K$ its ring of integers, 
\[
\Cl(\spec \OO_K) = 1 \iff \OO_K \text{ is a UFD}
,\]
and this coincides with $\Cl(\OO_K)$ from number theory.
:::

:::{.example title="A geometric non-example"}
If $X\da V(xy-z^2) \subseteq \AA^3\slice k$, then $\Cl(X) \neq 1$ since $xy = zz$ in $A$, exhibiting failure of unique factorization.
How to find an irreducible subscheme:

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We use that $\gens{x, z} \in k[x,y,z] / \gens{xy-z^2}$ is not principal.
Note that if $Y=V(p)$ then $2[Y] = 0$ in $\Cl(X)$: show that $x\in A\localize{p}$, then $v_p(x) = 2$.
:::