Class group and localization in number fields

Question

In The Rising Sea, Vakil says that $A = \mathbb{Z}[\sqrt{-5}]$ shows that the property "being the spectrum of a UFD" is not an affine-local property. Concretely, he points out that $D(2) = \operatorname{Spec} \mathbb{Z}[\sqrt{-5}]_2$ and $D(3)$ cover $\operatorname{Spec} A$, but claims that $\mathbb{Z}[\sqrt{-5}]_2$ and $\mathbb{Z}[\sqrt{-5}]_3$ are UFDs.

He sketches this second fact.

Show the class group of $\mathbb{Z}[\sqrt{-5}]$ is $\mathbb{Z}/2$ . . . Then show that the ideals $(1 + \sqrt{-5}, 2)$ and $(1 + \sqrt{-5}, 3)$ are not principal, using the usual norm.

I understand proofs of both of both facts he says here, and understand that these not-principal ideals become principal when we localize. Why does this then imply that the localization is a PID? That is, why does ideal class group $\mathbb{Z}/2$ imply that an arbitrary ideal is principal just because this particular ideal becomes principal in the localization?

Is there some general classification of what happens to the class group when we localize, say, at an integer?

Answer 1

Let $R$ be an arbitrary Dedekind domain, and choose some multiplicative set $S\subset R$. Let $\newcommand{\Pic}{\operatorname{Pic}}\Pic(R)$ denote the class group of $R$. Note that we have a map $\Pic(R)\to\Pic(S^{-1}R)$ sending a fractional ideal $M$ to its localization $\newcommand{\sinv}{S^{-1}}\sinv M$. We claim that this map is surjective, which suffices to show that $\Pic(A_2)=\Pic(A_3)=0$ in your example.

Pick $[M]\in\Pic(\sinv R)$, so $M$ is some finitely-generated $\sinv R$-submodule of $F=\operatorname{Frac}(R)$. We wish to show that $M=\sinv N$ for some finitely-generated $R$-submodule of $F$. Well, let $e_1,\dots,e_n\in M$ be a generating set, and take $N=\sum_{i=1}^nR e_i\subset M$. Then, $N$ is visibly a finitely-generated $R$-submodule of $F$, and we have $\sinv N=M$ by construction, so the map $\Pic(R)\to\Pic(\sinv R)$ is surjective and we win.