How to show $\mathbb{Z}[\sqrt{-5}]_2$ is a UFD?

Question

I would like to know how to show $\mathbb{Z}[\sqrt{-5}]_2$ is a UFD.

I am actually given hints that $\mathbb{Z}[\sqrt{-5}]$ has class group $\mathbb{Z}/2$ and that $(1 + \sqrt{-5},2)$ is not principal, but I can't quite figure how to use these hints. Thank you!

Answer 1

Nagata's theorem says that if you have a Krull domain (in particular, a Dedekind domain) $R$ and $S\subset R$ a multiplicative set, then $\operatorname{Cl}(R)\to\operatorname{Cl}(S^{-1}R)$ is surjective and its kernel is generated by the classes of prime ideals that meet $S$.
In our case the only prime that meets $S$ is $P=(1 + \sqrt{-5},2)$, so the kernel is generated by the class of $P$. Moreover, $P^2=(2)$, so the class of $P$ has order two. This shows that the class of $P$ not only generates the kernel, but $\operatorname{Cl}(\mathbb Z[\sqrt{-5}])$ as well. In particular, in our case the map is trivial, so its image is zero.

Answer 2

It’s been a while, but I think you may be able to use this strategy: Your two rings are $R=\Bbb Z[\sqrt{-5}]$ and $R'=\Bbb Z[\sqrt{-5}][1/2]$—a notation that I find less ambiguous. I think you should be able to show that the class group of $R$ maps onto that of $R'$, and that certainly the ideal you mentioned, representing a nonzero element of the class group of $R$, definitely becomes trivial in that of $R'$.