# Localization of a PID $R$ is always a DVR. What about the converse?

Suppose $$R$$ to be a PID and let $$M$$ be a maximal ideal of $$R$$. It is well-known that the localization $$R_M$$ is a valuation ring. Furthermore, it is Noetherian. Then $$R_M$$ is a discrete valuation ring. Now, my question is the following: if we assume that $$R_M$$ is a DVR and $$M$$ is a maximal ideal of $$R$$, is it true that $$R$$ is a PID?

• Even for noetherian domains the converse does not hold since there are Dedekind domains which are not PIDs. Jul 6, 2021 at 16:26
• To see some examples, take the ring of integers in a number field with class number $\neq 1$ (the class number is $1$ exactly when the ring of integers is a PID). In fact, there are only $9$ imaginary quadratic fields of class number $1$, and the rest give rise to non-PIDs. Jul 7, 2021 at 18:21

Now, there are almost Dedekind domains that aren't Dedekind (and hence certainly are not PIDs.) For example, the group ring $$\mathbb Q[\mathbb Q]$$ (using the underlying additive group of $$\mathbb Q$$, of course) is such an example, as given in
1. $$\mathbb Z[\sqrt{-5}]$$ and
2. $$\mathbb R[x,y]/(x^2+y^2-1)$$