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?

  • 2
    $\begingroup$ Even for noetherian domains the converse does not hold since there are Dedekind domains which are not PIDs. $\endgroup$
    – user26857
    Jul 6, 2021 at 16:26
  • $\begingroup$ 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. $\endgroup$ Jul 7, 2021 at 18:21

1 Answer 1


There's a type of ring called an almost Dedekind domain which is defined by the condition you are asking about, that the localization at every maximal ideal produces a DVR.

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

K. A. Loper. Almost Dedekind domains which are not Dedekind. (2006) Theorem 36 (item 2) p 291

As for almost Dedekind domains that are merely not PID's, one may add

  1. $\mathbb Z[\sqrt{-5}]$ and
  2. $\mathbb R[x,y]/(x^2+y^2-1)$

(Apologies for my earlier abortive answer. I vacillated halfway through and apparently forgot you were focused on DVR's.)


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