I understand that for any nonempty set $S$ of characteristics, there exists a PID $R$ such that the set of characteristics of residue fields of $R$ (i.e. quotients by of $R$ by maximal ideals -- I'm not including the residue field at the generic point. Thanks to Steven Landsburg for pointing out this terminological ambiguity in the comments below) is precisely $S$. I learned this from a paper of Heitmann, PID’s with specified residue fields (which proves much more), which I originally found at Exotic principal ideal domains.

**Question:** What is a "nice" example of a PID $R$ such that $R$ has a residue field of characteristic 0 and a residue field of finite characteristic?

By "nice", I'd ideally mean that $R$ is not just custom-built for the purpose of providing such an example, and might be a ring I'd meet on the street one day. Failing that, I'd settle for a streamlined description of such a ring $R$ (in order to understand Heitmann's example one must wade through several layers of extra generality related to his more ambitious aims).

If we only require $R$ to be Noetherian, then YCor gave a simple example in the comments (1 2 3) on If a PID has no nonzero divisible elements, then is the same true of its finitely-generated modules?: $R = \mathbb Z_p[t]$ has residue fields $\mathbb F_p$ and $\mathbb Q_p$ (the latter obtained by modding out by $(1-pt)$). Similarly, $\mathbb Z_{(p)}[t]$ has residue fields $\mathbb F_p$ and $\mathbb Q$. It would be nice if there were an example of a PID with this property just as "nice" as $\mathbb Z_p[t]$.

`https://mathoverflow.net/questions/373535/example-of-a-pid-with-a-residue-field-of-finite-characteristic-and-a-residue-fie#comment945342_373535`

can be trimmed to`https://mathoverflow.net/questions/373535#comment945342_373535`

. $\endgroup$ – LSpice Oct 7 '20 at 20:443more comments