# Example of a PID with a residue field of finite characteristic and a residue field of characteristic 0?

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 $$\mathbb Z_p[t]$$.

• Note to self: the URL for a direct link to an MO comment can be found in the timestamp for the comment. – Tim Campion Oct 7 at 18:37
• Further note to @TimCampion: if you're trying to cram a bunch of comment URLs into a characters-limited space like a comment box, you can strip the slug: e.g., 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. – LSpice Oct 7 at 20:44
• Am I missing something? ${\mathbb Z}_p$ has a residue field of finite characteristic at the closed point and a residue field of zero characteristic at the generic point. – Steven Landsburg Oct 7 at 20:56
• @StevenLandsburg By "residue field", I mean "quotient by a maximal ideal", as opposed to "fraction field of quotient by a prime ideal". I'd better check that's what Heitmann means.... – Tim Campion Oct 7 at 20:57
• At any rate, even if Heitmann is using the more general meaning of "residue field", in a PID the only prime which is not maximal is 0. Heitmann's theorem actually allows us to specify a collection of residue fields up to isomorphism, and there can be multiple residue fields of each characteristic. So if we apply his theorem with a set of two distinct fields of characteristic 0 and another of characteristic $p$, then only one of the fields of characteristic 0 can be the field of fractions, leaving the other to be a quotient by a maximal ideal, along with the field of characteristic $p$. Whew! – Tim Campion Oct 7 at 21:11

You can take the ring of fractions $$\frac{a}{b}$$ with $$a,b \in \mathbb Z[x]$$, where $$b$$ is nonzero mod $$p$$ and nonzero mod $$px-1$$.

Given any polynomial $$a$$, we can remove all factors of $$p$$ and remove all factors of $$px-1$$, obtaining a polynomial that is nonzero mod $$p$$ and nonzero mod $$px-1$$. So every polynomial is a power of $$p^i (px-1)^j$$ times a unit for natural numbers $$i,j$$.

Because the ideal generated by $$p$$ and $$px-1$$ contains $$1$$, the ideal generated by $$p^{i_1} (px-1)^{j_1}$$ and $$p^{i_2} (px-1)^{j_2}$$ is also generated by $$p^{ \min(i_1,i_2)} (px-1)^{\min(j_1,j_2) }$$. So every ideal is generated by a single element of the form $$p^i (px-1)^j$$.

There are two maximal ideals, $$(p),$$ and $$(px-1)$$, whose quotients $$\mathbb F_p(x)$$ and $$\mathbb Q$$ have characteristics $$p$$ and $$0$$ respectively.

• I'm so happy with this example that I can't resist recording my own train of thought about it: Let $R$ be the ring in question. We have $R/p = \mathbb F_p(x)$ and $R/(1-px) = \mathbb Q$, so these ideals are maximal. If $R \to k$ is a homomorphism to a field and neither $p$ nor $1-px$ are killed, then they are both inverted, and the map factors through $(1/p-x)^{-1}\mathbb Q[x]_{(1/p - x)} = \mathbb Q(x)$, the fraction field (where we used that $\mathbb Q[x]$ is a PID to conclude the last equality). So the only other prime is $0$.... – Tim Campion Oct 8 at 15:22
• ... $R$ is a localization of $\mathbb Z[x]$ and hence a UFD. To see it's a PID, it suffices to see that it's a Dedekind domain, i.e. that it's Noetherian (check) and that the localization at each maximal ideal is a DVR. To see that $R_{(p)}$ and $R_{(1-px)}$ are DVR's, we just need to know that their maximal ideals $(p)$ and $(1-px)$ are principal, which they are. To conclude: we can verify that this is an example without worrying about what Gauss' Lemma tells us about factoring polynomials in $\mathbb Z[x]$ (the sort of thing which always makes me a bit uncomfortable). – Tim Campion Oct 8 at 15:24
• In fact, let $A$ be any commutative ring, and let $p_1,\dots, p_n$ be prime ideals in $A$. Let $S = (A\setminus p_1) \cap \dots \cap (A\setminus p_n)$. Then by the prime avoidance lemma, the maximal ideals of $S^{-1} A$ are precisely $p_1,\dots, p_n$. If $A$ is a Noetherian UFD, then so is $S^{-1} A$. If in addition $p_1,\dots, p_n$ are principal, then the $A_{p_i}$ are DVR's, so that $S^{-1} A$ is a PID. – Tim Campion Oct 8 at 16:36
• @TimCampion For the last version, I think one needs to know that ideals like $p_i + p_j$ are $1$ because otherwise those will be additional prime ideals. – Will Sawin Oct 8 at 16:41
• Do we? This isn't the case in your example! The localization is going to kill any prime ideals containing the $p_i$'s, so conditions on the sum of the $p_i$'s seem superfluous, I think. And the criterion "Noetherian domain + localization at each maximal ideal is a DVR $\Rightarrow$ Dedekind domain" doesn't require consideration of non-maximal primes... – Tim Campion Oct 8 at 16:54