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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 $\mathbb Z_p[t]$.

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    $\begingroup$ Note to self: the URL for a direct link to an MO comment can be found in the timestamp for the comment. $\endgroup$
    – Tim Campion
    Commented Oct 7, 2020 at 18:37
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    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Oct 7, 2020 at 20:44
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    $\begingroup$ 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. $\endgroup$
    – Steven Landsburg
    Commented Oct 7, 2020 at 20:56
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    $\begingroup$ @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.... $\endgroup$
    – Tim Campion
    Commented Oct 7, 2020 at 20:57
  • $\begingroup$ 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! $\endgroup$
    – Tim Campion
    Commented Oct 7, 2020 at 21:11

1 Answer 1

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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.

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  • $\begingroup$ 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 $0$.... $\endgroup$
    – Tim Campion
    Commented Oct 8, 2020 at 15:22
  • $\begingroup$ ... $R$ 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). $\endgroup$
    – Tim Campion
    Commented Oct 8, 2020 at 15:24
  • $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Commented Oct 8, 2020 at 16:36
  • $\begingroup$ @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. $\endgroup$
    – Will Sawin
    Commented Oct 8, 2020 at 16:41
  • $\begingroup$ 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... $\endgroup$
    – Tim Campion
    Commented Oct 8, 2020 at 16:54

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