Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

There are a lot of compact (Hausdorff) groups, whereas every compact field is finite. What about rings? Is there a classification theorem for compact rings? If you take a cofiltered limit of finite rings, you get a compact ring; for example, the $p$-adic integers $\mathbb{Z}_p$ are obtained as a limit of $$ \cdots \twoheadrightarrow \mathbb{Z}/p^{n+1}\mathbb{Z} \twoheadrightarrow \mathbb{Z}/p^n\mathbb{Z}\twoheadrightarrow \cdots \twoheadrightarrow \mathbb{Z}/p\mathbb{Z}\twoheadrightarrow 0. $$ Can every compact ring be obtained as a cofiltered limit of finite rings?

For a counterexample, a compact ring that is not totally disconnected would suffice. In the other direction, proving that such a ring has to be totally disconnected wouldn't suffice a priori: It would show the the additive group is profinite, but not that the ring is a cofiltered limit of rings.

Remark: By "compact," I consistently mean "compact Hausdorff."

share|cite|improve this question
... and by "ring" you mean something associative with a unit, since the zero multiplication law would make any non-profinite compact group a counterexample. – S. Carnahan Dec 9 '10 at 6:49
Assume that the compact ring $A$ is an integral domain. Then its field of fractions $K$ is a locally compact field, which have all been classified. – Chandan Singh Dalawat Dec 9 '10 at 6:57
Google claims this is true, and is proved for example in theorem 26.10 of the book "Locally compact groups" by Markus Stroppel. – Gjergji Zaimi Dec 9 '10 at 7:21
Gjergji: Why don't you put this as an answer? This way the question remains unanswered. I also do not understand why you write that "Google claims something", if you found it in a textbook of the EMS. – Andreas Thom Dec 9 '10 at 9:18
Andreas: I wrote so, to indicate that the proof appears somewhere but I can not see it at the moment. Perhaps someone who knows the proof can write an answer with a quick sketch/main idea etc. Plus it appears that the result is old and possibly due to Kaplansky, so I'd like to research that a little more. – Gjergji Zaimi Dec 9 '10 at 9:32

1 Answer 1

up vote 5 down vote accepted

Every compact topological ring has "enough" open ideals and is thus profinite. See for example Sect. 5.1 in

Luis Ribes, Pavel Zalesski, Profinite Groups, Ergebnisse der Mathematik und ihrer Grenzgebiete, 3. Folge

share|cite|improve this answer
Yes, this appears to be the argument in Stroppel's book as well (see comments). Thanks for the reference! – Gene S. Kopp Dec 9 '10 at 10:27
For discussion of this result, see here:… . In particular, Todd Trimble gives a very nice, short… . I don't know whether it's the same as Ribes and Zalesski's. – Tom Leinster Sep 5 '14 at 13:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.