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## Is every compact topological ring a profinite ring?

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."

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... 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 2010 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 2010 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 2010 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 2010 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 2010 at 9:32
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