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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."

flag	asked Dec 9 2010 at 5:43 Gene S. Kopp 1,408●6●19

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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

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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

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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

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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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Martin Brandenburg · Accepted Answer · 2010-12-09 09:33:22Z UTC

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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

answered Dec 9 2010 at 9:33

Martin Brandenburg
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Yes, this appears to be the argument in Stroppel's book as well (see comments). Thanks for the reference! – Gene S. Kopp Dec 9 2010 at 10:27

Is every compact topological ring a profinite ring?

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