most recent 30 from http://mathoverflow.net 2013-05-23T18:18:42Z http://mathoverflow.net/feeds/question/48718 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48718/is-every-compact-topological-ring-a-profinite-ring Gene S. Kopp 2010-12-09T05:43:20Z 2010-12-09T09:33:22Z

<p>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?</p> <p>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 <em>a priori</em>: It would show the the additive <em>group</em> is profinite, but not that the ring is a cofiltered limit <em>of rings</em>.</p> <p>Remark: By "compact," I consistently mean "compact Hausdorff."</p>

http://mathoverflow.net/questions/48718/is-every-compact-topological-ring-a-profinite-ring/48735#48735 Martin Brandenburg 2010-12-09T09:33:22Z 2010-12-09T09:33:22Z

<p>Every compact topological ring has "enough" open ideals and is thus profinite. See for example Sect. 5.1 in</p> <p>Luis Ribes, Pavel Zalesski, <em>Profinite Groups</em>, Ergebnisse der Mathematik und ihrer Grenzgebiete, 3. Folge</p>