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I haven't received any substantial responses to a similar question on math.stackexchange, so let me try here.

Let $R$ be a profinite ring (that is a projective limit of finite rings). Assume that the ring $R$ is local, of maximal ideal $m$, and assume that $m$ is open for the pro-finite topology. Then is $R$ necessarily $m$-adically separated ? $m$-adically complete?

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    $\begingroup$ Since $R$ is local, it is an inverse limit of finite local rings, all with the same residue field, and its maximal ideal is the inverse limit of those maximal ideals, so $m$ is closed in $R$ with finite index, hence automatically open. The same analysis shows that $R$ is $m$-adically separated since each artinian quotient from the inverse-limit description is max-adically separated. Of course higher powers of $m$ need not be closed (let alone open). Since some power of $m$ vanishes in each artin local quotient of $R$, $m$-adic completeness holds too. $\endgroup$
    – user76758
    Dec 17, 2013 at 21:22
  • $\begingroup$ @user76758: I don't understand the proof of completeness. Is this true always? I think it's OK if $R$ is noetherian (since then the powers of $m$ are closed; the map to the completion $i : R \to \hat R$ is injective with dense image, $R$ is compact, and by the assumption $\hat R$ is Hausdorff, hence $i$ is an isom). $\endgroup$
    – fherzig
    Jul 31, 2019 at 20:04

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