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Let $G$ be a group, and let $\mathbb{Z}[G]$ denote its group ring. Its profinite completion is the inverse limit over all ideals of finite index. By Benjamin Steinberg's answer here, this profinite completion agrees with the completed group algebra $\widehat{\mathbb{Z}}[[\widehat{G}]]$, where $\widehat{G}$ denotes the profinite completion of $G$.

I say that a profinite ring $\Lambda$ is Noetherian if every closed submodule of a topologically finitely generated $\Lambda$-module is (topologically) finitely generated. (This seems to be weaker than the ACC for closed ideals!)

Now assume that $\mathbb{Z}[G]$ is Noetherian (As @YCor points out, this implies that $G$ is noetherian and amenable). When can we conclude that its profinite completion (equivalently $\widehat{\mathbb{Z}}[[\widehat{G}]]$) is also Noetherian?

I'm especially interested in the case where $G$ is finitely generated and abelian. Even the case $G = \mathbb{Z}^r$ would be interesting to me.

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    $\begingroup$ The only groups $G$ for which $Z[G]$ is known to be noetherian are virtually polycyclic groups. (Conversely, if $Z[G]$ is noetherian, it is known that $G$ is noetherian and amenable. See this MO answer) $\endgroup$
    – YCor
    Commented Aug 7, 2021 at 9:07
  • $\begingroup$ How can we describe that completed ring for $G = \mathbb{Z}$ more concretely? $\endgroup$
    – Martin Brandenburg
    Commented Aug 8, 2021 at 22:58
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    $\begingroup$ @MartinBrandenburg The completed ring for $G = \mathbb{Z}$ is described here. It is the product of $\mathbb{Z}_q[[t]]$, where each $q$ is a prime power and the corresponding direct factor appears multiple times (number of times is equal to the number of Frobenius orbits on $\mathbb{F}_q^\times$). It certainly does not satisfy ACC, but this doesn't necessarily mean that there exists an infinitely generated closed ideal, since the union of the ascending chain may not be closed. $\endgroup$
    – stupid_question_bot
    Commented Aug 9, 2021 at 3:10

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