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Let $A$ be a complete noetherian local ring, and $B \subset A$ a closed subring. Is there any chance that $B$ is also automatically noetherian?

I believe the answer is no, and that the same no-topology example of a nonnoetherian subring of a noetherian ring as given here works in the complete local case. Namely, let $A = k[[x,y]]$ for some field $k$ and let $B$ be the closed $k$-algebra of $A$ topologically generated by $\{x, xy, xy^2, xy^3, \ldots \}$. But I am a little anxious that I am overlooking something and $y$ is a limit of elements of $B$ in some way so that $A = B$. Or possibly $B$ happens to be noetherian in some other way. Can you put my mind to rest?

sibilant
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  • Oh, and if what I think is true is indeed true, does anyone know any out-of-left-field algebra-only criteria to guarantee that such a is noetherian? Apart from, like, $\mathbb m_B^2 = (\mathbb m_A \cap B)^2$ being open in the subspace topology from $A$ / containing $\mathbb m_A^n \cap B$ for some $n$, I mean. – sibilant Mar 10 '21 at 20:48
  • Trying first with $A=k[[x_1,\ldots,x_n]]$ I would check with $B$ a subring containing $k$, which is closed in $A$'s topology, and such that $(\mathfrak{m}_A)^N \subset B$ for some $N$ – reuns Mar 10 '21 at 21:16

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