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Let $A$ be a local ring with a maximal ideal $\mathfrak{m}$ finitely generated (not principal). Is there a sufficient condition for $A$ to be noetherian?

For example, we know that the completion $\hat{A}$ will be noetherian(http://stacks.math.columbia.edu/tag/05GH).

How can we "descend"?

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  • $\begingroup$ Well, by Cohen's theorem you could just stipulate that the rest of the prime ideals are finitely generated too.. or that $A$ is a 1-dimensional domain. $\endgroup$
    – Neil Epstein
    Commented Mar 4, 2015 at 16:40
  • $\begingroup$ asserting that all prime ideals are finitely generated is almost the definition of noetherian, not very interesting. $\endgroup$
    – prochet
    Commented Mar 4, 2015 at 16:47
  • $\begingroup$ @prochet: would you remind something about whether $A$ is noetherian without further hypotheses? I think I already saw a counterexample on MO but I can't find it. $\endgroup$
    – YCor
    Commented Mar 4, 2015 at 16:59
  • $\begingroup$ Make lots of counterexamples with ordered abelian semi-groups. Here's one: $G=Z⊕Z$ with lexicographic order, $S=\{(i,j)≥(0,0)\},M=\{(i,j)≥(0,1)\},P=∩\{(i,j)>(0,n):n∈Z\},R=k[T^S]$. Then $m=(T^M)$ is a maximal ideal of $R$ with one generator $T^{(0,1)}$, $(T^P)$ is a non-finitely generated prime ideal, and $A=R_m$ is the local ring. $\endgroup$
    – David Lampert
    Commented Mar 4, 2015 at 23:22

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