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Motivated by this question, I would like to ask:

If all non-prime ideals in a ring are finitely generated, then is the ring Noetherian? Can we at least say anything in the local case?

Note that for zero-dimensional rings, the answer is yes by the linked question. So we only need to think about rings of positive dimension.

NOTE: All our rings are commutative with unity.

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    $\begingroup$ you have erased your subsequent question mathoverflow.net/questions/304492/… after it's been answered (with a significant effort for the answerer to reply it); this is considered as incorrect behavior and you should undelete it. $\endgroup$ – YCor Jul 8 '18 at 10:10
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No.

If $\mathbb Z_{p^{\infty}}$ is a Prufer group for prime $p$, then its endomorphism ring is isomorphic to the ring $\mathbb Z_p$ of $p$-adic integers. Hence $\mathbb Z_{p^{\infty}}$ is a $\mathbb Z_p$-module, and we can form the idealization $R:=\mathbb Z_p\oplus \mathbb Z_{p^{\infty}}$. The ideals of this ring are known: they are of the form $0\oplus H$ for a subgroup $H\leq \mathbb Z_{p^{\infty}}$ or of the form $(p^k)\oplus \mathbb Z_{p^{\infty}}$ for some nonnegative integer $k$. (See Example 1 of Paul A. Froeschl, Chained rings. Pacific J. Math. Volume 65, Number 1 (1976), 47-53.) In particular, $R$ is a local ring with exactly one non-finitely generated ideal $I:=0\oplus \mathbb Z_{p^{\infty}}$. This ideal is prime because $R/I$ is isomorphic to $\mathbb Z_p$, a domain.

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  • $\begingroup$ I see ... thanks ... do you have any local domain counterexample ? Or to ask more ... a Valuation ring counterexample ... ? $\endgroup$ – user111492 Jul 3 '18 at 12:00
  • $\begingroup$ It would be reasonable to ask about a domain in a separate question. $\endgroup$ – YCor Jul 3 '18 at 12:12
  • $\begingroup$ There is no domain counterexample. I will explain this at the other question. $\endgroup$ – Keith Kearnes Jul 3 '18 at 16:42
  • $\begingroup$ yes ... the answer is trivially silly for domains ... $\endgroup$ – user111492 Jul 3 '18 at 17:36
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    $\begingroup$ In domains, $ r \notin rI$ for every $0\ne r \in R$ and every proper ideal $I$ of $R$. Now if $0\ne P$ is prime ideal and $0\ne r \in P$ then $r^2 \in rP$ but $r \notin rP$ , so $rP$ is not a prime ideal hence f.g. hence $rP \cong P$ (as $R$-modules) is f.g. $\endgroup$ – user111492 Jul 3 '18 at 17:41

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