# Rings with all non-prime ideals finitely generated

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.

• 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.
– YCor
Jul 8 '18 at 10:10

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.

• I see ... thanks ... do you have any local domain counterexample ? Or to ask more ... a Valuation ring counterexample ... ?
– user111492
Jul 3 '18 at 12:00
• It would be reasonable to ask about a domain in a separate question.
– YCor
Jul 3 '18 at 12:12
• There is no domain counterexample. I will explain this at the other question. Jul 3 '18 at 16:42
• yes ... the answer is trivially silly for domains ...
– user111492
Jul 3 '18 at 17:36
• 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.
– user111492
Jul 3 '18 at 17:41