# local ring all whose non-maximal ideals are finitely generated

Let $(R, \mathfrak m)$ be a commutative local ring such that every non-maximal ideal is finitely generated. Then, is $R$ Noetherian i.e. is $\mathfrak m$ finitely generated ideal ?

It is easy to see that the answer is yes when $R$ is integral domain by considering an ideal $r\mathfrak m$ for $r\in \mathfrak m$ and noting $r\mathfrak m\ne \mathfrak m$ and $r\mathfrak m \cong \mathfrak m$ (as $R$-modules) .

If $\mathfrak m$ is not finitely generated in a ring as above, then $\mathfrak m$ is an $R$-module all whose proper submodules are finitely generated, so from 1 (Proposition 1.1, Proposition 1.3 ) and 2 (Proposition 1.2) , one can see the following : $ann_R (\mathfrak m)$ is a prime ideal, $\mathfrak m$ is divisible as $R/ann_R (\mathfrak m)$-module and it is either torsion-free or every element is a torsion. The ring (possibly non-commutative) $End_R (\mathfrak m)$ is local i.e. the set of non-units forms an ideal. Also, $Ass_R (\mathfrak m)=\{P\}$ is singleton and that single associated prime is the set of all zero-divisors of $\mathfrak m$ . If the associated prime ideal is $P=ann_R (\mathfrak m)$, then $\mathfrak m$ is torsion-free over $R/ ann_R (\mathfrak m)$ and in that case $\mathfrak m$ is isomorphic to the fraction field of $R/ann_R (\mathfrak m)$ as $R$-modules and $ann_R (\mathfrak m)$ is not maximal and $\mathfrak m$ is not Artinian as $R$-module. If $ann_R(\mathfrak m) \notin Ass_R(\mathfrak m)$, then the associated prime ideal is maximal and every proper non-maximal ideal of $R$ has finite length, so in particular $\mathfrak m$ is an Artinian $R$-module. In any case, $ann_R (\mathfrak m)$ is not a maximal ideal.

• @YCor Please note this preprint: arxiv.org/abs/1806.03735 submitted to the arxiv on Sun, 10 Jun 2018 22:23:38 GMT — just a few hours after your answer here. I'm not sure what's going on...? Jun 14, 2018 at 1:11
• @ZachTeitler thanks very much for pointing out. The OP possibly ignores that copying others' arguments without a reference is considered as plagiarism.
– YCor
Jun 14, 2018 at 7:13
• @users solving a mathematical problem when you know the answer beforehand is often easier. In this case nothing is hard and indeed there are many ways to reach the conclusion, and visibly you didn't manage to do so by yourself. Second, part of the argument in the arxiv paper is on the non-local case, where you essentially copy my account of Keith's argument.
– YCor
Jun 14, 2018 at 16:05
• @YCor: Note also that the references in the arxiv preprint are essentially the same as the references in the problem statement on the current page. Looks very fishy. Jul 8, 2018 at 10:17
• @ZachTeitler thanks very much for your worthwhile information. A discussion can be found on meta: meta.mathoverflow.net/questions/3798/…
– YCor
Jul 8, 2018 at 22:21

There exists no such non-noetherian local ring.

Below I assume by contradiction that we have such a ring.

(a) The first observation is a particular case Proposition 1.2(a) in your reference to Armendariz: for every $r\in R$, we have $r\mathfrak{m}\in\{\mathfrak{m},\{0\}\}$. Indeed, $r\mathfrak{m}$ is a quotient of $\mathfrak{m}$; if it's nonzero, it's quotient by a proper submodule and hence is also not noetherian, which implies $r\mathfrak{m}=\mathfrak{m}$.

(b) case when $R$ is a domain (I just expand your argument). Choose $r\in\mathfrak{m}\smallsetminus\{0\}$. Since $R$ is a domain, by (a) we have $r\mathfrak{m}=\mathfrak{m}$, and in particular so $r\in r\mathfrak{m}$, and since $R$ is a domain this implies $1\in\mathfrak{m}$, a contradiction.

(c) let us check that $R$ is necessarily of Krull dimension 0. Indeed if $P$ is a nonmaximal prime ideal, then $P\neq\mathfrak{m}$ and hence is finitely generated, so $R/P$ is a counterexample to (b).

(d) Now we use Proposition 1.2(b) in Armendariz: $P=\mathrm{Ann}_R(\mathfrak{m})$ is prime (as you've already mentioned). The argument is easy: indeed, for $x,y\notin P$, by (b) we have $x\mathfrak{m}=y\mathfrak{m}=\mathfrak{m}$, which implies $xy\mathfrak{m}\neq 0$, so $xy\notin P$.

(e) Combining (c) and (d), the only option for $\mathrm{Ann}_R(\mathfrak{m})$ is that it's equal to $\mathfrak{m}$. Thus $\mathfrak{m}^2=\{0\}$. Then $\mathfrak{m}$ is an infinite-dimensional vector space over the field $R/\mathfrak{m}$, and all its hyperplanes are ideals. In particular they fail to be finitely generated, and this is a contradiction.

Edit 1: the argument can be extended to show that there is no commutative ring at all with these conditions (non-noetherian such that all non-maximal ideals are finitely generated). That is, assuming $R$ local is unnecessary. In other words, a commutative ring is noetherian if and only if all its non-maximal ideals are finitely generated ideals.

Indeed, (a),(b),(c),(d) work with no change for every given infinitely generated maximal ideal $\mathfrak{m}$. Let us adapt (e):

(e') for every infinitely generated maximal ideal $\mathfrak{m}$, by combining (c) and (d), its annihilator is another maximal ideal $\mathfrak{m}'$. Then $\mathfrak{m}$ can be viewed as a $R/\mathfrak{m}'$-vector space. Hence the lattice of ideals contained in $\mathfrak{m}'$ can be identified to the lattice of $R/\mathfrak{m}'$-vector subspaces of $\mathfrak{m}$. In particular, the condition that all its elements except whole $\mathfrak{m}$ are noetherian, implies that $\mathfrak{m}$ has finite dimension (as vector space over $R/\mathfrak{m}'$), hence is finitely generated as an ideal, a contradiction. We deduce every maximal ideal is finitely generated, and hence (since all non-maximal ones are also finitely generated by assumption) that $R$ is noetherian.

Edit 2:

As mentioned by Keith in the comments, the non-local case can be handled in an even easier way:

let $R$ be a ring (commutativity is unnecessary) in which every non-maximal left ideal is finitely generated, and having at least two maximal left-ideals. If $\mathfrak{m}$ is a maximal left-ideal and $\mathfrak{m}'$ is another one, then $\mathfrak{m}\cap \mathfrak{m}'$ is finitely generated, and $\mathfrak{m}/(\mathfrak{m}\cap \mathfrak{m}')\simeq (\mathfrak{m}+\mathfrak{m}')/\mathfrak{m}'=R/\mathfrak{m}'$ is a simple module, so $\mathfrak{m}$ is also finitely generated.

• Ah sorry I didn't get why is $\mathfrak m$ infinite dimensional over $R/\mathfrak m$ and the subsequent hyper-plane argument you make ... you could also use Proposition 1.3 of Weakley's paper that if $\mathfrak m$ is not finitely generated then $ann_R (\mathfrak m)$ is not maximal , but we know it is prime , so there can be no zero-dimensional counterexample.
– user111524
Jun 10, 2018 at 15:27
• The argument for domains I had in mind is like this (very similar but not by contradiction ) : If $\mathfrak m=0$ we are done. O.w. take $0\ne r \in \mathfrak m$ . Then $r \mathfrak m=\mathfrak m$ implies $r=rx$ for some $x \in \mathfrak m$ , then $x=1$, impossible ! So $r \mathfrak m$ is not maximal , hence finitely generated and moreover since $R$ is a domain and $r \ne 0$, we get $r\mathfrak m \cong \mathfrak m$ is f.g.
– user111524
Jun 10, 2018 at 15:29
• @users right, taking the square is useless, I removed it. It's the same argument up to frills.
– YCor
Jun 10, 2018 at 17:11
• Curious about the meaning of the downvote.
– YCor
Jun 10, 2018 at 20:10
• Can drop the assumption that $R$ is local: If $R$ is a commutative ring whose non-maximal ideals are f.g., then $R$ is Noetherian. The local case is handled above. If not local, and $M, N\lhd R$ are distinct maximal ideals, then $M$ is f.g. over $M\cap N$, which is f.g. Jun 10, 2018 at 22:52