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](https://ac.els-cdn.com/0021869383900753/1-s2.0-0021869383900753-main.pdf?_tid=40dd23e0-afc5-4c52-b27f-df58928e4322&acdnat=1528634237_54e7eb164e9628fbab29c68b9a8c7e8a) (Proposition 1.1, Proposition 1.3 ) and [2](http://www.mscand.dk/article/viewFile/11699/9715) (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.


1. [Modules Whose Proper Submodules Are
Finitely Generated, 
WILLIAM D. WEAKLEY](https://ac.els-cdn.com/0021869383900753/1-s2.0-0021869383900753-main.pdf?_tid=40dd23e0-afc5-4c52-b27f-df58928e4322&acdnat=1528634237_54e7eb164e9628fbab29c68b9a8c7e8a)
2. [Rings with an almost Noetherian ring of fractions,Efraim Armendariz](http://www.mscand.dk/article/viewFile/11699/9715)



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




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*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.




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*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. 


