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.

12more comments