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.