$\newcommand{\Rad}{\operatorname{Rad}}$Let $M$ be a left $R$-module with $\Rad(M)\neq 0$ where $\Rad(M)$ is defined as the intersection of all maximal submodules of $M$. Thus, if $K = \bigcap_{i\in I} N_i$ where $N_i$ is a maximal submodule of $N$, then it is clear that $\Rad(M)\subseteq K$. But what about the converse of this statement ? That is, if $0\neq \Rad(M)\leq K\leq M$, is there a family $\lbrace N_i \rbrace_{i\in I}$ of maximal submodules of $M$ such that $K = \bigcap_{i\in I} N_i$ ?
In $\mathbb{Z}_{60}$ for example, every submodule containing the radical $\Rad(M) = 30\mathbb{Z}_{60}$ is an intersection of maximal submodules. It seems to me that in $\mathbb{Z}_{n}$ in general, every submodule containing the radical is an intersection of maximal submodules. If this property is true (in $\mathbb{Z}_{n}$, artinian and noetherian modules), how to prove this ?
Any help (proofs, indications, references) about modules satisfying this property would be appreciated. Thanks in advance.
Rings over which every submodule of every right module is an intersection of maximal submodules of the module are called right V-rings in honor of Villamayor.
Commutative $V$ rings are just von Neumann regular rings, of which $\Bbb Z_{30}$ is an example. But $\Bbb Z_{8}$ is not a V-ring, since the ideal generated by 4 is not an intersection of maximal ideals.
The reason you can't find any among quotients of $\Bbb Z$ is that quotienting by a semiprime ideal leaves a semisimple ring, which is a V ring.
To get an example of a ring that doesn't have this property, take the ring of continuous functions on the interval $[0,1]$ into $\Bbb R$ (pointwise addition and multiplication, regular topology.) Its Jacobson radical is zero, but it is not von Neumann regular, hence it has an ideal which isn't an intersection of maximal ideals.
For that matter, the the module $\Bbb Z_\Bbb Z$ is a Noetherian module which lacks the property.
This ring can be used to create a ring with nonzero radical having the same property. For example, take $\Bbb Z\times \Bbb Z$ with addition $(n,m)+(n',m')=(n+n',m+m')$ and multiplication $(n,m)(n'm')=(nn',nm'+mn')$. The ideal $\{0\}\times \Bbb Z$ is nilpotent, so the radical isn't zero. The other ideals are exactly the same structure as $\Bbb Z$, so the ideal generated by $(4,0)$ is not an intersection of maximal ideals, for example.