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$\newcommand{\Rad}{\operatorname{Rad}}$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.

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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.

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  • $\begingroup$ thanks for your answer. The $V-$ rings are an interesting case, but i think they present a strong condition in comparison with what is required in my questions. May i add one futher point: $\mathbb{Z}_{60}$ as a $\mathbb{Z}$ module has a submodule ($4\mathbb{Z}_{60}$) wich is not an intersection of maximal submodules and also $4\mathbb{Z}_{60}$ does not contain the radical. $\endgroup$ – Demba Mar 17 '15 at 1:48
  • $\begingroup$ @Demba oh yes, you're right. Somehow I was doing 30 in my head instead of 60 $\endgroup$ – rschwieb Mar 17 '15 at 2:27
  • $\begingroup$ Thanks a lot Mr rschwieb, effectively $\mathbb{Z}_{\mathbb{Z}}$ does not verify the property and you can remark that his radical is zero and it is the same with $\mathbb{Z_{30}}$. Is there a $V$-ring with nonzero radical ? $\endgroup$ – Demba Mar 17 '15 at 12:40
  • $\begingroup$ Thanks a lot Mr rschwieb, effectively $\mathbb{Z}_{\mathbb{Z}}$ does not verify the property and you can remark that his radical is zero and it is the same with $\mathbb{Z_{30}}$. Is there a $V$-ring with nonzero radical ? i recall that one of the conditions of the property above is tha radical of $M$ must be non zero. And my formal conjecture is if $M$ is artinian and noetherian with non zero radical, then the property is true. $\endgroup$ – Demba Mar 17 '15 at 12:51
  • $\begingroup$ @Demba Oh, sorry I overlooked the condition about the radical being nonzero. It's a little unnatural, and at any rate, I can modify the example around this requirement. $\endgroup$ – rschwieb Mar 17 '15 at 16:46

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