Prove that $M$ is finitely generated and semisimple, if Artinian and the intersection of all maximal submodules is zero

Question

Let $M$ be a $R$-module. It is given that intersection of all maximal submodules of $M$ is the zero module. Moreover, the module is given to be Artinian.

Prove that $M$ is finitely generated and it is a semisimple module.

I feel I should somehow try to construct a non-empty collection of submodules of $M$ and then use the Artinian condition. But I could not construct such a collection.

Any help.

Answer 1

Let $\mathcal{M}$ be the set of maximal submodules of $M$.

Lemma: There exist a finite subset $\mathcal{N}$ of $\mathcal{M}$ such that $\bigcap \mathcal{N} = 0$.

Proof: Let $\mathcal{C}$ be the collection of the intersection finite number of maximal submodules. As $M$ is artinian, $\mathcal{C}$ must have a minimal element $N = \bigcap \mathcal{N}$. If it is not zero then there is a nonzero element $x \in N$ and a maximal submodule $M'$ of $M$ not containing $x$ (from the hypothesis $\bigcap \mathcal{M} = 0$). Then $M' \cap \bigcap \mathcal{N} < \bigcap \mathcal{N}$, which contradicts to the minimality of $\bigcap\mathcal{N}$.

Now for a proof of your question.

Proof: From the above lemma, the kernel of map $M \to \bigoplus_{N \in \mathcal{N}} M/N$ is $\bigcap \mathcal{N} = 0$. So it is semisimple (as $M$ is isomorphic to a submodule of $\bigoplus M/N$) and finitely generated (as $M/N$ is simple and hence $M/N = R\bar{x}$ for some $\bar{x} \in M/N$).

Answer 2

From the artinian hypothesis you may deduce that the zero submodule is an intersection of finitely many maximal submodules. Therefore $M$ is isomorphic to a submodule of a finitely generated semisimple module, hence it is semisimple and finitely generated.