2

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.

Jendrik Stelzner
  • 17,081
Learnmore
  • 31,675
  • 10
  • 110
  • 250
  • 1
    A minor comment: writing $R-$ module means "$R$ minus module", whereas $R$-module means $R$-module. (Putting the character - inside MathJax makes it a minus sign.) – Zev Chonoles Jun 27 '15 at 06:20
  • Do you know of radical and Artin-Wedderburn theorem? – Orat Jun 27 '15 at 06:28
  • YES what to do next @Orat – Learnmore Jun 27 '15 at 06:31
  • Sorry, I had a misconception about finitely generated part (So forget about AW theorem). Semi simple part should be proven from the fact that radical is zero. May I ask your definition of semi simple module? – Orat Jun 27 '15 at 06:45
  • which can be written as direct sum of simple modules – Learnmore Jun 27 '15 at 07:03

2 Answers2

5

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 $\bar{x} \in M/N$).

Orat
  • 4,120
  • How do you know the collection $C$ is non-empty – Learnmore Jun 27 '15 at 08:46
  • If $M$ has a maximal submodule $U$ then $U$ is an element of $\mathcal{C}$. If not then it means $M = 0$ and it is obviously semisimple and finitely generated. – Orat Jun 27 '15 at 08:52
  • @Orat,, If $M$ has a maximal submodule $U$, then the collection is non-empty otherwise either $M=0$ or $M$ is simple and hence semisimple and finitely generated. – permutation_matrix Dec 25 '22 at 13:58
1

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.

arienda
  • 393