Prove that if $M$ is finitely generated then it is Artinian.

Question

Let $M $ be a semisimple $R$-module. Prove that if $M$ is finitely generated then it is Artinian.

To show this we have to prove that every non-empty collection of sub-modules of $M$ has a minimal element. Let $C$ be a non-empty collection of sub-modules of $M$.

Also $M$ is finitely generated say by $\{x_1,x_2,\ldots,x_n\}$. How to arrive at the proof?

Answer 1

Write $M$ as the direct sum of its irreducible submodules. Each element of the generating set can be written as a sum of elements from finitely many of the irreducible submodules. Then $M$ is a direct sum of finitely many irreducible submodules. This implies $M$ is Artinian.

Answer 2

Without any assumptions about $R$, this is false. For example, consider $\mathbb{Z}$ as a $\mathbb{Z}$-module. It's finitely generated (the set $\{1\}$ is a generating set), but the collection of submodules $$\mathbb{Z}\supset 2\mathbb{Z}\supset 4\mathbb{Z}\supset 8\mathbb{Z}\supset\cdots$$ has no minimal element.

Answer 3

I am not sure I fully understand the question. As Hoot said, "is $R$ artinian?" to which you replied as "yes". But you also have the finitely generated assumption. In which case the assumption that $M$ is semisimple is completely unnecessary. A finitely generated module over an Artinian ring is Artinian. Because, we have the exact sequence, $$ 0 \to K \to R^n \to M \to 0 $$ Since $R^n$ is Artinian (your assumption) then so are both $K$ and $M$.