# Module of finite length $\implies$ finite direct sum of indecomposable modules

I'm trying to solve the following question.

Let $M$ be an $R$-module of finite length (i.e, both Artinian and Noetherian). Prove that it is isomorphic to a finite direct sum of indecomposable submodules.

I was thinking I could induct on the length, though I'm having a difficulty in the induction step.

Base case holds trivially. If for all modules of length $<n$, the proposition holds, consider a module of length $n$, which has the following decomposition:$$M \supset M_1 \supset \ldots \supset M_n \supset (0)$$

By the Induction hypotheses, $M_1$ is isomorphic to a finite direct sum of indecomposable modules. Now, if I show that the following short exact sequence splits, then I'm done. $$0 \rightarrow M_{1} \rightarrow M \rightarrow M/M_1 \rightarrow 0$$ But I'm unable to do this. Any help will be appreciated.

Hint: The size of a decomposition as a sum of nontrivial modules is bounded by the length of the module. Do you see why, and can you proceed from here? The crucial thing is that the length of any decomposition series bounds the size of any other nontrivial filtration - have you already proved this?

• Oh, by finite length, I meant the module is both an Artinian as well as a Noetherian Module. – Karthik May 20 '16 at 6:00
• @Karthik: Yes - but I think the example is valid nonetheless? – Hanno May 20 '16 at 6:02
• Yes, I do apologize. Not simple, but indecomposable. I'll edit the question. – Karthik May 20 '16 at 6:06
• Ah! That does change matters :) – Hanno May 20 '16 at 6:07
• @Karthik Do you have any questions? – Hanno May 21 '16 at 19:05

If $M$ is indecomposable, you're done. Otherwise, $M=X\oplus Y$, with $X$ and $Y$ both nonzero.

Now prove that the lengths of $X$ and $Y$ are less than the length of $M$.

Hint: build composition series for $X$ and $Y$ and paste them up.