# The length of a semisimple module is finite if it is finitely generated

I'm trying to prove the next proposition:

Let $$M$$ be a semisimple module over a ring R. Then $$L(M)\in \mathbb{N}$$ if and only if $$M$$ is finitely generated.

Where $$L(M)$$ is the length of $$M$$, which by definition is the length of a proper composition series.

For the $$( \Longrightarrow )$$ implication I use that have finite lenght implies that the module is artinian and noetherian and since being noetherian implies finitely generated, it's done.

But I'm stuck in the $$(\Longleftarrow)$$ implication; Any help is appreciated, thank you.

• My algebra isn't great, but I would try a proof by contradiction. Suppose that $M$ is not finitely generated. Take your infinitely many generators and create a composition series of infinite length in the obvious way. Contradiction. Would that work? – Matt Jun 29 '20 at 8:03
• In fact, your question has an answer here: math.stackexchange.com/questions/403070/… – Matt Jun 29 '20 at 8:05
• I'm trying to prove that if $M$ is finitely generated and semisimple, then $M$ have finite length. I think you and the link have proved the implications that I've proved too. – José Daniel Castilla del Valle Jun 29 '20 at 8:42
• Oh, I'm sorry. I misunderstood entirely. Of course you're right. – Matt Jun 29 '20 at 8:59

It depends on what you know about semisimple modules.

The $$\Rightarrow$$ direction is OK: finite length implies the module is Noetherian, hence finitely generated.

For the $$\Leftarrow$$ part, consider the collection $$\mathscr{S}$$ of all finite length submodules of the finitely generated semisimple module $$M$$. The sum of any finite family of finite length submodules has finite length (easy proof) and the sum of all members of $$\mathscr{S}$$ is $$M$$, because every simple module has finite length and $$M$$ is the sum of its simple submodules.

Since $$M$$ is finitely generated, one of the members of $$\mathscr{S}$$ equals $$M$$.

• Now I have proved that, since M is finitely generated and the direct sum of simple submodules, this last one must be finite, and of course every simple submodule has finite length, but I don't see how to prove what it's supposed to be easy. Don't you have a hint about it? Thanks. – José Daniel Castilla del Valle Jun 29 '20 at 10:26
• @JoséDanielCastilladelValle I'm not sure what is the point where you're stuck at. – egreg Jun 29 '20 at 10:33

Suppose your finite generating set is $$\{g_1,\ldots, g_n\}$$, and $$M=\oplus_{i\in I}S_i$$ where all the $$S_i$$ are simple.

Then $$g_i\in \oplus_{i\in F_i}S_i$$ where $$F_i$$ is a finite subset of $$I$$, and $$\cup_1^n F_i=F$$ is still a finite set such that $$g_1,\ldots, g_n\in \oplus_{i\in F}S_i$$.

Therefore $$\oplus_{i\in I}S_i=\oplus_{i\in F}S_i$$, and the thing on the right hand side obviously has finite length, since modules of the form $$\oplus_{i=1}^jS_i$$ for $$j\in \{1,\ldots, n\}$$ obviously form a finite composition series for $$M$$.

• Thanks for answering. The $F_{i}$ comes of $M$ being finitely generated? – José Daniel Castilla del Valle Jun 30 '20 at 2:55
• @JoséDanielCastilladelValle No, that is true in any direct sum. Finite generation helps us say you only need finitely many pieces to generate the whole. – rschwieb Jun 30 '20 at 3:42

Suppose that $$M$$ is a finitely generated semisimple module.

Then, $$M$$ must be the direct sum of finitely many simple modules, each of which is trivially Artinian and Noetherian, and hence itself be Artinian and Noetherian.

Hence, $$M$$ must have finite length, and in fact, the length must equal the number of summands in the decomposition of $$M$$ as a direct sum of finitely many simple modules.