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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.

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  • $\begingroup$ 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? $\endgroup$ – Matt Jun 29 '20 at 8:03
  • $\begingroup$ In fact, your question has an answer here: math.stackexchange.com/questions/403070/… $\endgroup$ – Matt Jun 29 '20 at 8:05
  • $\begingroup$ 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. $\endgroup$ – José Daniel Castilla del Valle Jun 29 '20 at 8:42
  • $\begingroup$ Oh, I'm sorry. I misunderstood entirely. Of course you're right. $\endgroup$ – Matt Jun 29 '20 at 8:59
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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 $M$.

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  • $\begingroup$ 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. $\endgroup$ – José Daniel Castilla del Valle Jun 29 '20 at 10:26
  • $\begingroup$ @JoséDanielCastilladelValle I'm not sure what is the point where you're stuck at. $\endgroup$ – egreg Jun 29 '20 at 10:33
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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 $M$.

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

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