How to prove finitely many? which route is easier

abstract-algebrafinitely-generatedhomological-algebramodulessemi-simple-rings

I want to prove that: If $N$ is finitely generated semi-simple $R-$module, then $N$ is a sum of finitely many simple submodules.

I know that if $N$ is a finitely generated $R-$module, then that the following three conditions are equivalent(I have seen its proof before).

  1. $N$ is a sum of simple modules.

  2. $N$ is a direct sum of simple modules.

  3. Given any submodule $M \subset N,$ there exists a unique submodule $M' \subset N$ such that $N = M \oplus M'.$

And any module satisfying these conditions is called a semi-simple module.

Then, to prove my first statement above, I know that I can adjust my proofs either for $2 \implies 1$ in the statement above or for $3 \implies 1.$ I prefer to prove my first statement to prove that: $N$ is a direct sum of simple modules implies $N$ is a sum of finitely many simple modules.

Here is my trial:

Assume that $N$ is a direct sum of simple modules, then, by definition of direct sum, $N$ is a sum of simple modules. But then how can I prove that it is a finite sum? could anyone help me in proving this please?

EDIT:
Maybe proving that every sequence of submodules of $N$ is stationary is a better way of proving my first statement but in that case I also do not see the proof of finitely many.

Any help is appreciated!

EDIT:

Since $N$ is semi-simple by assumption, then $N$ is a direct sum of (possibly infinitely many) simple modules. This infinitely many sum can come from either one of those two cases:

$(1)$ $N$ is infinitely generated i.e. generated by an infinite set. But this case is excluded because by assumption our $N$ is finitely generated.

$(2)$ at least one of the generators of $N$ is infinitely generated, which is not our case because each of the generators is simple i.e., it has only two submodules.

Is that reasoning correct or there is a more succinct proof?

Best Answer

A direct sum of infinitely many nonzero (right) modules is not finitely generated.

Indeed, if you're given a family of nonzero module $M_\lambda$, for $\lambda\in\Lambda$, and $x_1,\dots,x_n$ are elements of the direct sum thereof, there exists a finite set $F\subseteq\Lambda$ such that $$ x_1R+x_2R+\dots+x_nR\subseteq \bigoplus_{\lambda\in F}M_\lambda $$ and this finite direct sum is a proper submodule of the whole direct sum.

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