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$R$ ring and $M$ a left $R$-module. Call $\mathrm{Soc}\;M$ the sum of all the simple submodules of $M$. Then

$M$ is artinian if and only if $\lambda_R(\mathrm{Soc}(M))<\infty$ and for very $0\neq Q\subset M$ we have $Q\cap\mathrm{Soc}\;M\neq0$

Could you help me solve this exercise?

($\lambda_R$ denotes the composition length)

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    @AlexM Hi, thanks for clarifying, but next time please @ me in. I didn't realize the question was fixed, so it delayed my answer :)2012-06-03

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If $M$ is artinian, its socle $\newcommand\soc{\operatorname{soc}}\soc M$, which is a submodule, is also artinian. Since $\soc M$ is semisimple, being artinian it is also noetherian, and therefore has finite length. Moreover, if $Q$ is a non-zero submodule of $M$, then it is also artinian, and has a non-zero socle: to check this, one has to show that $Q$ contains a simple module, but if it didn't we could easily construct a non-stopping decreasing sequence of submodules. Finally, $\soc Q$, being a non-zero sum of simple submodules of $Q$ (so of $M$) intersects $\soc M$ non-trivially.

Can you do the converse using similar ideas?

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    I took a descending chain $\ldots\subset M_2\subset M_1\subset M$ with $M_i\neq 0$, then I intersecated with $\mathrm{Soc}\;M$, and since it is artinian we have for some $n$ that $M_n\cap\mathrm{Soc}\;M=M_{n+1}\cap\mathrm{Soc};M$, I want to prove that $M_n\subset M_{n+1}$. I proved that every simple submodule of $M_n$ is contained in $M_{n+1}$ but I don't know how to continue, any help?2012-05-28
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With the clarification made that we are talking about composition length, and that the socle should have finite length, the proposed statement is false.

A module is called finitely cogenerated if it has a finitely generated essential socle. (Here the "finitely generated" may be swapped for "finite composition length", since the two notions are identical for a semisimple module.)

So, the question amounts to asking "Show a module is Artinian iff it is finitely cogenerated." From Mariano's answer, we know that Artinian modules are finitely cogenerated, but the converse is false. A correct statement would be that "a module $M$ is Artinian iff $M/N$ is finitely cogenerated for every submodule $N$ of $M$."

There exists a non-Artinian commutative ring $R$ such that $R_R$ is a finitely cogenerated module. The first example I located is due to Osofsky and can be found in Lam's Lectures on Modules and Rings at the top of page 514.

If you request, I can duplicate it here.