On the sum of all the simple submodules of a module $R$ 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)
 A: 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?
A: 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.
