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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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    Is this homework? Are you conjecturing this is true? You picked the statement from a book? Have you tried any approach? Can you do at least one of the two implications? (You've asked almost a 100 questions by now!) Probably telling us what $\lambda_R$ means would also be a good idea :D2012-05-18
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    I'm doing exercises by myself, I found this in an old exam. $\lambda_R(M)$ is the length of $M$ as an $R$-module2012-05-18
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    @AlexM Please define "length" specifically. There is more than one notion of length. The statement is false for the first two notions that came to my mind. It obviously can't mean "composition length" because there are Artinian modules over commutative rings with infinite composition length. It also can't be [uniform dimension](http://en.wikipedia.org/wiki/Uniform_dimension) because there exists a commutative non-Artinian uniserial ring with a minimum ideal which serves as a counterexample.2012-05-18
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    you're right, it should be $\lambda_R(\mathrm{Soc}(M))$ and the notion is the "composition length"2012-05-18
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    Please add all this information to the question itself, so that people do not have to read all the comments! :D2012-05-18
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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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