It is easy to show that if $M$ is a Noetherian $R$-module then $R/\mbox{ann}(M)$ is a Noetherian ring. Is there a similar (or dual) result for Artinian modules?
If $M$ is an Artinian $R$-module then $R/\mbox{ann}(M)$ is an Artinian ring?
3
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commutative-algebra
modules
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1The title you picked for your question is almost completely unrelated to the question itself! – 2011-07-08
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0I hope there is no mathematical term for "relevant", but what I meant was that if any Artinian module can be reduced to an Artinian module over an Artinian ring (as is the case for Notherian modules), then there is no point considering Artinian modules over non-Artinian rings. – 2011-07-08
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0simple modules, and finite length modules are intensively studied for all rings, including non-artinian rings. That's what representation theory mostly does! – 2011-07-08