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It is easy to show that if $M$ is a Noetherian $R$-module then $R/\mbox{ann}(M)$ is a Noetherian ring (my thanks to all those who enlightened me on this). Is there a similar (or dual) result for Artinian modules?

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The title you picked for your question is almost completely unrelated to the question itself! –  Mariano Suárez-Alvarez Jul 8 '11 at 3:37
    
I 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. –  ashpool Jul 8 '11 at 19:17
    
simple modules, and finite length modules are intensively studied for all rings, including non-artinian rings. That's what representation theory mostly does! –  Mariano Suárez-Alvarez Jul 8 '11 at 22:24

1 Answer 1

If $M$ is an Artinian $R$-module, then so is any submodule and any quotient of $M$. Thus if $M$ is finitely generated, then $R/\mathrm{Ann}(M)$ is Artinian. But there exists non finitely generated Artinian $\mathbb{Z}$-module $\mathbb{Z}[1/p]/\mathbb{Z}$. And $\mathbb{Z}$ is not Artinian. Thus if $M$ is Artinian $R$-module, then $R$ is not necessary Artinian. (See the article of wiki about Artinian module.)

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In fact, every ring has artinian modules, becaue it has maximal left ideals. –  Mariano Suárez-Alvarez Jul 9 '11 at 20:59
    
right. if we donot require the ring is commutative. then,the maximal left ideal is not necessary a two side ideal. –  wxu Jul 10 '11 at 6:04
    
I don't see how that is related... –  Mariano Suárez-Alvarez Jul 10 '11 at 6:07
    
what i mean is that if ann(M) is an two side ideal, we may view M as a R/ann(M) module. anyway, since the op taged the post as commutative algebra, i think he prefer to thinking commutative rings. –  wxu Jul 10 '11 at 6:21

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