# If $M$ is an Artinian $R$-module then $R/\mbox{ann}(M)$ is an Artinian ring? [duplicate]

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?

## marked as duplicate by rschwieb, Eric Wofsey, user26857, Claude Leibovici, Tom-TomSep 22 '15 at 8:25

• The title you picked for your question is almost completely unrelated to the question itself! – Mariano Suárez-Álvarez 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-Álvarez Jul 8 '11 at 22:24

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 $\mathbb{Z}[1/p]/\mathbb{Z}$ is a non finitely generated Artinian $\mathbb{Z}$-module and $\mathbb{Z}$ is not Artinian. Thus if $M$ is an Artinian $R$-module, then $R$ is not necessary Artinian. (See the article of wiki about Artinian module.)
• Why is $R/\mathrm{Ann}(M)$ Artinian exactly? – D_S Sep 22 '15 at 4:12