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
$\begingroup$
commutative-algebra
modules
-
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
1 Answers
5
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.)
-
3Why is $R/\mathrm{Ann}(M)$ Artinian exactly? – 2015-09-22