# Prove or Disprove: Finitely generated Artinian module is Noetherian.

Prove or Disprove: Finitely generated Artinian module is Noetherian.

I think it is true and I am trying to prove it. I am considering reducing the case to Artinian rings. Say $M$ is finitely generated Artinian $R$-module. Then $R/\mathrm{Ann}(M)$ is an Artinian $R$-module. Thus $R/\mathrm{Ann}(M)$ is Artinian as an $R/\mathrm{Ann}(M)$-module. That is, $R/\mathrm{Ann}(M)$ is an Artinian ring. We know that by Hopkins, $R/\mathrm{Ann}(M)$ is a Noetherian ring. Thus $M$ when seen as an $R/\mathrm{Ann}(M)$-module is Noetherian. I wonder how I can prove that $M$ is Noetherian as an $R$-module?

• See here for perhaps a partial answer. Not clear that the full answer is known. – rogerl Mar 18 '15 at 0:48
• In short: induct on the number of generators, and for cyclic modules use a simpler version of the argument given here. – Slade Mar 18 '15 at 0:52
• Can you elaborate? – user_m Mar 18 '15 at 1:02

The answer is trivially yes: If $M$ is finitely generated and artinian, then $R/\mathrm{Ann}(M)$ is an artinian ring, hence noetherian. Now just notice that the $R/\mathrm{Ann}(M)$-submodules of $M$ coincide with the $R$-submodules of $M$.

• Is that for the reason why we can treat any $R$-module M as an $R/Ann(M)$-module by defining $\bar{r}m=rm$? – user_m Mar 18 '15 at 2:15

Note that for any module element $m \in M$, there is an injection of $R$ modules $R/\operatorname{Ann}(m) \hookrightarrow M: r \mapsto r \cdot m$. If $M$ is finitely generated, I have the map $$R/\operatorname{Ann}(m_1) \oplus \dotsb \oplus R/\operatorname{Ann}(m_n) \twoheadrightarrow M$$ which is the sum of the inclusions. If $M$ is Artinian, then each $R/\operatorname{Ann}(m_i)$ is Artinian because it's a submodule. The structure of $R/\operatorname{Ann}(m_i)$ as an $R$ module is the same as its structure as an $R/\operatorname{Ann}(m_i)$ module, so we can say that $R/\operatorname{Ann}(m_i)$ is an Artinian ring, so a Noetherian ring, so a Noetherian $R$ module.

Now $M$ receives a surjection from a Noetherian module, so it is Noetherian.

I'm not really sure why you'd use $R/\operatorname{Ann}(M)$ here. Perhaps someone could enlighten me.

• Maybe since $\mathrm{Ann}(M)=\bigcap_{i=1}^n\mathrm{Ann}(m_i)$, and $R/\mathrm{Ann}(M)\hookrightarrow R/\operatorname{Ann}(m_1) \oplus \dotsb \oplus R/\operatorname{Ann}(m_n)$? – user26857 Jul 5 '17 at 21:54
• @user26857 Yes, but why is that necessary to the proof? And why is it easy to see that $R/\operatorname{Ann}(M)$ is Artinian, unless you set up the above map? – Eric Auld Jul 5 '17 at 21:57
• Who said it's necessary? It's convenient since the module structures are the same and then $M$ can be supposed faithful. – user26857 Jul 5 '17 at 21:57
• Assume $M$ is faithful. @user26857 left the proof of the fact that $R$ is artinian to the reader. The proof that came to my mind was to observe that $R$ embeds in $M^n$ for $n$ large enough. I don't know if it's the proof user26857 had in mind, or if there are simpler arguments... – Pierre-Yves Gaillard Jul 5 '17 at 22:44
• @EricAuld - Map $r$ to $(rg_1,\dots,rg_n)$ where $g_1,\dots,g_n$ are generators. – Pierre-Yves Gaillard Jul 6 '17 at 9:51