Prove or Disprove: Finitely generated Artinian module is Noetherian.

Question

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?

Answer 1

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 $M$.

Answer 2

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.