# $M_p$ is artinian for all primes p, then M is artinian

Let $M$ be a finite $R$-module that is noetherian and such that $M_{\mathfrak{p}}$ is artinian for each $\mathfrak{p}\in \text{Spec}(R)$. Then $M$ is an artinian $R$-module.

I have tried using a proof with Zorn's lemma: Let $\Gamma = \{ N: N\subseteq M \text{ is artinian } \}$. Since $0\in \Gamma$, and $M$ is noetherian, $\Gamma$ contains a maximal element, call it $N$. But then $N_{\mathfrak{p}}$ is a maximal artinian submodule of $M_{\mathfrak{p}}$ (I don't think this is true (a priori); I think proving this is equivalent to proving the wanted statement) for all $\mathfrak{p}$, so $N_{\mathfrak{p}} = M_{\mathfrak{p}}$ for all $\mathfrak{p}$, implying $N=M$.

I have also tried showing Jac$(R)M = \mathfrak{m}_1\cap \dots \cap \mathfrak{m}_t M$ where $\mathfrak{m}_t$ are finitely many maximal ideals, with which I have had no luck.

• Consider a chain of submodules in M. If you localize at each prime the resulting chain stabilizes. Show that finiteness allows you to show that all the chains corresponding to all primes stabilize at the same point. Feb 1, 2017 at 3:01
• First you can reduce to cyclic modules, then you can reduce to $M=R$, and then noetherianness lets you replace artinian by zero-dimensional. The dimension of a ring can of course be tested locally.
– MooS
Feb 1, 2017 at 6:12

The sum of two artinian submodules is artinian. Hence - by induction on the number of generators - you can assume that $M$ is cyclic, i.e. $M = R/\operatorname{Ann}(M)$. Of course $R/\operatorname{Ann}(M)$ is artinian as a $R$-module if and only if it is artinian as a $R/\operatorname{Ann}(M)$-module (i.e. as a ring).
So we have the following situation: We have a ring $A$, which is noetherian and locally artinian. We want to prove that it is artinian.
Because of the noetherianess of $A$ and all its localizations, we have artinian = zero-dimensional and then the claim follows from $$\dim A = \sup\limits_{\mathfrak p \in \operatorname{Spec} A} \dim A_{\mathfrak p}.$$