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Let $R$ be a commutative Ring, $M$ an $R$-module, we define $\text{Ann} _R(M) := \{ r \in R : rm = 0 \text{ }\forall m \in M\} $. As an exercise we have to show:

"$M$ noetherian $\iff$ $M$ finitely generated, $R/\text{Ann}_R(M)$ noetherian "

I was able to prove the "=>" implication, however i am struggling with the other direction. I do not really know how I can use the condition "$R/\text{Ann}_R(M)$ noetherian ". What does this tell me about the submodules of $M$? I do not see the connection. My first guess was to consider finitely generated $R/\text{Ann}_R(M)$-modules, which would automatically be noetherian, maybe there could exist such a module that is isomorphic to $M$ (or a submodule of $M$?), however, my attempt didn't succeed.

My second guess was to use induction over the number of M-generating elements. Here, I was unable to go back to the inductive hypothesis. What else could I try here? Does somebody have some advice for me? Thanks in advance!

LinearAlgebruh
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  • Note that $M$ is an $R/I$ module for $I$ the annihilator ideal. If it is f.g. over $R$ then so it is over this quotient. Can you proceed? – Pedro Apr 17 '21 at 14:27
  • How do you prove that $M$ noetherian implies $R/\mathrm{Ann}_R(M)$ noetherian? Edit. This answers it: https://math.stackexchange.com/questions/763433/if-m-is-a-noetherian-r-module-then-r-textannm-is-a-noetherian-ring – Olivier Bégassat Apr 25 '21 at 19:02

1 Answers1

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Let us put $R_M=R/\mathrm{Ann}_R(M)$. Note that

  1. $M$ is canonically an $R_M$-module and
  2. $\Big\{R$-submodules of $M\Big\}=\Big\{R_M$-submodules of $M\Big\}$.

The lattices of $R$-submodules of $M$ and that of $R_M$-submodules of $M$ are equal. This is relevant because the noetherian property you are after is a property of the lattice of submodules: $M$ noetherian $R$-module means that every nonempty collection of submodules has a maximal element. Thus $M$ is noetherian as an $R$-module iff it is noetherian as an $R_M$-module.

Note furthermore that:

  1. A subset of $M$ is $R$-generating iff it is $R_M$-generating.

In particular, $M$ is finitely generated over $R$ if and only if it is finitely generated over $R_M$.

Proof of $(\!\!\impliedby\!\!)$ Suppose $M$ is finitely generated over $R$. By the preceding point it is finitely generated over $R_M$. If we furthermore assume $R_M$ to be noetherian, then so is the finitely generated $R_M$-module $M$. By the point in bold, $M$ is noetherian over $R$.

Olivier Bégassat
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