Is the dual of a Noetherian module Noetherian?

Question

Say $M$ is a Noetherian module over $R$. I wanna see how far I can get without the axiom of choice, so let me clarify that I mean "satisfies the Ascending Chain Condition" by Noetherian.

Does the module $M' = \lbrace \varphi : M \rightarrow R \mid \varphi \text{ is linear}\rbrace$ also satisfy this condition?

Answer 1

Here is a counterexample with $R$ commutative. Let $k$ be a field, let $V$ be an infinite-dimensional vector space over $k$, and let $R=k\oplus V$ with the product of any two elements of $V$ being $0$. Then $R/V$ is a Noetherian (indeed, simple) $R$-module. However, $\operatorname{Hom}(R/V,R)\cong V$ (the generator of $R/V$ can be set to any element of $V$) is not Noetherian, since any vector subspace of $V$ is an $R$-submodule.

Answer 2

I won't assume that $R$ is commutative, so let's say that $M$ is a left $R$-module, and thus $M'$ is naturally a right $R$-module via $(\varphi r)(m) = \varphi(m)r$.

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If $R$ is right Noetherian, the answer is "yes".

First, since $M$ is Noetherian, it is finitely generated. Let $X$ be a finite generating subset of $M$. Each element of $M'$ is determined by its behavior on $X$, i.e. the restriction map $\rho : M' \to \operatorname{Hom}_{\mathsf{Set}}(X,R)$ is injective.

We have a natural right $R$-module structure on $\operatorname{Hom}_{\mathsf{Set}}(X,R)$ via the canonical bijection with $R^X$. If $R$ is right Noetherian, then $R^X$ is a Noetherian right $R$-module (since $X$ is finite). To be explicit, the action of $R$ on $\operatorname{Hom}_{\mathsf{Set}}(X,R)$ is given by $(fr)(x) = f(x)r$.

Finally, the map $\rho$ is a homomorphism of right $R$-modules:

$$\rho(\varphi r) = (\varphi r)|_X = (x \mapsto \varphi(x)r : X \to R) = (\varphi|_X)r.$$

So, $\rho$ identifies $M'$ with a right $R$-submodule of the Noetherian right $R$-module $\operatorname{Hom}_{\mathsf{Set}}(X,R)$. We conclude that $M'$ is Noetherian.

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This also shows us how to get a counterexample in the case that $R$ is not right Noetherian. Let $R$ be any ring which is left Noetherian but not right Noetherian. Then let $M = R$. Now $M$ is Noetherian because $R$ is left Noetherian, but $M' = R$ is not Noetherian because $R$ is not right Noetherian.

I'm not sure how to find a counterexample in the commutative case!