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If $M$ is finitely generated left module over a left noetherian ring $R$, then $M$ is a noetherian module.

I need to show that an arbitrary ascending chain of submodules of $M$ must satisfy ACC, or equivalently, every submodule of $M$ is finitely generated.

Since $R$ is a left noetherian ring, every left ideal in $R$ is finitely generated. I am trying to find a 1-1 correspondence between left ideals of $R$ and submodules of $M$. Is it a correct way, or how can I show the assertion by a different way? Thanks.


marked as duplicate by rschwieb abstract-algebra Jun 20 '17 at 10:41

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In general, there will not be a 1-1 correspondence between submodules of $M$ and $R$. Just consider $R = k$ a field, and $M = k^n$ a $n$-dimensional $k$-vector space. Then $k$ has only two subspaces, but $M$ has quite a lot more if $n > 1$.

I would try the following. First of all, note that finite direct sums of noetherian modules are noetherian and that quotients of noetherian modules are noetherian, too. These are special cases of the more general result mentioned in cat's answer. Now, note that the assumption that $M$ is finitely generated gives you a surjective homomorphism $R^n \to M$ for some $n$.


One way, first show this fact: if $0 \to A \to B \to C \to 0$ are $R$-modules, then $A$ and $C$ are Noetherian if and only if $B$ is Noetherian.

Then $M$ finitely generated as a $R$-module means there is an exact sequence of (left) $R$-modules $0 \to K \to R^n \to M \to 0$ for some $n$. Since $R$ is Noetherian, $R^n$ is Noetherian (this is another fact you might have to show). By the above fact $M$ is Noetherian.

  • $\begingroup$ $R$ Noetherian => $R^n$ Noetherian follows by induction using your fact about exact sequences. $\endgroup$ – Justin Young Jun 20 '17 at 10:35

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