# Consequence of epimorphism from Noetherian $R$-module

Let $$R,S$$ be a commutative rings with $$1_R,1_S$$ respectively. In the most commutative algebra one can find the following proposition.

Proposition. Let $$φ:R\twoheadrightarrow S$$ be a ring epimorphism. Then, if $$R$$ is Noetherian/Artinian, $$S$$ is also Noetherian/Artinian.

I was wondering if this is still valid for $$R$$-modules.

That is, let $$R$$ be a commutative ring with $$1_R$$, $$M,N$$ two $$R$$-modules and $$φ:M \twoheadrightarrow N$$ a module epimorphism. If $$R$$-module $$M$$ is Noetherian/Artinian, then $$R$$-module $$N$$ is Noetherian/Artinian.

Proof. Trying to follow the steps of the proposition above, we consider an arbitrary ascending chain of $$R$$-submodules of $$N$$, $$N_1\subseteq N_2 \subseteq N_3 \subseteq \dotsb.$$ We set $$M_i=φ^{-1}(N_i),\ \forall i=1,2,3,\dots.$$ The preimage of $$R$$-module is again an $$R$$-module. So, taking the preimages on the chain above, we can constract the ascending chain of $$R$$-submodules of $$M$$, $$M_1\subseteq M_2 \subseteq M_3 \subseteq \dotsb.$$ But $$M$$ is Noetherian $$R$$-module, so there is an index $$m\in \Bbb Z^+$$ s.t. \begin{alignat*}{2} M_m \quad = & \quad M_k,\ && \forall k\geq m \iff\\ φ^{-1} (N_m) \quad = & \quad φ^{-1} (N_k),\ && \forall k\geq m \implies \\ φ(φ^{-1} (N_m)) \quad = & \quad φ(φ^{-1} (N_k)),\ && \forall k\geq m \iff \\ N_m \quad = & \quad N_k,\ && \forall k\geq m \end{alignat*} and $$\iff"$$ is valid because $$φ$$ is surjective.

Is this proof correct and complete?

Thank you.

• Looks ok to me. – jgon Mar 8 at 3:27

Yes it's correct.

I guess most commutative algebra courses/books show that:

Proposition. If $$0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$$ is an exact sequence of $$R$$-modules, then:

$$M$$ is artinian/noetherian iff $$M',M''$$ are artinian/noetherian.

So you might want to try proving this result.

Your proposition is the special case $$0 \rightarrow \ker\varphi \rightarrow M \rightarrow N \rightarrow 0$$.

Note that being an epimorphism is equivalent to being surjective for modules, but this is not the case for rings, as $$\mathbb Z\to\mathbb Q$$ is epi - I'm not sure if your version for ring epimorphisms is true, although I don't have a counter-example on hand.

• I think for ring epimorphisms the Noetherianity does not behave well. See here. – user26857 Mar 8 at 12:50
• Btw, when I read the question I thought the OP refers to ring epimorphisms in the category of commutative rings. – user26857 Mar 8 at 12:55
• Yes, I thought of commutative rings in my asnwer as well. – lush Mar 8 at 13:34