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.
Best Answer
Yes it's correct.
I guess most commutative algebra courses/books show that:
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.