1
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ Looks ok to me. $\endgroup$
    – jgon
    Commented Mar 8, 2019 at 3:27

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I think for ring epimorphisms the Noetherianity does not behave well. See here. $\endgroup$
    – user26857
    Commented Mar 8, 2019 at 12:50
  • $\begingroup$ Btw, when I read the question I thought the OP refers to ring epimorphisms in the category of commutative rings. $\endgroup$
    – user26857
    Commented Mar 8, 2019 at 12:55
  • $\begingroup$ Yes, I thought of commutative rings in my asnwer as well. $\endgroup$
    – lush
    Commented Mar 8, 2019 at 13:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .