Let $$ 0 \to M' \to M \to M'' \to 0 $$ be an exact sequence of $R$-module. We know that $M$ is Noetherian if and only if $M'$ and $M''$ are Noetherian. Here, to prove that they use its ascending chain.
I want to prove that $M$ is $S$-Noetherian if and only if $M'$ and $M''$ is $S$-Noetherian. How to prove it? Should I also use ascending chain?
Definition Let $R$ be a ring and $S\subseteq R$ a multiplicative subset. A right $S$-module $M$ is said to be $S$-finite if $Ms \subseteq F$ for some $s \in S$ and some finitely generated submodule $F$ of $M$, and $M$ said to be $S$-Noetherian if every submodule of $M$ is an $S$-finite module.