Noetherian module over two rings

Question

Let $f : R → S $ be a surjective ring homomorphism.

How can I show the following statement:

An $S$-module $M$ is Noetherian if and only if it is Noetherian as $R$-module.

Answer 1

We have $S \cong R/I$ where $I$ is the kernel of the homomorphism. Now, the $R$-submodules of $M$ are precisely the $R/I$-submodules of $M$, which means that $M$ satisfies the ascending chain condition as an $R$-module if and only if it satisfies the ascending chain condition as an $R/I$-module.

Note that the isomorphism $S \cong R/I$ is sufficient for the result to follow, because of the way we define the action of $S$ on $M$.