Let $N \to M$ be an abelian subgroup. Use surjectivity of $R\to S$ to show that $N$ is an $S$-submodule if and only if it is an $R$-submodule.

Question

Let $M$ be an $S$-module, and let $R \to S$ be a surjective ring homomorphism. Then

(a) $M$ is a simple $S$-module if and only if $M$ is a simple $R$-module.

(b) $M$ is a noetherian $S$-module if and only if $M$ is a noetherian $R$-module.

(c) $M$ is an artinian $S$-module if and only if $M$ is an artinian $R$-module.

Proof: Hint: Let $N \to M$ be an abelian subgroup. Use surjectivity of $R\to S$ to show that $N$ is an $S$-submodule if and only if it is an $R$-submodule.

How do I use the surjectivity?

Answer 1

If $\varphi\colon R\to S$ is the homomorphism, the action of $R$ on elements of $M$ is $(r,x)\mapsto \varphi(r)x$.

Suppose $N$ is a subgroup of $M$ with respect to addition.

If $N$ is also an $S$-submodule of $M$, then for $r\in R$ and $x\in N$, we have $rx=\varphi(r)x\in N$.

Suppose instead $N$ is an $R$-submodule; if $s\in S$ and $x\in N$, you have to prove that $sx\in N$. Since $\varphi$ is surjective, $s=\varphi(r)$ for some $r\in R$. Then $$ sx=\varphi(r)x=rx\in N $$ by assumption.