Summand Noetherian is Noetherian?

Question

I want to prove if $S$ is a Noetherian ring, $R\subset S$ a subring, and there is an $R$-module homomorphism $\pi:S\to R$ such that $\pi$ is surjective and every element of $R$ is fixed by $\pi$ ($R$ is a summand of $S$), then $R$ is also Noetherian.

My attempt: It suffices to show that the kernel of $\pi$ is a submodule of $S$. Then since submodules and quotients of Noetherian modules are Noetherian, $\frac{S}{\ker\pi}\cong R$ is Noetherian.

$\ker\pi$ is an $R$-module, so we just need to show that it is also an $S$-module. This is where I'm not sure what to do since I can't just say $\pi(sr)=\pi(s)\pi(r)$ for some $r\in\ker\pi$ since $\pi$ doesn't necessarily preserve the multiplication by $s\in S$ action.

Am I missing something?

Answer 1

Assuming the rings are commutative, one can show with the assumption on $\pi$ that if $I \subseteq R$ is any ideal, and $SI$ is the ideal in $S$ generated by $I$, then $\pi(SI) = I$. From there, one can use the ascending chain condition for $S$-ideals in $S$ to get the a.c.c. for $R$-ideals in $R$.

Answer 2

you have a sequence $0\to ker\pi \to S \to R \to 0 $ then $S$ is noetherian if and only if $ker\pi $ and $R$ noetherian.