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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?

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    1) Are your rings commutative and unital? 2) What exactly do you want to show? "R Noetherian" as an R- or S-module? 3) In the general situation you describe, ker $\pi$ will not be an S-module. Example: $S = k((x)), R = k, \pi(\sum a_i x^i) = a_0$. – 2017-01-24
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    Yes, all the rings are unital. The question is to show that $R$ is Noetherian as a ring. So is this actually trivial since that just means $R$ is Noetherian as an $R$-module since the kernel is an $R$-module and the quotient of an $R$-module is an $R$-module? (And then Noetherian preserved under submodule and quotients) – 2017-01-24
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    Well, it would be trivial if we had that $S$ is Noetherian *as an $R$-module*. But so far, "S is a Noetherian ring" means, if we have consistent definitions, that it is Noetherian as $S$-module. – 2017-01-24
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    Any suggestions? I'm trying to use the ascending chain condition, but having no luck – 2017-01-24
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    Could you figure it out or is there something missing? By the way, the example in my first comment is one for the situation where $S$ is a noetherian ring, but not noetherian as an $R$-module. – 2017-01-25

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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 $R$.

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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.

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    ... as $R$-modules. I understand the hypothesis differently, namely, $S$ is Noetherian as an $S$-module. As was already noted in the comments, if we assume $S$ is Noetherian as $R$-module, then by your argument the assertion is easy. – 2017-01-24