If a ring is noetherian, then so is its subring

Question

Suppose $A\subset R$ are rings so that $R$ is also an $A$-module. Assume further that there is an $A$-module homomorphism $R\to A$ that is identity on $A$. How to deduce that $A$ is noetherian if $R$ is?

Let $I_1\subset I_2\subset \dots$ be an increasing sequence of ideals of $A$. We need to prove that it stabilizes eventually. The only data we have is the $A$-module homomorphism $R\to A$. I do know that the preimage of an ideal is an ideal under a ring homomorphism; but in our case there is no ring homomorphism. How do I get hold of the corresponding sequence of ideals in $R$? (I think that's what I need to do.)

Answer 1

Let $J_n$ be the ideal of $R$ generated by $I_n$. The sequence $J_1,...,J_n,...$ stabilizes. Let $f:R\rightarrow A$ whose restriction to $A$ is the identity. Remark that for every $r\in R, a\in A, f(ra)=f(r)f(a)=f(r)a\in A$. This implies that $f(J_n)=f(RI_n)=f(R)I_n=I_n$, thus $I_1,...,I_n,...$ stabilizes too.

Answer 2

If $f:R \to A$ restricted to $A$ is the identity, then $f$ is surjective, and so $A$ is isomorphic to $R/N$, for some $N$ $A$-submodule of $R$. Now clearly $R/N$ is noetherian if $R$ is so.