This is a problem of Rotman Homological Algebra 5.35. I have a few questions.
Define $Mod_R$ ACC and prove if $Mod_S\cong Mod_R$ as category and $Mod_R$ has acc, then $Mod_S$ has acc. More specifically $S$ is noetherian.
I think it suffices to define noetherian objects in $Mod_R$. Noetherian objects $M\in Mod_R$ are given any finite $n$ such that if there is exact sequence $R^n\to M\to 0$, then $M$ has the standard definition of ACC condition(any ascending chain of $M$ stabilizes).
It is clear that $Mod_S\cong Mod_R$ implies there are two exact functors $F:Mod_S\to Mod_R$ and $G:Mod_R\to Mod_S$ by adjoint functor theorem. By adjoint functor theorem, both $F,G$ preserve direct limit and products.
If I can show $S$ noetherian, I am done. It is not clear that $F(S)$ would be a finitely generated object(i.e. there will be $R^m\to F(S)\to 0$ exact sequence).
The following are my questions.
Does $F$ or $G$ preserve direct sum or direct products? I think they do by adjoint functor theorem unless I made big mistake somewhere. Since both $FG,GF$ are natually isomorphic to $1_R,1_S$, it does not matter which one is treated as left or which one is treated as right?
Why $F(S)$ would be a f.g object in $Mod_R$ category? I think I need some how identify $R^m\to F(S)\to 0$ as exact here in order to conclude that $S$ is noetherian. How do I argue $m<\infty$?
If $F:A\to B$ is an equivalence of categories, show that for all objects $x$ in $A$ the functor induces an isomorphism from the poset of subobjects of $x$ to that of $F(x)$.
Now in your situation, the equivalence maps $R$ to $F(R)$, which is a projective generator in $Mod_S$. If $R$ is noetherian, then $F(R)$ is noetherian. Since $F(R)$ is a generator, there is a surjection from a direct sum of copies of $F(R)$ to $S$, and since $S$ is generated by $1$, there is in fact sich a map on a finite direct sum of copies of $F(R)$. It follows that $S$ is noetherian, as it is a quotient of a noetherian $S$-module.