Let $R \subseteq S$ be unital rings and let $G \leq \mathrm{Aut}_R(S)$ be a finite group of automorphisms of $S$ as an $R$-algebra. We define the invariant subring: $$S^G = \{a \in S \mid \forall \sigma \in G, \sigma a = a\}$$ Suppose now, that $S$ is finitely generated as an $R$-algebra. I am to show that there exists a finitely generated $R$-subalgebra $A \subseteq S^G$, such that $S$ is an integral extension of $A$.
What I've got so far is this: If $S$ is generated by $\{s_1, ... , s_n\}$, let $A$ be the algebra generated by the subset of those generators, such that $s_j \in S^G$. Then we have $A \subseteq S^G$. To complete the proof, it would suffice to show that $S^G$ is also integral over $A$, as we've already shown in a previous excercise that $S$ is integral over $S^G$. By the transitive property of integral extensions, $S$ would be integral over $A$.
Is this the right line of thinking? Can it be proven in general that given a finitely generated $R$-algebra $S$, a subring $T \subseteq S$ and a finitely generated $R$-subalgebra $A \subseteq T$, that then $T$ is an integral extension of $A$? I've tried to prove this myself, but was unable to do so. It seems almost too good to be true, but I was also unable to find a counterexample.
Any help with a different solution to the above problem would also be appreciated.
There might not be any elements such that $s_j\in S^G$. In general, generators don't have to belong to a subalgebra. And even if there are, they might not generate the whole algebra $S^G$.
You mentioned that you already know that $S$ is integral over $S^G$. Usually this is proved by defining the following polynomial for each $s\in S$:
$f_s(x)=\prod\limits_{g\in G}(x-g(s))\in S^G[x]$
Then $s$ is its root. So now let $A$ be the subalgebra of $S$ generated by the coefficients of the polynomials $f_{s_1}, f_{s_2},...,f_{s_n}$. Can you finish from here?