Let $X$ be an affine variety, $A(X)$ be a coordinate ring of $X$, and let $G$ be a finite group. Let's assume that $G$ acts on $X$ i.e. we have morphisms $g:X\to X$ for all $g\in G$ s.t. $e:X\to X$ is the identity morphism and $(g\circ h)(x)=g(h(x))$ for all $x\in X$ and $g,h\in G$.
I want to show that $A(X)^G=\{f\in A(X):f(g(a))=f(a),\text{ for all }a\in X, g\in G\}$ is a finite generated subalgebra of $A(X)$.
My idea is the following: since $k[x_1,\dots,x_n]$ is finitely generated, then $A(X)=k[x_1,\dots,x_n]/I(X)$ is finitely generated i.e. let's say that $A(X)=(f_1,\dots,f_k)$.
Also, we can consider the operator $T:A(X)\to A(X)^G$ which is given by $T(f)=\overline{f}$ where $\overline{f}=\frac{1}{|G|}\sum_{g\in G}f(g)$ and $$\overline{f}(a)=T(f)(a)=\frac{1}{|G|}\sum_{g\in G}f(ga).$$ We can easy to check that $\overline{f}\in A(X)^G$ and $T|_{A(X)^G}=id_{A(X)^G}$.
So, I want to show that $A(X)^G=(\overline{f_1},\dots,\overline{f_k})$. But, this would follow from the fact that $T$ is an algebra homomorphism i.e. if $g\in A(X)^G\subset A(X)$, then $g=p(f_1,\dots,f_k)$ and $g=T(g)=p(\overline{f_1},\dots,\overline{f_k})$ where $p\in k[f_1,\dots,f_k]$.
However, $T$ is just a linear map. Is there another approach showing this?
By the Artin-Tate lemma, it suffices to show that $A(X)$ is finitely generated as an $A(X)^G$-module. To prove this, note that for each $f\in A(X)$, $f$ is a root of the monic polynomial $p(t)=\prod_{g\in G}(t-f\circ g)\in A(X)[t]$. But this polynomial is actually in $A(X)^G[t]$, since the action of $G$ just permutes the factors. Thus every element of $A(X)$ is a root of a monic polynomial with coefficients in $A(X)^G$, i.e. $A(X)$ is an integral extension of $A(X)^G$. Since $A(X)$ is finitely generated as an algebra, this implies it is finitely generated as an $A(X)^G$-module.