Let $G$ be a finite group and $K$ be a field. Let $R$ be a $K$-algebra of finite type with a $G$-action leaving $K$ fixed. Define $R^G:=\{r\in R | gx=x \ \forall x\in G\}$. Show that $R$ is finitely generated as $R^G$-module.
Attempt. First of all, we know that $R$ is integral over $R^G$.
- Since $R$ is a $K$-algebra with a $G$-action leaving $K$ fixed, there exists a ring homomorphism $f: K\to R$ such that $gk:=gf(k)=f(k)$ for all $g\in G,k\in K$, i.e. $f(k)\in R^G$ for all $k\in K$.
- Since it is of finite type, there exist $r_1,\dots,r_n\in R$ such that every element of $R$ can be written as a polynomial in $r_1,\dots,r_n$ with coeficients in $f(K)$.
Clearly, the products $r_1^{a_1}r_2^{a_2}...r_n^{a_n}$ generate $R$ as $R^G$-module, where $a_1,...,a_n$ are nonnegative integers. But, how can we conclude that $R$ is finitely generated as $R^G$-module? Thanks!