Suppose we have a normal domain $R$ (i.e. integrally closed in its field of fractions) with a group $G$ acting on it by ring homomorphisms.
I was wondering how one could prove that the ring of $G$-invariants $R^{G}=\{r \in R \mid gr = r , \; \forall g \in G \}$ is also a normal domain.
Let $k$ denote the field of fractions of $R^G$. Then if $z\in k$ and $z$ is integral over $R^G$, then $z$ lies in the field of fractions of $R$ and is integral over $R$, so $z\in R$.
Thus we have $x,y\in R^G$ with $y\neq 0$ such that $\frac xy=z$, with $z\in R$. Thus $x=yz$.
For any $g\in G$, we may apply $g$ to both sides and obtain $x=yg(z)$. Thus $$yg(z)=x=yz.$$ As $y\neq 0$, we may conclude $g(z)=z$.
Thus $z\in R^G$ as required.