In Lemma 5.0.4 of Toric Varieties by Cox, Little and Shenck I don't understand which part of the proof uses that $R^G$ is finitely generated. Can someone please help me?
Lemma : Let $G$ act on $X=\operatorname{Spec}(R)$ such that $R^G$ is a finitely generated $\mathbb C$ - algebrs and let $\pi:X\rightarrow Y=\operatorname{Spec}(R^G)$ be the morphism of affine varieties induced by $R^G\subseteq R$ Then :
Given any morphism of affine varieties $\phi:X\rightarrow Z$ such that $\phi(g\cdot x)=\phi(x)$ for $g\in G$ and $x\in X$ there is a unique morphism $\bar{\phi}:Y\rightarrow Z$ such that $\bar\phi\circ\pi=\phi$ .
If $X$ is irreducible then so is $Y$.
If $X$ is normal the $Y$ is normal.
Proof :
Suppose that $Z=\operatorname{Spec}(S)$ and that $\phi$ is induced by $\phi^*:S\rightarrow R$. Then $\phi^*(S)\subseteq R^G$ follows easily from $\phi(g\cdot x)=\phi(x)$ for $g\in G$ and $x\in X$. Thus $\phi^*$ factors uniquely as $\require{AMScd}$ \begin{CD} S @>{\bar{\phi^*}}>> R^G @>{\pi^*}>> R \end{CD}
The induced map $\bar\phi:Y\rightarrow Z$ clearly has the desired properties.
Part 2 is immediate as $R^G$ is a subring of $R$. For part 3, let $K$ be the field of fractions of $R^G$. If $aa\in K$ is integral over $R^G$ then it is also integral over $R$ and hence lies in $R$ since $R$ is normal. It follows that $a\in R\cap K$ which obviously equals $R^G$ sune $G$ acts trivially on $K$.
Terminology -
Let $G$ act on a variety $X$ such that every $g\in G$ defines a morphism $\phi_g:X\rightarrow X$ given by $\phi_g(x)=g\cdot x$. If $X=\operatorname{Spec}(R)$ is affine then $\phi_g$ comes from a map $\phi_g^*:R\rightarrow R$. We define the induced action of $G$ on $R$ by $$g\cdot f=\phi_{g^{-1}}^{*}(f)$$ for $f\in R$
We define the ring of invariants $R^G=\{f\in R:g\cdot f=f \text{ for all }g\in G\}$
Thank you.