# algebraic geometry – Commutation of integral closure and group invariance

We work over $$mathbb{C}$$. Let $$G$$ be a reductive group acting on a normal affine variety $$X$$: we have an induced $$G$$-action on $$k(X)$$ given by
$$gcdot p(x)=p(g^{-1}cdot x).$$
Thanks to the GITwe know the good quotient $$Y$$ is affine and given by $$Y=text{Spec} k(X)^G$$.

I would like to prove that $$Y$$ is still normal, that is $$k(X)^G=k(Y)=overline{k(Y)}=overline{k(X)^G}$$ (where the closure is taken in the field $$k(X)$$, that is, the operation of integral closure and $$G$$-invariance is commutative

The inclusion $$k(X)^Gsubset overline{k(X)^G}$$ is obvious, so let us prove the reverse one. By the normality, it suffices to show that

$$overline{k(X)^G}subset overline{k(X)}^G,$$
I take $$fin overline{k(X)^G}$$, so that I can write

$$f^n(x)+a_{1}(x)f^{n-1}(x)+ldots+a_{n-1}(x)f(x)+a_n(x)=0$$
with $$a_iin k(X)^G=overline{k(X)}^G$$. The $$G$$-invariance allow me to see that the above formula works also
$$f^n(g^{-1}cdot x)+a_{1}(g^{-1}cdot x)f^{n-1}(g^{-1}cdot x)+ldots+a_{n-1}(g^{-1}cdot x)f(g^{-1}cdot x)+a_n(g^{-1}cdot x)=0$$
for any $$gin G$$, but now I’m a bit stuck as I don’t know how to continue (I can subsract them both but I don’t see what this imply).