Quotient variety

Question

Is the following statement true.

Suppose $X$ is an affine variety and $G$ is a finite group which acts on $X$ as automorphisms. Suppose $Y$ is an affine variety and there is a surjective map $\pi:X\to Y$ such that

• $\pi(x)=\pi(y)$ iff $ \exists \ g\in G$ such that $g.x=y$

Then is it true that $A(Y) \cong A(X)^G$ i.e the ring of regular functions of $Y$ is isomorphic to the ring of $G$-invariant regular functions on $X$.

Answer 1

Consider the following example: $k$ is a field, $k\subset K\subset L$ distinct finite separable extensions. Consider the zero dimensional $k$-variety $Spec(k), Spec(K)$ and $Spec(L)$. $Gal(L:K)$ acts on $Spec(L)$ and $A(L)^{Gal(L:K)}=L^{Gal(L:K)}=K$. Let $\pi:Spec(K)\rightarrow Spec(k)$ defined by $k\rightarrow L$. Since $Spec(k)$ and $Spec(L)$ are singleton, for every $g\in Gal(L:K), g.x=x$ and $ \pi(g.x)=\pi(x)$ , but $k\neq K$.

Answer 2

This is true if your base is algebraically closed. Indeed, the quotient of an affine variety by a finite group is a categorical quotient and therefore unique up to isomorphism. Hence, there is a canonical isomorphism $Y\cong\operatorname{Spec}(A(X)^G)$. Furthermore, the coordinate rings of two affine varieties are isomorphic if and only if the varieties are isomorphic - hence, $A(Y)\cong A(X)^G$.