Quotient of a normal quasi-projective variety by a finite group I'm a topologist and not an algebraic geometer, but the following question arose in my work.
Let $X$ be a quasiprojective algebraic variety over $\mathbb{C}$ and let $G$ be a finite group acting on $X$.  Since $X$ is quasiprojective, we have the quotient variety $X/G$.
Question: if $X$ is smooth, must $X/G$ be normal?  Even better, does this hold if $X$ is assumed not to be smooth, but only normal?
I've googled and searched through various books, but I can't find this anywhere (though as I said, I am not a specialist in algebraic geometry, so it could be the case that something far more general is true and I just don't know the how to translate the general statement into the above rather simple-minded one).
 A: This is true because $\mathrm{X}/\mathrm{G}$ is a 'categorical quotient' - the projection $\pi:\mathrm{X}\rightarrow \mathrm{X}/\mathrm{G}$ is $\mathrm{G}$-invariant (in the sense that $\pi(gx)=\pi(x)$  for all $x\in \mathrm{X}$ and $g\in \mathrm{G}$) and in fact universal with this property. If $\mathrm{X}$ is normal, consider the normalisation $\widetilde{\mathrm{X}/\mathrm{G}}\rightarrow \mathrm{X}/\mathrm{G}$ of $\mathrm{X}/\mathrm{G}$. Then $\pi$ lifts to a map $\mathrm{X}\rightarrow \widetilde{\mathrm{X}/\mathrm{G}}$, which is $\mathrm{G}$-invariant, and in fact is also a categorical quotient of $\mathrm{X}$ - hence $\widetilde{\mathrm{X}/\mathrm{G}}\rightarrow \mathrm{X}/\mathrm{G}$ must be an isomorphism, and $\mathrm{X}/\mathrm{G}$ is normal.
A: Just an algebraic interpretation of @Simpleton's answer in the case of finite group actions.  Let $B$ an integrally closed domain with the field of fractions $L$. Let $G$ be a finite subgroup of ${\rm{Aut}}(L)$. Then the extension $L/L^G$ is Galois and $B^G=B\cap L^G$ is a domain (the superscript denotes the invariant subfield/subring) inside $L^G$. It is easy to see that $L^G$ is the field of fractions of $B^G$. An element of $L^G$ integral over $B^G$ is integral over $B$ and hence lies in $B\cap L^G=B^G$. So $B^G$ is integrally closed.   Now recall that normality is a local property which for an affine and irreducible variety $X$ translates to the integral closedness of $\mathcal{O}(X)$. Applying the algebraic result above, if $\mathcal{O}(X)$ is integrally closed, then so is $\mathcal{O}(X/G)=\mathcal{O}(X)^G$.
