TITLE: Finite surjective morphism of normal varieties and Galois coverings
QUESTION [7 upvotes]: Let $f:X\to Y$ be a finite surjective morphism of normal varieties. We say that $f$ is a Galois covering if the extension $k(X)/k(Y)$ is Galois.
I have the following questions:
1.- Why we need to assume normality of the varieties for the definition of a Galois covering?
2.- Is it true that a finite surjective morphism between normal varieties is always flat? 
3.- Is it true that $f$ is a Galois covering if and only if the group of automorphism $Aut(X/Y)$ acts transitively on all fibers of $f$?

REPLY [9 votes]: I'm just going to make a quick additional comment on #2 (just to add on to Timo Keller's explanations and Sasha's counter example).
Theorem (see Matsumura, page 179) If $R \subseteq S$ is a finite extension of Noetherian rings with $R$ regular and $S$ Cohen-Macaulay, then $S$ is a flat $R$ module.
So you just need $Y$ to be regular and $X$ to be Cohen-Macaulay (Matsumura actually has a more general statement which also works in the non-finite case).
On the other hand, if $Y$ is not regular then $X/Y$ is very rarely flat.  Indeed, since the map is finite, flat is the same as $f_* O_Y$ being a locally free $O_X$-module and we can ask how many (local) $O_X$-summands $f_* O_Y$ has as an $O_X$-module.
Fact  If $R \subseteq S$ is a finite extension of normal local rings with the same residue field which is etale in codimension 1 (ie, the sort of group quotient that Sasha wrote above), then $S$ has at most one $R$-summand as an $R$-module (in particular, it is nowhere near flat).  See the question I asked earlier: Number of free summands of finite local extensions