Finite surjective morphism of normal varieties and Galois coverings

Question

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 $f$?

Answer 1

The simplest counterexample to question 2 is the map $$ X := \mathbb{A}^2 \to \mathbb{A}^2/\{ \pm 1 \} =: Y. $$ Here $Y$ is a quadratic cone in $\mathbb{A}^3$. Both $X$ and $Y$ are normal, and the map is a Galois covering, but $X$ is not flat over $Y$.

Answer 2

(I assume by a covering, you mean a finite surjective étale morphism.)

1. This is needed as in your definition of Galois covering you only look at the extension of the function fields. Using the more general definition from [SGA1, Exp. V, Remarque 5.11], this works for any connected scheme. For this, see also [Stacks Project], http://stacks.math.columbia.edu/download/pione.pdf, p. 6, after Lemma 3.7: $X/Y$ is Galois (this works for any Galois category with fibre functor $\mathrm{Fib}_\bar{y}$) iff $X$ is connected and $|\mathrm{Aut}(X/Y)| = \mathrm{Fib}_\bar{y}X$ iff $X$ is connected and $\mathrm{Aut}(X/Y)$ acts transitively on $\mathrm{Fib}_\bar{y}X$. The étale fundamental group is defined in Definition 6.1 there. In section 10, the fundamental group of normal schemes is considered. See also http://websites.math.leidenuniv.nl/algebra/GSchemes.pdf, p. 41, 3.14 Galois objects. For the question what happens in the non-normal case, see The étale fundamental group in the non-normal case.

2. If $f: X \to Y$ is a finite surjective morphism of regular schemes (or more generally, if $Y$ is regular and $X$ is Cohen-Macaulay), it is flat, see [Liu], Remark 4.3.11, or [EGAIV${}_3$], p. 230, Proposition (15.4.2).

3. I think this is true, but I do not have a reference. I would show that $\mathrm{Aut}(X/Y)$ acting transitively on the fibres is equivalent to the usual definition of a Galois cover from [SGA1], and then that this is equivalent to the generic fibre being a Galois cover. I think this follows from the fact that one has a surjection $G_{k(Y)} \twoheadrightarrow \pi_1^{\mathrm{\acute{e}t}}(Y,\bar{y})$ and under surjective group homomorphisms, normal subgroups correspond to normal subgroups.

Answer 3

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