Here are four possible definitions for an etale, finite, surjective map $X\rightarrow Y$ between integral schemes to be considered Galois:


1. There exists a finite group $G$, and an action $\varphi: G\times X\rightarrow X$, so that the induced map $\varphi\times p\_2: G\times X\rightarrow X\times\_Y X$ is an isomorphism.
2. There exists a finite group $G$, and an action $\varphi: G\times X\rightarrow X$, so that $G$ acts freely an transitively on the geometric fibers of $X\rightarrow Y$.
3. There exists a finite group $G$, and an action $\varphi: G\times X\rightarrow X$, so that the extension of function fields $\kappa(X)/\kappa(Y)$ is Galois with group isomorphic to $G$.
4. The extension of function fields $\kappa(X)/\kappa(Y)$ is Galois.


Obviously, 1 implies 2 implies 3 implies 4. Note that the difference between 3 and 4 is that in 4 you are not guaranteed that the action extends from the generic point to $X$. I am inclined to believe that 4 does not imply 3, but I can't think of a counterexample.


I *think* that definitions 1 and 2 are probably equivalent. This is because definition 2 is probably the same as saying that $\varphi\times p\_2$ is a bijection on geometric points, which, at least if $X$ and $Y$ are varieties, should imply that it is an isomorphism.


I am completely in the dark about whether or not definitions 2 and 3 are equivalent. I'm not sure what I should believe...


So:


### Questions


1. Is 1 equivalent to 2?
2. Is 2 equivalent 3?
3. What is a counterexample for the equivalence of 3 and 4?
4. Is it at all helpful to ask that $X$ and $Y$ be regular for any of these definitions? Any other adjectives that need adding?



Which of these 4 definitions of Galois coverings of integral schemes are equivalent?
I can prove $(3) \Rightarrow (1)$ when $X$ and $Y$ are irreducible (**Lemma 1** below; no integrality assumptions needed. Irreducible is needed for the question to make sense), and I can prove $(4) \Rightarrow (1)$ when $X$ and $Y$ are normal, connected, and Noetherian (in particular integral, see [Tag 033M](http://stacks.math.columbia.edu/tag/033M)) (**Lemma 2** below).


I do not know how to approach the general case if $Y$ is not normal; see the remark at the end. Maybe you can weaken it to something like unibranch? The property that I need is that a connected finite étale cover of $Y$ is always integral; I don't know in what generality this is true.


**Reference.** A good introductory reference to some of the stuff I'm using is the chapter *Fundamental groups of schemes* of the Stacks project. (I make precise references as I go along.)


**Notation.** Let $\eta\_X$ (resp. $\eta\_Y$) be the generic point of $X$ (resp. $Y$). I will use the following definition.


**Definition.** Let $\phi \colon X \to Y$ be a morphism of schemes, let $G$ be a finite group (viewed as constant étale group scheme), and let $G \times X \to X$ be an action of $Y$-schemes. Then $\phi$ is *Galois with group $G$* if the induced map $G \times X \to X \times\_Y X$ is an isomorphism.


If $F \colon \operatorname{FÉt}\_Y \to \operatorname{Set}$ is a fibre functor (notably, $F = F\_{\bar y}$ associated to taking the geometric fibre over a geometric point $\{\bar y\} \to Y$), there is the following equivalent notion of being Galois.



> 
> **Auxiliary lemma.** Let $\phi \colon X \to Y$ be a finite étale morphism of connected schemes. Then $|\!\operatorname{Aut}\_Y(X)| \leq |F(X)|$, and equality holds if and only if $\phi$ is Galois.
> 
> 
> 


*Proof.* Let $G = \operatorname{Aut}\_Y(X)$, with the natural action on $X$, and consider the induced map
$$\psi \colon G \times X \to X \times\_Y X.$$
Applying the fibre functor $F = F\_{\bar y}$ gives
\begin{align\*}
F(\psi) \colon G \times F(X) &\to F(X) \times F(X)\\
(\sigma, s) &\mapsto (\sigma(s),s).
\end{align\*}
By [Tag 0BN0](http://stacks.math.columbia.edu/tag/0BN0) (7), this map is injective, and it is bijective if and only if $|\!\operatorname{Aut}\_Y(X)| = |F(X)|$.


Thus we need to show that $\psi$ is an isomorphism if and only if $F(\psi)$ is a bijection. This follows since $F$ reflects isomorphisms (see [Tag 0BMY](http://stacks.math.columbia.edu/tag/0BMY) and [Tag 0BNB](http://stacks.math.columbia.edu/tag/0BNB)). $\square$




---



> 
> **Lemma 1.** Let $\phi \colon X \to Y$ be a finite étale morphism of irreducible schemes. Assume there exists a finite group $G$ and an action $G \times X \to X$ whose base change to $\{\eta\_X\}$ makes $\{\eta\_X\} \to \{\eta\_Y\}$ a Galois extension with group $G$. Then $X \to Y$ is Galois with group $G$.
> 
> 
> 


*Proof.* By the auxiliary lemma, we only have to show that $X$ has enough automorphisms. This is now a simple degree/counting argument (the elements of $G$ providing the automorphisms). $\square$



> 
> **Lemma 2.** Let $\phi \colon X \to Y$ be a finite étale morphism of connected normal Noetherian schemes, and assume that the extension $\{\eta\_X\} \to \{\eta\_Y\}$ is Galois. Then $\phi$ is Galois.
> 
> 
> 


*Proof.* We want to apply [Tag 0BN6](http://stacks.math.columbia.edu/tag/0BN6) $(3) \Rightarrow (2)$ to the pullback functor
\begin{align\*}
H \colon \operatorname{FÉt}\_Y &\to \operatorname{FÉt}\_{\eta\_Y}\\
Z &\mapsto Z\_{\eta\_Y}.
\end{align\*}
Assume $Z$ is a connected scheme finite étale over $Y$. Then $Z$ is normal ([Tag 025P](http://stacks.math.columbia.edu/tag/025P)), connected, and Noetherian; hence integral ([Tag 033M](http://stacks.math.columbia.edu/tag/033M)). Hence, it only has one generic point $\eta\_Z$, and this is the only point lying over $\eta\_Y$. Thus, $H(Z)$ is connected, so [Tag 0BN6](http://stacks.math.columbia.edu/tag/0BN6) $(3) \Rightarrow (2)$ implies that $H$ is fully faithful.


Note that $H$ also reflects isomorphisms, since $F\_{\overline{\eta\_Y}}$ does, which factors through $H$. For a fully faithful functor that reflects isomorphisms, we have
$$\operatorname{Aut}\_Y(X) = \operatorname{Aut}\_{H(Y)}(H(X)),$$
so $X$ has enough automorphisms since $H(X)$ does so. $\square$


**Remark.** We see that $H$ is fully faithful if and only if for every connected étale $Y$-scheme $Z$, the generic fibre $Z\_{\eta\_Y}$ is just a point. This is certainly the case if $Z$ is integral, and I think the converse holds as well. (See below for an example where $Z$ has multiple components.)


Also, we might not need the full strength of $H$ being fully faithful: we only need to get back all *automorphisms*.


**Example.** Let $A = \mathbb R[x,y]/(x^2+y^2)$. Then $A$ is a domain. It has a finite étale extension
$$B = \mathbb C[x,y]/(x^2+y^2) = \mathbb C[x,y]/(x+iy) \times \mathbb C[x,y]/(x-iy),$$
which is connected but not integral. The fibre above the generic point consists of two points. Correspondingly, if $Y = \operatorname{Spec} A$, the functor
$$H \colon \operatorname{FÉt}\_Y \to \operatorname{FÉt}\_{\eta\_Y}$$
is not fully faithful, by [Tag 0BN6](http://stacks.math.columbia.edu/tag/0BN6) again. I do not know whether there exists an integral finite étale $A$-algebra $B$ for which $\operatorname{Frac}(A) \to \operatorname{Frac}(B)$ is Galois, but $A \to B$ is not.


