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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?
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    $\begingroup$ For curves at least, it's more or less standard that all are definitions equivalent… not sure about the general case... $\endgroup$
    – user40276
    Jun 18, 2016 at 4:34

1 Answer 1

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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) (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 (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 and 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 $(3) \Rightarrow (2)$ 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), connected, and Noetherian; hence integral (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 $(3) \Rightarrow (2)$ 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 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.

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  • $\begingroup$ The quoted result "Tag 0BN6" is a purely categorical result. I would like to see the reason why an isomorphism over the generic point can be lifted to the whole isomorphism. This is still unclear for me... $\endgroup$
    – Li Yutong
    Jun 6, 2022 at 5:09
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    $\begingroup$ @LiYutong Oh yeah, it doesn't need to be as complicated as what I wrote. The point is that any $\eta_Y$-automorphism $f \colon \{\eta_X\} \to \{\eta_X\}$ comes from a (unique) $Y$-automorphism $f \colon X \to X$, because $X$ is the integral closure of $Y$ in $\{\eta_X\}$ and integral closure is functorial for isomorphisms. (Also I'm just realising that any fully faithful functor reflects isomorphisms, so I didn't need to mention that separately.) $\endgroup$ Jun 6, 2022 at 13:12
  • $\begingroup$ I had rethought (4) and felt that the lift may not always be possible: let $p \in X$ be a closed point, then $X-\{p\} \to Y$ also satisfies (4) (for most cases), but a lift is not always possible (because some point $q \in f^{-1}(f(p))$ may have to be sent to $p$). Certainly, if $Y$ is defined to be the normalization of $X$ in $K(Y)$, there is no such problem and your comments work. $\endgroup$
    – Li Yutong
    Jun 7, 2022 at 5:41

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