Is a cover Galois if and only if it is geometrically Galois

Question

Let $K$ be a number field and let $\pi:X\to \mathbf{P}^1_K$ be a finite morphism, where $X$ is a smooth projective geometrically connected curve.

Is $\pi$ a Galois cover if and only if the base change $\pi_{\overline K} : X_{\overline K} \to \mathbf{P}^1_{\overline K}$ is a Galois cover?

Answer 1

If $\pi$ is Galois, then so is $\pi_{\overline{K}}$. This is more or less obvious.

But the converse is false. Consider the cover $X:=\mathbb P^1_{K} \to \mathbb P^1_{K}$, $x\mapsto x^3$. On the level of function fields, it corresponds to the extension $K(t)\subset K(s)$, $s^3=t$. This extension is not Galois if $K=\mathbb Q$, but becomes cyclic over $\overline{\mathbb Q}$.

Answer 2

QiL has given what has to be the simplest example of a geometrically Galois but not Galois extension of function fields. I want to give different -- so, necessarily, less simple -- example, but one that shows how important this distinction is in arithmetic geometry.

Let $K$ be a field -- say of characteristic zero -- let $E_{/K}$ be an elliptic curve, and let $\varphi: E \rightarrow E'$ be an isogeny of degree $n$. Then $C = \operatorname{ker} \varphi$ is a cyclic subgroup of order $n$, and we may view $E' = E/C$. From this one can see that $\varphi$ is an unramified covering map and that the deck transformations are precisely $\tau_P: x \mapsto x + P$ with $P \in C(\overline{K})$.

This means the cover is geometrically Galois: $\# C(\overline{K}) = \operatorname{deg} \varphi$. Moreover, it is Galois iff the Galois action on $C(\overline{K})$ is trivial: i.e., rather than being stabilized by the Galois group of $\overline{K}/K$ (which is necessary for the isogeny $\varphi$ to be defined), in order for $K(E)/K(E')$ to be Galois we need the action of $\operatorname{Gal}(\overline{K}/K)$ to be trivial. In the case of cyclic $C$, this is the distinction between a rationally defined subgroup and a rational torsion point.

The second condition is definitely more stringent than the first. For instance, suppose $K = \mathbb{Q}$ and $C$ is a cyclic subgroup of prime order $p$. Then by work of Mazur, the largest prime $p$ such that there exists an elliptic curve $E$ and a rationally defined subgroup $C$ is $p = 163$, whereas the largest prime such that there exists a rational torsion point of order $p$ is $p = 7$.

Similarly, taking $E$ to be a generic elliptic curve over $K = \mathbb{Q}(t)$, the distinction between geometrically Galois and Galois can be viewed as responsible for the difference between the modular curves $X_1(N)$ and $X_0(N)$.

Answer 3

Here is another example of a more arithmetic nature. Let $K$ be a field and $L/K$ finite extension that is not Galois. Let $X=\mathbb{P}^1_L = \mathbb{P}^1_K\times_K L$ and $\pi\colon X\to \mathbb{P}^1_K$ the projection map. Then $\pi$ is not Galois. On the other hand, since $X_{\bar{K}}=\mathbb{P}^1_{\bar{K}}$, so $\pi_{\bar{K}}$ is the trivial map, hence Galois.