5 Galois Theory

The usual proof of the usual Fundamental Theorem of Galois Theory can pretty much be carried over to the general case if we can say the word “Galois”. Fortunately, the word is not too difficult to utter.

The following proposition should be reassuring, but will not be used.

The diagonal map $Y \to Y \times _X Y$ gives an isomorphism between $Y$ and a component of $Y \times _X Y$. We fix a geometric point $z \in F(Y)$, and then we have a diagram

By definition of the pullback, the lifts of $z$ to $Y \times _X Y$ bijects with the lifts of $x$ to $Y$, and this preserves the action of $\operatorname{Aut}(Y)$. Since $\operatorname{Aut}(Y)$ acts freely on the lifts of $z$, it must act freely on $Y \subseteq Y \times _X Y$. So $Y \times _X Y$ splits as $n$ copies of $Y$, and there is nothing left since $Y \times _X Y \to Y$ has degree $n$.

In Galois theory, we can construct Galois closures. We can do the same here.

In Galois theory, the Galois closure is constructed by taking the field generated by all field embeddings into the algebraic closure. The construction here is similar.

Let $y_1, \ldots , y_n \in F(Y)$, which gives rise to a geometric point $y = (y_1, \ldots , y_n) \in Y^n$. Let $Y'$ be the component of $y \in Y^n$, mapping to $Y$ by first projection. This is an étale cover since it is a connected component of one. We claim that this works.

If $\pi _{i, j}: Y^n \to Y^2$ is the projection onto the $i, j$ coordinates, then this is the same as saying $\pi _{ij}^{-1}(\Delta (Y)) \cap Y' = \emptyset$. But since $\Delta (Y)$ is open and closed and $\pi _{ij}^{-1}$ is continuous, we know $\pi _{ij}^{-1}(\Delta (Y)) \cap Y'$ is either $Y'$ or empty. But it does not contain $(y_1, \ldots , y_n) \in Y'$. So it is empty.

Any element in $F(Y^n)$ is of the form $(x_{i_1}, \ldots , x_{i_n})$, and by the above, if it is in $Y'$, then it is of the form $(x_{\sigma (1)}, \ldots , x_{\sigma (n)})$ for some permutation $\sigma \in S_n$. This $\sigma$ also acts on $X^n$ by permuting the coordinates, and $\sigma (Y') \cap Y'$ is non-empty since it contains $(x_{\sigma (1)}, \ldots , x_{\sigma (n)})$. So $\sigma (Y') = Y'$, and hence $\sigma \in \operatorname{Aut}(Y')$. So $\operatorname{Aut}(Y')$ acts transitively on $F(Y')$.

Fix a lift $z \in F(Z)$. Then by composing with automorphisms of $\operatorname{Aut}(Z)$, we can pick maps $q_1, \ldots , q_n: Z \to Y$ such that $q_i(z) = y_i$. Then the image of $(q_1, \ldots , q_n): Z \to Y^n$ includes a point in $Y'$, namely $y$, and hence maps into $Y'$ by connectedness of $Z$.

This is indeed a pro-system since the pullback of two Galois covers is yet another Galois cover of $X$. There is a natural transformation

$$\operatorname*{colim}\operatorname{Hom}(X_\alpha , Y) \to F(Y)$$

that sends a map $q: X_\alpha \to Y$ to $q(x_\alpha )$. Conversely, given any $y \in F(Y)$, we may restrict to the component of $y$ and assume $Y$ is connected. Then since $X_\alpha \to Y$ is surjective and $\operatorname{Aut}(X_\alpha )$ acts transitively on $F(X_\alpha )$, precomposing with an automorphism gives a morphism $X_\alpha \to Y$ that sends $x_\alpha$ to $y$, and such a map is unique since $\operatorname{Aut}(X_\alpha )$ acts freely. This gives the desired inverse map.

The functor is given by $F$. Conversely, given a continuous finite $\pi _1(X, x)$ set $S$, pick $X_\alpha$ such that $\operatorname{Aut}(X_\alpha )$ acts transitively with stabilizer $H$. Then $X \cong F(X_\alpha )/H \cong F(X_\alpha /H)$ ¹ . Functions can be constructed similarly.