The Étale Fundamental GroupGalois Theory

# 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.

Definition (Galois cover)

A Galois cover of $X$ is an element $Y \in \mathscr {FE}t_{X}$ such that $\operatorname{Aut}(Y)$ acts transitively on $F(Y)$.

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

Let $Y \in \mathscr {FE}t_{X}$ be Galois. Then $Y \times _X Y$ splits as finitely many disjoint copies of $Y$, and $\operatorname{Aut}(Y)$ acts freely and transitively over the copies, where $\operatorname{Aut}(Y)$ acts as an automorphism of $\pi _1: Y \times _X Y \to Y$ by pullback.

Proof
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 $n$.

Proof

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

Lemma

Let $Y \in \mathscr {FE}t_{X}$ be connected. Then there is a Galois cover $Y'$ with a map $Y' \to Y$ such that if $Z \in \mathscr {FE}t_{X}$ is any Galois cover, then any map $Z \to Y$ factors through $Y'$.

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.

Proof
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.

Claim

If $(x_1, \ldots , x_n) \in Y'$, then $x_i \not= x_j$ for all $i$ and $j$.

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.

Claim

$Y' \to X$ is Galois.

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 $F(Y')$.

Claim

If $Z$ is Galois and $q: Z \to Y$, then it factors through $Y'$.

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 $Z$.

Proof

Theorem

Let $(X_\alpha )_\alpha \subseteq \mathscr {FE}t_{X}$ be the poset of Galois covers of $X$, ordered under $X_\alpha \leq X_\beta$ if there is a morphism $X_\beta \to X_\alpha$. For each $X_\alpha$, pick a point $x_\alpha \in F(X_\alpha )$. Then if $X_\alpha \leq X_\beta$, we pick the morphism that sends $x_\alpha$ to $x_\beta$ for use in the pro-system. Then this pro-system pro-represents $F$.

Proof
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.

Proof

Theorem

$\mathscr {FE}t_{X}$ is naturally equivalent to the category of continuous finite $\pi _1(X, x)$-sets.

Proof
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.
Proof