# Galois Theory and Geometry

In this post I want to talk a bit about how one can extend some of the ideas of Galois Theory to varieties. To begin, we remind ourselves how Galois Theory works over some field $K$.

Galois Theory over a Field

We’ll be working in the category $\mathsf{Sep}_K$ of finite separable extensions of $K$. If we pick some algebraic closure $\overline{K}$ then we can define a functor $F: \mathsf{Sep}_K \to \mathsf{Set}$ given by $F(L)=\mathrm{Hom}(L, \overline{K})$. $\mathrm{Aut}(F)$ (the group of natural isomorphisms of $F$) is a profinite group and the fundamental theorem is that $F$ gives an equivalence of categories between $\mathsf{Sep}_K$ and transitive $\mathrm{Aut}(F)$-sets.

Maybe the above description is a bit unfamiliar so lets break it down super carefully. Let $L/K$ be a finite separable extension. Then if $[L:K]=n$ we know that $L$ has precisely $n$ embeddings into $\overline{K}$ and as defined above, $F(L)$ is precisely this set of embeddings. $\mathrm{Aut}(F)$ is definitionally the set of natural isomorphisms of the functor $F$ which is the same thing as a collection of automorphisms $\sigma_L$ of $F(L)$, one for each $L \in \mathsf{Sep}_K$, which form a compatible system in the sense that if we have a map $f: L \to L'$ of $K$-Algebras then

$\begin{matrix}F(L) & \xrightarrow{\sigma_L} & F(L)\\ \downarrow^{f} & & \downarrow^f\\ F(L') & \xrightarrow{\sigma_{L'}} & F(L') \end{matrix}$

is a commutative diagram. In particular, for any $L$, the automorphisms of an embedding $L \to \overline{K}$ are all induced by automorphisms of $\overline{K}$ (exercise) and so a compatible system of such automorphisms is exactly an element of $\mathrm{Gal}(\overline{K}/K)$.

For certain special extensions $L/K$ we have that the elements of $\mathrm{Aut}(F)$ restricted to their actions on $F(L)$ are also induced via pre-composition by automorphisms of $L$. These are precisely the Galois extensions and of note is the fact that they are cofinal in the category $\mathsf{Sep}_K$ in the sense that any finite separable extension of $K$ is contained within some Galois extension. If one sits down and traces through what this implies about $\mathrm{Aut}(F)$, it is that $\mathrm{Aut}(F)$ can be computed as an inverse limit of the automorphism groups of all the Galois extensions of $K$.

The reader who was excited by my first paragraph but immediately disgruntled by my second should have started to feel happier now.

Galois Theory over a Scheme

By being rather abstract in my constructions above, my aim was to hint that this theory might be generalized to other contexts. This can be very satisfactorily done and anyone interested in the details should read Lenstra’s online notes Galois Theory for Schemes.

The point is that one can define a special type of category called a Galois category equipped with a “fiber” functor $F: G \to Set$. The fundamental theorem one proves is that $F$ gives an equivalence of categories between $G$ and $\mathrm{Aut}(F)$-sets. The astute reader will notice that in the case of fields we got an equivalence between $\mathsf{Sep}_K$ and transitive $\mathrm{Aut}(F)$-sets. This came about because we didn’t consider the full Galois category associated to a field $K$. This is the category of finite separable $K$-algebras which we call $\mathsf{F\acute{E}t}_K$. So for instance the full category also contains products of fields and this is how we get actions with multiple orbits. As in the case of fields we can consider Galois objects in a general Galois category and $Aut(F)$ will be the limit of the automorphism groups of such objects.

Before I delve into schemes, I want to briefly give another example of a Galois category. Given a sufficiently nice topological space $X$ we can consider the category of covering spaces of $X$. In this case the fiber functor $F$ is given by picking a base point $x \in X$ and then mapping a cover $Y$ of $X$ to its fiber over $x$. Any undergrad has probably noticed the intense similarities between the theory of covering spaces and Galois theory. These similarities can be explained by the fact that both categories are Galois categories. Now onto schemes!

The field $K$ should be thought of geometrically as a single point corresponding to its unique prime ideal. Then the Galois category $\mathsf{F\acute{E}t}_K$ which we have constructed is what is called the finite etale site over $\mathrm{Spec}(K)$. More generally, given some scheme $X$ we define the finite etale site over $X$ to be the category whose object are pairs $(Y, \pi_Y: Y \to X)$ where $\pi_Y$ is finite etale and morphisms are maps $f: Y \to Y'$ satisfying $\pi_{Y'} \circ f= \pi_Y$. A finite etale morphism is a morphism which is both finite and etale. An etale morphism is defined to be flat and unramified where we remind the reader that an unramified morphism $f: Y \to X$ is defined to be locally of finite type and for each $y \in Y$, we have $\mathscr{O}_{Y, y} / m_x \mathscr{O}_{Y, y}$ is a finite separable field extension of $\mathscr{O}_{X,x} / m_x \mathscr{O}_{X,x}$ where $m_x$ is the maximal ideal in the stalk $\mathscr{O}_{X,x}$. Maybe on a first pass, though, it’s best not to worry too much about the technical definition. The point is that an etale map is the geometric analog of a finite cover of topological spaces. In other words if $f: Y \to X$ is etale, then $X, Y$ can’t look so different. In particular, at the very least they have to be the same dimension.

So lets check that this agrees with our understanding of $\mathsf{F\acute{E}t}_K$. In particular, lets figure out for which schemes $X$ we have an etale map $\pi_X: X \to \mathrm{Spec}(K)$ for $K$ a perfect field. $\pi_X$ is finite so for starters $X$ is some finite $K$-algebra. Consider $\mathrm{Spec}\left(K[x]/(x^2)\right)$. It has one point $(x)$ and the local ring there is $K[x]/(x^2)$ (since this ring is already local!). But the maximal ideal of $K$ is just $(0)$ and so the relevant ring in the definition of unramified is $K[x]/(x^2)$ which is not even a field! A similar argument shows that $X$ will have to be reduced and so will be a product of fields.

So to recap, what we’ve been talking about is how, given a scheme $X$, we can construct a Galois category $\mathsf{F\acute{E}t}_X$. In the Galois category of covers of a topological space, the group $\mathrm{Aut}(F)$ was exactly the fundamental group of $X$. So in the case of a scheme $X$, we intuitively think of $\mathrm{Aut}(F)$ similarly as some sort of fundamental group and so we call $\mathrm{Aut}(F)$ the “etale fundamental group of $X$”.

In fact one can make a concrete connection between the two concepts. If $X$ is a complex variety, then we can give $X$ the analytic topology coming from the natural topology on $\mathbb{C}$. We call the resulting topological space $X^{an}$. Then one can prove that the profinite completion of the fundamental group of $X^{an}$ is exactly the etale fundamental group of $X$.

Galois Groups acting on Geometric Spaces

We are now going to temporarily switch gears. (Don’t worry everything will come together in the end).  Take a field $K$ and pick an algebraic closure $\overline{K}$. Let $X, Y$ be two schemes over $K$ and let $X_{\overline{K}}, Y_{\overline{K}}$ be $X \times_{\mathrm{Spec}(K)}\mathrm{Spec}(\overline{K}), Y \times_{\mathrm{Spec}(K)} \mathrm{Spec}(\overline{K})$ respectively. Given a map $f: X \to Y$ defined over $K$ we can uniquely lift $f$ to a map $\hat{f}: X_{\overline{K}} \to Y_{\overline{K}}$ defined over $\overline{K}$. Explicitly, a map $X_{\overline{K}} \to Y_{\overline{K}}$ over $\overline{K}$ is the same thing as maps  $X_{\overline{K}} \to Y, X_{\overline{K}} \to \overline{K}$ which agree over $K$. $X_{\overline{K}}$ comes  equipped with projections to $X, \overline{K}$ and so composing the first of these with $f$ gives the desired maps.

The question we’re interested in is when we can do the converse. In other words, given a map $\hat{f}: X_{\overline{K}} \to Y_{\overline{K}}$ over $\overline{K}$, when does it descend to a map $f: X \to Y$ over $K$? To answer this question, we consider the action of $\mathrm{Gal}(\overline{K}/K)$ on $\hat{f}$. Take $\sigma \in \mathrm{Gal}(\overline{K}/K)$. Then acting with $\sigma$ on $X \times_K \overline{K}$ on the right component gives us a map $X_{\overline{K}} \to X_{\overline{K}}$ and so composing with $\hat{f}$ we get a different map $X_{\overline{K}} \to Y_{\overline{K}}$. Careful though! This new map is not defined over $\overline{K}$. To get this, we also need to have $\sigma^{-1}$ act on $Y_{\overline{K}}$. So we define $\hat{f}^{\sigma}$ as

$\begin{matrix} X_{\overline{K}} & \xrightarrow{\sigma} & X_{\overline{K}} & \xrightarrow{\widehat{f}} & Y_{\overline{K}} & \xrightarrow{\sigma^{-1}} & Y_{\overline{K}}\\ \downarrow & & \downarrow & & \downarrow & & \downarrow\\ \overline{K} & \xrightarrow{\sigma} & \overline{K} & \to & \overline{K} & \xrightarrow{\sigma^{-1}} & \overline{K}\end{matrix}$

Then the maps $\hat{f}$ that descend are precisely those that are invariant under this Galois action. After all, any map that descends will clearly have to be invariant under the Galois action, but we can argue that an invariant map will locally be defined by equations with coefficients in $K$ and so this will allow us to construct $f$.

The Etale Fundamental Group of an Elliptic Curve

Lets work through a careful example of the above ideas. Let $E$ be an elliptic curve defined over some number field $K$. If we then base change to $\overline{K}$, we get some elliptic curve $E_{\overline{K}}$ which we can think of as a torus. Since tori have fundamental group $\mathbb{Z} \times \mathbb{Z}$, we expect $E_{\overline{K}}$ to have a etale fundamental group that is $\hat{\mathbb{Z}}^2$. In fact, we recall that $E_{\overline{K}}$ has a group structure and that the multiplication by $n$ map $[n]: E_{\overline{K}} \to E_{\overline{K}}$ is etale.  The kernel of $[n]$ has exactly $n^2$ elements and for any $x \in \ker[n]$, we have that the translation by $x$ map $T_x: E_{\overline{K}} \to E_{\overline{K}}$ is an automorphism of $E_{\overline{K}}$ which commutes with $[n]$. Thus we have found $n^2$ automorphisms of $E_{\overline{K}}$ over itself via $[n]$. It follows that the pairs $(E_{\overline{K}}, [n]: E_{\overline{K}} \to E_{\overline{K}})$ are all Galois over $E_{\overline{K}}$ and the inverse limit of their automorphism groups is $\hat{\mathbb{Z}}^2$. To confirm that this is in fact the full etale fundamental group of $E_{\overline{K}}$ we need to show that the maps $[n]$ are cofinal in the etale site over $E_{\overline{K}}$. This can be done by showing that the etale maps are isogenies and noting that given an isogeny $f: E'_{\overline{K}} \to E_{\overline{K}}$, we get a dual isogeny $\hat{f}: E_{\overline{K}} \to E'_{\overline{K}}$ and their composition is $[\deg f]$.

Now, fix some $[n]: E_{\overline{K}} \to E_{\overline{K}}$ and consider the automorphisms $T_x:E_{\overline{K}} \to E_{\overline{K}}$ for $x \in \ker [n]$. From our discussion in the previous section, we know that $\mathrm{Hom}_{\overline{K}}( E_{\overline{K}}, E_{\overline{K}})$ admits a $\mathrm{Gal}( \overline{K}/K)$ action. But $E$ is defined over $K$ and therefore so are $[n]$. Thus, $[n]$ is fixed by this Galois action and so the upshot is that $\mathrm{Gal}(\overline{K}/K)$ acts on the set of automorphisms $T_x$, $x \in \ker E_{\overline{K}}$. It’s clear that these actions are compatible over all $[n]$ and so we get a $\mathrm{Gal}(\overline{K}/K)$ action on the etale fundamental group.  The etale fundamental group we’ve constructed is precisely the product of the Tate module of $E$ over each prime, and the Galois action is indeed the usual one.

Lets conclude by sketching a slightly different way to see this action. Returning to $E$ defined over $K$, we now try to compute its etale fundamental group. The map $E_{\overline{K}}\to E$ gives a functor of their etale sites and therefore etale fundamental groups. Similarly, we have the structure map $E \to \mathrm{Spec}(K)$. One shows that inside $E$, one still has all the multiplication by $[n]$ maps and also a new class of etale morphisms given by base-changing $E$ by some finite extension of $K$. Using these morphisms, one shows that the maps $\pi_1(E_{\overline{K}}) \to \pi_1(E)$ and $\pi_1(E) \to \pi_1(K)$ (where $\pi_1(X)$ denotes the etale fundamental group of $X$) are injective and surjective respectively. Thus we have an exact sequence

$1 \to \pi_1(E_{\overline{K}}) \to \pi_1(E) \to \pi_1(K) \to 1$

But in any short exact sequence of groups $1 \to F \to G \to H \to 1$, we can take lifts of $H$ into $G$ and these lifts give an action of $H$ on $F \hookrightarrow G$ by conjugation. This is again our desired action.

I mentioned a few grievances to you in person, but I thought I’d add the two most important ones here. 🙂 First, I forget what the definition of Galois category Lenstra uses, but it shouldn’t be true that the Galois group (automorphism of the fiber functor) is profinite. At least not if you want to allow all discrete covers to be a Galois category which has Galois group $\pi_1$ (opposed to finite covers which has Galois group $\widehat{\pi_1}$).
Secondly, I think it’s misleading to say that geometrically $\mathrm{Spec}(K)$ is a point. Geometry happens over $\overline{K}$. In particular, points should be contractible, and so shouldn’t have non-trivial homotopy/cohomology, whereas $\mathrm{Spec}(K)$ does. It’s better to think about it as a point ($\mathrm{Spec}(\overline{K})$) and a group action (the action of $G_K$).