# A quick intro to Galois descent for schemes

This is a very quick introduction to Galois descent for schemes defined over fields. It is a very special case of faithfully flat descent and other topos-descent theorems, which I won't go into at all. Typically, if you look up descent in an algebraic geometry text you will quickly run into all sorts of diagrams and descent data. In my opinion, that is a very counterintuitive way to present the basic idea.

## What is the descent theorem?

Here is the main topic of this post:

Theorem. Let $E/F$ be a finite Galois extension of fields with Galois group $G$. Then the functor
\begin{align*} \{\text{quasprojective F-schemes}\}&\longrightarrow \{\text{quasprojective E schemes with compatible G action} \} \\ X&\longmapsto X\otimes_F E\end{align*}
where $X\otimes_F E$ is given an Galois action via the canonical action on $E$, is an equivalence of categories.

This is the basic theorem of Galois descent. What does it mean, and how does it work? First, I have to tell you what a compatible Galois action is. Well, if $X$ is an $E$-scheme, then there is a map $X\to{\rm Spec}(E)$, and there is the usual action of $G$ on ${\rm Spec}(E)$. Compatible just means that for each $\sigma\in G$, the square

commutes.

## How do you use the descent theorem?

Let's start with a basic example. If $E/F$ is the extension $\C/\R$ of complex numbers over the real numbers, and $A$ is the $\C$-algebra defined by $\C[x,y]/(x-iy)$, then the action of $G \cong \Z/2$ on $\C$ does not give by itself a compatible Galois action on $A$, because the ideal $(x-iy)$ is not fixed by $G$, but the extension of the action of $G$ on $\C$ to $\C[x,y](x+y)$ where $G$ fixes $x$ and $y$ is a compatible Galois action. The descent theorem in this case just says that $\C[x+y]/(x+y)$ comes from a scheme defined over $\R$, namely $\R[x,y]/(x+y)$ whereas the other one does not.

In other words, for a Galois extension of fields $E/F$, the descent theorem is essentially a fancy way of saying that if polynomial with coefficients in $E$ has roots permuted by the Galois group, then that polynomial actually has coefficients in $F$.

The most important part is how you can use the descent theorem. Why should we even care about it? It should help us understand the fundamental objects of mathematics. And by that, I mean groups. Algebraic groups, in this case.

Let's consider a more interesting example. Consider the group $\G_m$ over $E$. As we know, it is represented by the algebra $E[x,x^{-1}]$. Of course, the Galois group $G$ acts on this algebra via its action on $E$ and by fixing $x$. The descent theorem here is no surprise: it just says that $E[x,x^{-1}]$ is obtained by applying the functor $-\otimes_F E$ to some $F$-algebra, and that algebra is of course $F[x,x^{-1}]$.

Things get way more interesting if you change the action of the Galois group. Let us suppose that $E/F$ is a quadratic extension of fields. Let $\sigma\in G$ be the nontrivial $F$-automorphism of $E$. Before, we just let $\sigma$ act on $E[x,x^{-1}]$ by letting it act on $E$ through the Galois action and we set $\sigma(x) = x$. But, of course, we can get $\sigma(x) = x^{-1}$. Now we've really got something.

This action is obviously a compatible Galois action. So there exists an $F$-algebraic group that represents it. What is that algebraic group? The last thing you ever want to do is write down an algebra that represents it. Instead, we are only interested in its functor of points. So, let $X$ be this mysterious algebraic group. What is its functor of points? The descent theorem tells us how to calculate it. That is, for an $F$-algebra $B$, the group $X(B)$ is defined as the set of $G$-equivariant morphisms $\Spec(B\otimes_F E)\to \G_m$. Here, $\Spec(B\otimes_F E)$ has the $G$ action via $G$ acting on $E$ in the tensor product, and $G$ acts on $\G_m$ through our "special" action of $G$, defined on $E[x,x^{-1}]$ by the Galois action on $E$ together with the nontrivial element of $G$ sending $x\mapsto x^{-1}$. On the group side, this means inversion.

So, concretely,
$$X(B) = \{ t\in (B\otimes_F E)^\times : t^{-1} = \sigma(t) \}.$$
So, $X$ is the so-called "norm-one torus". In particular,
$$X(F) = \{ t\in \G_m(E) : t\sigma(t) = 1 \}.$$

## In closing

Here are step-by-step instructions for using the descent theorem:

1. Pick a Galois extension $E/F$ with Galois group $G$.
2. Pick an $E$-scheme represented by an $E$-algebra $E[x_1,\dots,x_n]/I$.
3. Define an action on $G$ on $E$ via the Galois action of $G$ on $E$ together with a "custom" action on the variables $x_1,\dots,x_n$, which fixes $I$.
4. The descent theorem tells you that $E$-scheme with Galois action comes from an $F$-scheme $X$.
5. Calculate the functor of points $X(B)$ for an $F$-algebra $B$ via the descent theorem as a Hom-set in the category of $E$-schemes with compatible Galois action.

Not so bad, right? Of course, you can also do the same with projective varieties if required.