Martin's Blog : Galois descent for morphisms of functors of points

Galois descent for morphisms of functors of points

Posted by martin on Saturday, 20 February 2010 at 21:58

I was disappointed in my last post that I was unable to prove any results about Galois descent for morphisms of functors. I have now tracked down a fairly mild condition on the functors that you need for this descent to work, which I shall explain below. Importantly, this condition is satisfied automatically by the functors of points of a scheme (though I won’t prove this).

This tells us that if you have two $k$ -functors satisfying the Galois exactness property, and a morphism of their restrictions to $K\textbf{-Alg}$ which commutes with the action of $\mathop{\mathrm{Gal}}(K/k)$ , then it comes from a unique morphism of $k$ -functors.

I shall not discuss descending functors, only morphisms. But a small modification to the Galois exactness condition should allow you to descend functors themselves.

I shall use notation from my previous post setting up the framework for Galois descent of functors.

Throughout this post, $K/k$ will denote a Galois field extension, with Galois group $G$ . If $B$ is a $k$ -algebra, I shall write $B_K$ for $B \otimes_k K$ , given the structure of a $K$ -algebra through the right-hand factor.

Galois exactness

If $X$ is a $k$ -functor, then I shall write $X_K$ for the $K$ -functor obtained by composing $X$ with the forgetful functor $K\textbf{-Alg} \to k\textbf{-Alg}$ .

Recall that $X_K$ comes with a semilinear $\mathop{\mathrm{Gal}}(K/k)$ -action $(\phi_\sigma)$ – in fact this is just given by identity maps, but I would prefer to keep the action explicit as I tend to think about $(X_K, \phi_\sigma)$ as only defined up to a canonical isomorphism of $K$ -functors with $K/k$ -action.

For every $k$ -algebra $B$ , there is a homomorphism of $k$ -algebras $\Delta : B \to B \otimes_k K$ given by $b \mapsto b \otimes 1$ . There are also homomorphisms (of $K$ -algebras) $\sigma_B : B_K \to F_\sigma(B_K)$ given by $b \otimes x \mapsto b \otimes \sigma(x)$ for each $\sigma \in G$ .

These homomorphisms have the property that if we identify $B_K$ and $F_\sigma(B_K)$ in the “trivial” way (as $k$ -algebras but only semilinearly as $K$ -algebras), then $\Delta$ is the equalizer of the $\sigma_B$ in the category $k\textbf{-Alg}$ .

I shall call the $k$ -functor $X$ Galois exact if, for every $k$ -algebra $B$ , $X(\Delta)$ is the equalizer (in $\textbf{Set}$ ) of $\phi_\sigma \circ X(\sigma_B) : X(B_K) \to X(B_K)$ ; in other words

$X(\Delta)$ is injective, and
The image of $X(\Delta)$ is $X(B_K)^G := \{ x \in X(B_K) : \phi_\sigma(X(\sigma_B)(x)) = x \text{ for all } \sigma \in G \}$ .

An important fact is that the functor of points of a $k$ -scheme is always Galois exact. To prove this, you need to do some geometry to reduce to the affine case.

Descending morphims of functors

Now let $X$ , $Y$ be two $k$ -functors and $\tilde{f}$ a natural transformation $X_K \to Y_K$ . We need to assume that $Y$ is Galois exact.

I want to show that, if $\tilde{f}$ commutes with the Galois action, then there is a unique morphism of $k$ -functors $f : X \to Y$ extending $\tilde{f}$ .

I will say that $\tilde{f}$ commutes with the Galois action if the following diagram commutes for all $K$ -algebras $B$ and $\sigma \in G$ :

$\usepackage{xypic} \xymatrixcolsep{4pc} \xymatrix{ {X(F_\sigma(B))} \ar[d]^{X(\sigma_B)} \ar[r]^{\tilde{f}(F_\sigma(B))} & {Y(F_\sigma(B))} \ar[d]^{Y(\sigma_B)} \\ {X(B)} \ar[r]^{\tilde{f}(B)} & {Y(B)} }$

To construct $f$ , consider the following diagram for each $k$ -algebra $B$ :

$\usepackage{xypic} \xymatrixcolsep{3pc} \xymatrix{ {X(B_K)} \ar[r]^{\tilde{f}(B_K)} & {Y(B_K)} \\ {X(B)} \ar[u]^{X(\Delta)} \ar@{.>}[r]^{f(B)} & {Y(B)} \ar[u]^{Y(\Delta)} }$

Because $\tilde{f}$ commutes with the Galois action, $\tilde{f}(B_K) \circ X(\Delta)$ is unchanged by postcomposition with $\phi_\sigma \circ Y(\sigma_B)$ for each $\sigma \in G$ . So by the Galois exactness of $Y$ , there is a unique $f(B)$ making the diagram commute.

Proving that this really is descent

To check that $f_K = \tilde{f}$ , we need to check that if $B$ is a $K$ -algebra, then the above diagram commutes when $\tilde{f}(B)$ is put in place of the bottom arrow.

This is not an immediate consequence of the fact that $\tilde{f}$ is a natural transformation because $\Delta$ is not a homomorphism of $K$ -algebras. Instead consider the following diagram, where $\mu : B_K \to B$ is the $K$ -algebra homomorphism $b \otimes x \mapsto bx$ :

$\usepackage{xypic}\xymatrixcolsep{3pc}\xymatrix{ & {X(B_K)} \ar[d]^{X(\mu)} \ar[r]^{\tilde{f}(B_K)} & {Y(B_K)} \ar[d]^{Y(\mu)} \\ {X(B)} \ar[ur]^{X(\Delta)} \ar[r]^{\mathop{\mathrm{id}}} & {X(B)} \ar[r]^{\tilde{f}(B)} & {Y(B)} }$

The triangle commutes because $\mu \circ \Delta = \mathop{\mathrm{id}_B}$ , and the square commutes because $\tilde{f}$ is a natural transformation of $K$ -functors. Hence the outer trapezium commutes.

Since $Y$ is Galois exact, the restriction of $Y(\mu)$ to $Y(B_K)^G$ is a bijection $Y(B_K)^G \to Y(B)$ , with inverse $Y(\Delta)$ . Since $\tilde{f}$ is Galois exact, the image of $\tilde{f}(B_K) \circ X(\Delta)$ is contained in $Y(K_B)^G$ . Combining these, we get that the required diagram commutes.