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Maths > Algebraic geometry > Functor of points

Galois descent for morphisms of functors of points

Posted by Martin Orr 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

  1. X(\Delta) is injective, and
  2. 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.

Tags alg-geom, descent, maths, points-func

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