What is etale descent?

Question

What is etale descent? I have a vague notion that, for example, given a variety $V$ over a number field $K$, etale descent will produce (sometimes) a variety $V'$ over $\mathbb{Q}$ of the same complex dimension which is isomorphic to $V$ over $K$ and such that $V(K)=V'(\mathbb{Q})$. Is this at all right? How does one do such a thing?

Answer 1

Let $L/K$ be a Galois field extension and consider a variety $Y$ over $L$. The theory of (Galois) descent addresses the question whether $Y$ can be defined over $K$. More precisely, the question is: "does there exist a variety $X$ over $K$ such that $Y = X \times_{Spec(K)} Spec(L)$".

Now assume such $X$ does exist. In this case $Y$ is endowed with $Gal(L/K)$ action coming from an action on the second factor.

Conversely, if $Y$ has a Galois action compatible with the action on $Spec(L)$, then $Y$ descends to some $X$ defined over $K$. $X$ is actually a quotient of $Y$ by $Gal(L/K)$ (so that the conjugate points glue together to form one point on $X$). Note that the set of $K$-points of $X$ is the set of Galois fixed points.

Example. $K = \mathbf R$ and $L = \mathbf C$. For any real variety, the set of complex points admits the action of $ \mathbf Z/2$ by complex conjugation. Conversely, if a complex variety is endowed with conjugation, it descends to a real variety. This is in fact an exercise in Hartshorne.

Remarks. Theory of descent also classifies all possible $X$'s arising from $Y$. Such $X$'s are called forms of $Y$. They are in 1-1 correspondence with a certain Galois cohomology group.

Answer 2

As I write the question looks like a muddle of two distinct notions:

1) Restriction of scalars. Given $L/K$ finite and a variety $V/L$ there's a variety $W/K$ of dimension $(dim V)[L:K]$ with $W(K)=V(L)$ canonically. For example over the complexes the variety ${\mathbf C}^*$ is defined by the equation $z\not=0$ and its restriction of scalars to the reals is (isomorphic to) the subspace of affine 2-space defined by $x^2+y^2\not=0$.

2) Descent. $L/K$ finite again, but this time separable too, and let's even make it Galois for simplicity. Given $V/K$ one can imagine $V$ as a variety over $L$. Over $L$, $V$ is suspiciously isomorphic to its conjugates. Descent (vaguely) is the idea that conversely, given a variety over $L$ isomorphic to all its conjugates (in a good way), it's indeed the base change to $L$ of a variety over $K$.

Answer 3

When a functor happens to be a sheaf in étale topology you say it satisfies étale descent. So for example algebraic K-theory does not satisfy étale descent but Bott periodic algebraic K-theory modulo a suitable prime power does.

Answer 4

Let $K/k$ be a finite separable extension (not necessarily galois) and $Y$ a quasi-projective variety over $K$.

The functor $k-Alg \to Sets:A \mapsto Y(A\otimes_k K)$ is representable by a quasi-projective $k$-scheme $Y_0=R_{K/k}(Y)$. We have a functorial adjunction isomorphism $Hom_{k-schemes}(X,R_{K/k}(Y))=Hom_{K-schemes}(X\otimes _k K,Y)$

and the $k$-scheme $Y_0=R_{K/k}(Y)$ is said to be obtained from the $K$-scheme $Y$ by Weil descent. For example if you quite modestly take $X=Spec(k)$, you get $(R_{K/k}(Y))(k)=Y_0(k)=Y(K)$, a formula that Buzzard quite rightfully mentions. If $Y=G$ is an algebraic group over $K$, its Weil restriction $R_{K/k}(G)$ will be an algebraic group over $k$.

As the name says this is due (in a different language) to André Weil: The field of definition of a variety. Amer. J. Math. 78 (1956), 509–524.

Chapter 16 of Milne's online Algebraic Geometry book is a masterful exposition of descent theory, which will give you many properties of $(R_{K/k}(Y))(k)$ (with proofs), and the only reasonable thing for me to do is stop here and refer you to his wondeful notes.

Answer 5

You may be interested in Illusie's survey.

Answer 6

This used to be a comment, but as Kevin pointed out you might never have found out that I left one. So just in case this is still of any relevance, I will repeat it here.

I know, this thread is old so maybe you have already figured it out yourself, but in case this is not so, here goes: you asked in a comment whether there was any explicit way of thinking about Weil restriction of scalars and indeed there is. Let $L/K$ be a finite extension of fields and $V/L$ a variety, given by a set of equations $f_i(x_1,…,x_n)$ with coefficients in $L$. Fix a basis $u_1,…,u_m$ for $L/K$. Each variable $x_r$ is a variable in $L$ but you can instead write $x_r=\sum_s y_{r,s}u_s$, where now $y_{r,s}$ are variables in $K$. Do the same with the coefficients of the $f_i$. By comparing the coefficients of each $u_s$, you get $[L:K]$ equations over $K$ and this new system describes a variety over $K$ - the Weil restriction of scalars. You immediately check that its dimension is indeed $[L:K]$ times the dimension of the original variety. If you do this with an explicit simple example, like Weil restrict an elliptic curve from $\mathbb{Q}_i$ to $\mathbb{Q}$, then you will get a much better feel for what's going on.