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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?

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up vote 9 down vote accepted

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.

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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 $K$.

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Ah yes, you are right, I muddled these two things together. Is there a nice explicit way to think about Weil restriction of scalars in more complicated situations, for example over number fields? – David Hansen Dec 1 '09 at 0:11
@Alex: just to let you know, the mechanics of this site mean that it's highly likely that the only person who will see your comments are the person who wrote the answer that you're commenting on (i.e. me, in this case). Commenting on my answer does not bump the question to the top of the list, and there are no messages sent to the original poster or others involved. In short, what I'm saying is that if you want to say something to me then sure, submit a comment on my answer, but if you want to say something to the OP you're much better off commenting on the question orsubmittinganothranswr – Kevin Buzzard Sep 20 '10 at 19:55
Thanks for pointing this out, Kevin! I have now posted it as an answer, so I will just delete my comments, so as not to clutter up the thread. – Alex B. Sep 25 '10 at 9:32

You may be interested in Illusie's survey.

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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.

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See also: – S. Carnahan Sep 25 '10 at 11:49
Good point. Always worth checking Wikipedia, before posting. – Alex B. Sep 25 '10 at 13:23

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.

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I've never seen the word "descent" used to refer to Weil restriction (aka restriction of scalars) before. As far as I can tell, Weil restriction is a functor, and descent is a way to determine if an object is in the essential image of the left adjoint to that functor. – S. Carnahan Nov 28 '09 at 2:43
Yes,this is correct (as the very book I recommend confirms !) I was going to edit it, and I noticed you (probably) already did. How come "edited" didn't appear below the answer? – Georges Elencwajg Nov 28 '09 at 8:42

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