I am starting on a Phd program and am supposed to read Colliot Thelene and Sansuc's article on R-equivalence for tori. I find it very difficult and although I have some knowledge over schemes , I am completely baffled by this scalar restriction business of having a field extension $K/k$ , a torus over $K$ and "restricting" it to $k$. I would be very gratefull for a reference or even better by some explanation . I found nothing in my standard books (Hartshorne, Qing Liu, Mumford etc) so I hope this question is appropriate for the site. Thank you.

  • 2
    I believe you are looking for "Weil restriction" of a variety. For a start see Wikipedia: en.wikipedia.org/wiki/Weil_restriction There's also an example involving tori. – Lars Dec 3 '09 at 21:28
  • You can look at the book of "Neron Models" - BLR, for a more detailed description in somewhat more generality. – isildur Jul 10 '11 at 0:55
up vote 24 down vote accepted

As said, the sought after concept is also known as Weil restriction. In a word, it is the algebraic analogue of the process of viewing an $n$-dimensional complex variety as a $(2n)$-dimensional real variety.

The setup is as follows: let $L/K$ be a finite degree field extension and let $X$ be a scheme over $L$. Then the Weil restriction $W_{L/K} X$ is the $K$-scheme representing the following functor on the category of K-algebras:

$A\mapsto X(A \otimes_K L)$.

In particular, one has $W_{L/K} X(K) = X(L)$.

By abstract nonsense (Yoneda...), if such a scheme exists it is uniquely determined by the above functor. For existence, some hypotheses are necessary, but I believe that it exists whenever $X$ is reduced of finite type.

Now for a more concrete description. Suppose $X = \mathrm{Spec} L[y_1,...,y_n]/J$ is an affine scheme. Let $d = [L:K]$ and $a_1,...,a_d$ be a $K$-basis of $L$. Then we make the following "substitution":

$$y_i = a_1 x_{i1} + ... + a_d x_{id},$$

thus replacing each $y_i$ by a linear expression in d new variables $x_{ij}$. Moreover, suppose $J = \langle g_1,...,g_m \rangle$; then we substitute each of the above equations into $g_k(y_1,...,y_n)$ getting a polynomial in the $x$-variables, however still with $L$-coefficients. But now using our fixed basis of $L/K$, we can regard a single polynomial with $L$-coefficients as a vector of $d$ polynomials with $K$ coefficients. Thus we end up with $md$ generating polynomials in the $x$-variables, say generating an ideal $I$ in $K[x_{ij}]$, and we put $\mathrm Res_{L/K} X = \mathrm{Spec} K[x_{ij}]/I$.

A great example to look at is the case $X = G_m$ (multiplicative group) over $L = \mathbb{C}$ (complex numbers) and $K = \mathbb{R}$. Then $X$ is the spectrum of

$$\mathbb{C}[y_1,y_2]/(y_1 y_2 - 1);$$

put $y_i = x_{i1} + \sqrt{-1} x_{i2}$ and do the algebra. You can really see that the corresponding real affine variety is $\mathbb{R}\left[x,y\right]\left[(x^2+y^2)^{-1}\right]$, as it should be: see e.g. p. 2 of


for the calculations.

Note the important general property that for a variety $X/L$, the dimension of the Weil restriction from $L$ down to $K$ is $[L:K]$ times the dimension of $X/L$. This is good to keep in mind so as not to confuse it with another possible interpretation of "restriction of scalars", namely composition of the map $X \to \mathrm{Spec} L$ with the map $\mathrm{Spec} L \to Spec K$ to give a map $X \to \mathrm{Spec} K$. This is a much weirder functor, which preserves the dimension but screws up things like geometric integrality. (When I first heard about "restriction of scalars", I guessed it was this latter thing and got very confused.)

  • 2
    Pete: While this is generally a great answer, the restriction of scalars of (G_m)_C is not A^2_R - {0,0}. Rather, it is Spec R[x,y][ (x^2+y^2)^{-1} ]. These have the same real points, but they are not the same scheme! – David E Speyer Dec 6 '09 at 3:39
  • @David: you're right. In the back of my mind I was wondering about how the Weil restriction of an affine variety became quasi-affine. I'll change it. – Pete L. Clark Dec 6 '09 at 6:07
  • You should be reading my answers more carefully Pete ;-) mathoverflow.net/questions/6979/what-is-etale-descent/6986#6986 – Kevin Buzzard Dec 6 '09 at 16:52
  • @buzzard: yes, sir. Sorry, sir. :) – Pete L. Clark Dec 6 '09 at 17:06
  • Great answer: one of the clearer and more concise description of Weil restriction I have ever found. – Filippo Alberto Edoardo May 17 '14 at 22:10

There is another description for tori. The category of tori over a field F is equivalent to the category of finite-dimensional $G_F$-lattices. Now, there is an operation of induction for group representations that converts a $G_K$-lattice into a $G_k$-lattice; this is the lattice you need.

See http://en.wikipedia.org/wiki/Algebraic_torus

It's not that hard at all. Here is an example. Let $k = \mathbf R$ and $K = \mathbf C$. Consider a 1-dimensional torus $G_m$ over $\mathbf C$. It basically the group $\mathbf C^*$ over $\mathbf C$.

Now $G = Res_{\mathbf C/\mathbf R} G_m$ is the same group $\mathbf C^*$ considered as group over $\mathbf R$.

  • 11
    Giving the points over one field of a torus doesn't even begin to describe it. the comment "it's not hard at all" is useless and patronising. – user2292 Dec 5 '09 at 21:36
  • As other people commented, for a definition and properties you could check out wikipedia first. I gave you an example. And I do believe that this kind of stuff is easy. – Evgeny Shinder Dec 29 '09 at 5:15
  • 1
    I think Evgeny's example is a good one to keep in mind. (As Pete Clark wrote in his answer, restriction of scalars is just the process of thinking of a ${\mathbb C}$-variety, say, as an ${\mathbb R}$-variety instead.) – Emerton Jan 16 '10 at 4:33

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.