# Points of a variety defined by Galois descent

Let k be a perfect field. By a k-variety, I shall mean a geometrically reduced separated scheme of finite type over k. I think that is enough conditions that the following data determine an affine k-variety:

1. A subset $X(\bar{k})$ of $\bar{k}^n$ which is defined by polynomials
2. A continuous action of $\mathop{\mathrm{Gal}}(\bar{k}/k)$ on $X(\bar{k})$, such that each $\sigma \in \mathop{\mathrm{Gal}}(\bar{k}/k)$ acts as $\sigma \circ f$ where f is a $\bar{k}$-regular map

When I say that these data determine an affine k-variety, I mean that there is a unique affine k-variety X whose $\bar{k}$-points are $X(\bar{k})$ with the correct Galois action.

Given these data, I want to work out the functor of points of X (which I consider to have domain the category of k-algebras). You can do that by following through the proof that these data determine a k-variety: first construct the coordinate ring A of X, as the Galois-fixed points of the ring of regular functions $X(\bar{k}) \to \bar{k}$; then $X(R) = \mathop{\mathrm{Hom}}(A, R)$ for any k-algebra R.

But if L is an algebraic extension of k, then there is a much simpler way of working out the L-points of X: just take the subset of $X(\bar{k})$ fixed by $\mathop{\mathrm{Gal}}(\bar{k}/L)$.

If L is a transcendental extension of k (or even a k-algebra which is not a field), is there a direct way of writing down the L-points of X which does not require going through the coordinate ring (or essentially equivalently, going through defining equations for X)?

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Do you have any reason to expect that there would be? –  Qiaochu Yuan Jan 13 '10 at 22:41
No, it seems likely to me that there is not (but I don't have much idea why I think that). I should have said this when I wrote the question. –  Martin Orr Jan 14 '10 at 0:53

With that proviso, my preliminary answer is no. The data of the coordinate ring is of course equivalent to that of the set of polynomials $\{P_i\}$ in (1). Thus your question sounds to me like asking: is there some way to dispense with condition (1)? Of course not: just because the set is Galois invariant doesn't mean it has any kind of algebraic structure (e.g. take $k = \overline{k}$ and we are merely saying that not just any old subset of affine space defines an affine variety).
Moreover, I don't see any shortcut around actually using the data of (1) and (2) to compute the coordinate ring. This is a very basic Galois descent argument involving Hilbert 90 applied to the ideal of $\overline{k}[x_1,\ldots,x_n]$ defined via (1).
Here is a story related to my first paragraph above (and hence squarely off-topic; I couldn't resist). Once as a graduate student I found a $40 parking fee added to my term bill. I called and explained that this was a mistake. "Are you sure?" "Yes -- I don't have a car." They removed the charge, but then a few months later it happened again. That made me really nervous: what more can I say? Suppose they asked me to supply written documentation for my claim? How would I go about convincing an even mildly skeptical person? Luckily that was the last time it happened! – Pete L. Clark Jan 14 '10 at 1:46 The following seems to give a reasonable affirmative answer which avoids computing the coordinate ring directly, and replaces condition (2) with the more natural condition that the subset$\Sigma := X(\overline{k})$in (1) is stable under the action of the Galois group on$\overline{k}^n$. Let's be cleaner by working more generally over an arbitrary (not necessarily perfect) field$k$and with geometrically reduced closed subschemes$X$in a fixed separated$k$-scheme$Y$locally of finite type. (Note: now affine schemes are gone; can take$Y$to be an affine space, but this is irrelevant.) The${\rm{Gal}}(k_s/k)$-stable set$\Sigma = X(k_s)$in$Y(k_s)$recovers$X$as follows. For a$k$-algebra$A$,$X(A)$is the${\rm{Gal}}(k_s/k)$-invariants in$X(A_{k_s})$, so we just need to describe$X(A_{k_s})$as a${\rm{Gal}}(k_s/k)$-stable subset of$Y(A_{k_s})$. The description in this latter case will be in terms of$\Sigma$, and the${\rm{Gal}}(k_s/k)$-stability of$\Sigma$inside of$Y(k_s)$will ensure that the description we give for$X(A_{k_s})$is${\rm{Gal}}(k_s/k)$-stable inside of$Y(A_{k_s})$. That being noted, we rename$k_s$as$k$so that$k$is separably closed and$\Sigma$is simply a set of$k$-rational points of$Y$(so the notation is now marginally cleaner). First assume$A$is geometrically reduced in the sense that$A_K$is reduced for any extension field$K/k$. Since$X(A)$is the direct limit (inside$Y(A)$) of the$X(A_i)$as$A_i$varies through$k$-subalgebras of finite type in$A$(all of which are geometrically reduced), we may assume$A$is finitely generated over$k$. Then the$k$-points are Zariski-dense (as$k = k_s$) and so the condition on$y \in Y(A)$that it lies in$X(A)$is that$y(\xi) \in \Sigma$for all$k$-points$\xi$of$A$. That describes$X(A)$for any (possibly not finitely generated)$k$-algebra$A$that is geometrically reduced. In general, to check if$y \in Y(A)$lies in$X(A)$amounts to the same for each local ring of$A$, so we can assume$A$is local. Then the condition for$y$to be in$X(A)$is exactly that there is a local map of local$k$-algebras$B \rightarrow A$with$B$geometrically reduced such that$y$is in the image of$X(B)$under the induced map$Y(B) \rightarrow Y(A)$. I don't claim this formulation is the best way to think about it, but it "works". Of course, one can apply this process to any${\rm{Gal}}(k_s/k)$-stable subset$\Sigma$of$Y(k_s)$provided that we first replace$\Sigma$with with the set of$k_s$-points of its Zariski-closure in$Y_{k_s}$. Then we just obtain the Galois descent$X$of the Zariski closure in$Y_{k_s}$of$\Sigma$. In general$X(k_s)$may be larger than$\Sigma$, but nonetheless$\Sigma$is Zariski-dense in$X_{k_s}$. This is perfectly interesting in practice, regardless of whether or not$\Sigma$is equal to$X_{k_s}$, since it is what underlies the construction of derived groups, commutator subgroups, images, orbits, and related things in the theory of linear algebraic groups over a general field. For example, the$k$-group${\rm{PGL}}_n$is its own derived group in the sense of algebraic groups, but the commutator subgroup of${\rm{PGL}}_n(k_s)$is a proper subgroup whenever$k$is imperfect and${\rm{char}}(k)|n$. To give a nifty application, suppose one begins with an arbitrary closed subscheme$X'$in$Y$(such as$X' = Y$!), then forms the${\rm{Gal}}(k_s/k)$-stable set$X'(k_s)$(which could well be empty, or somehow really tiny), and then applies the above procedure to get a geometrically reduced closed subscheme$X$in$X'$. What is it? It is the maximal geometrically reduced closed subscheme of$X'$, and one can check its formation is compatible with products (as well as separable extensions$K/k$, such as completions$k_v/k$for a global field$k$). If$k$is perfect then$X = X'_{\rm{red}}$, so this is more interesting when$k$is imperfect. It is especially interesting in the special case when$X'$is equipped with a structure of$k$-group scheme. Then$X$is its maximal smooth closed$k$-subgroup, since geometrically reduced$k$-groups locally of finite type are smooth. So what? If one is faced with the task of studying the Tate-Shararevich set for such an$X'$(e.g., maybe$X'$is a nasty automorphism scheme of something nice) then all that really intervenes is$X$since it captures all of the local points, so for some purposes we can replace the possibly bad$X'$with the smooth$X$. (This trick is used in the proof of finiteness of Tate-Shafarevich sets for arbitrary affine groups of finite type over global function fields.) But beware: if the$k$-group$X'$is connected (and$k$is imperfect) then$X$may be disconnected and have much smaller dimension; see Remark C.4.2 in the book "Pseudo-reductive groups" for an example. - The condition (2) is there because I didn't want to be limited to cases where the Galois action is induced from an action on affine space. For example, I wanted to be able to take the subscheme of$\mathbb{A}^2_\mathbb{C}$defined by$X - iY = 0$, let$\mathop{\mathrm{Gal}}(\mathbb{C}\mathbb{R})$act on its$\mathbb{C}$points trivially, and say that this descends (to something isomorphic to$\mathbb{A}^1_\mathbb{R}$). Condition (2) was my attempt to write down a condition to do this, which might not be correct. I am not sure why I wanted this. – Martin Orr Feb 15 '10 at 21:36 Thanks for the answer, but there is one thing I don't understand: if$a \in Y(A)$and$x \in \Sigma = X(\bar{k})$, then what does$a(x)\$ mean? –  Martin Orr Feb 15 '10 at 21:38