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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
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2 Answers 2

up vote 4 down vote accepted

Despite what you learn in logic classes, there's truth to the conventional wisdom that you can't prove a negative. That is, it's a lot harder to explain why something necessarily can't work than why it can. (One almost invariably settles for: "it cannot work under the following explicit conditions, plus possibly others that I have left implicit.")

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).

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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
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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.

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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
    
Martin, I have revised it now to address that. –  BCnrd Feb 16 '10 at 6:17
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