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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 integral 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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${\rm{Gal}}(k_s/k)$-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). To detect if a point

First assume $y \in Y(A)$ lies A$is geometrically reduced in the sense that$X(A)$, it A_K$ is equivalent to do so with reduced for any extension field $K/k$. Since $X(A)$ is the local rings direct limit (inside $Y(A)$) of the $A$, so we now take X(A_i)$as$A_i$varies through$k$-subalgebras of finite type in$A$to be local(all of which are geometrically reduced), say with closed point we may assume$s$. A$ is finitely generated over $k$. Then the subset $X(A)$ in k$-points are Zariski-dense (as$Y(A)$consists of those k = k_s$) and so the condition on $y \in Y(A)$ such that for any local function it lies in $f$ near X(A)$is that$y(s)$which vanishes at y(\xi) \in \Sigma$ for all points $k$-points $\xi$ of $\Sigma$ near A$. That describes$y(s)$, necessarily X(A)$ for any (possibly not finitely generated) $f(y) = 0$ k$-algebra$A$that is geometrically reduced. In general, to check if$y \in Y(A)$lies in$A$. This formulation makes sense X(A)$ amounts to the same for separated algebraic spaces locally each local ring of finite type over $k$ (where "A$, so we can assume$A$is localfunction" . Then the condition for$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$is in the sense image of$X(B)$under the induced map$Y(B) \'etale topology) and rightarrow Y(A)$. I don't claim this formulation is the best way to think about itworks there too, even though$Y$is not generally covered by affine opens. but it "works". Of course, one can apply this process to any${\rm{Gal}}(k_s/k)$-stable subset$\Sigma$of$Y(k_s)$, and what is Y(k_s)$ provided that we first replace $\Sigma$ with with the meaning set of the resulting functor? It represents $k_s$-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 ${\rm{char}}(k)|n$.

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The following seems to give a reasonable affirmative answer which avoids computing the regular functionscoordinate ring directly, and replaces condition (2) (whose statement I don't understand: why is $f$ mentioned there?) 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 spaceif you wish, but this is irrelevant.) The ${\rm{Gal}}(k_s/k)$-stable set $\Sigma = X(k_s)$ in $Y(k_s)$ recovers $X$ by the following 2-step procedureas follows. First, if $A$ is For a finitely generated $k$-algebra then $A$, $X(A)$ is the set of points ${\rm{Gal}}(k_s/k)$-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 Y(A)$such that this latter case will be in terms of$a(x) = 0$\Sigma$, and the ${\rm{Gal}}(k_s/k)$-stability of $\Sigma$ inside of $Y(k_s)$ will ensure that the description we give for all $x \in \Sigma$.In generalX(A_{k_s})$is${\rm{Gal}}(k_s/k)$-stable inside of$Y(A_{k_s})$. That being noted, for any we rename$k$-algebra k_s$ as $L$ k$so that$k$is separably closed and$\Sigma$is simply a set of$k$-rational points of$Y$(so the subset notation is now marginally cleaner). To detect if a point$X(L)$y \in Y(A)$ lies in $Y(L)$ X(A)$, it is equivalent to do so with the directed union local rings of$A$, so we now take$A$to be local, say with closed point$s$. Then the sets subset$X(A)$where in$A$varies through Y(A)$ consists of those $y \in Y(A)$ such that for any local function $f$ near $y(s)$ which vanishes at all points of $k$-subalgebras \Sigma$near$A \subseteq L$y(s)$, necessarily $f(y) = 0$ in $A$. This formulation makes sense for separated algebraic spaces locally of finite type over $k$. Voila, no coordinate rings. k$(Remember: we didn't require where "local function" is in the sense of the \'etale topology) and it works there too, even though$Y\$ to be affine.) is not generally covered by affine opens.

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