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Denote $A=k[x_1,\dots,x_n]/I$, where $I$ is a radical ideal and suppose then $X=\Spec A$. The previous two observations imply that $X(k)\cong V(I)$ as sets. It is also not difficult to see that this bijection promotes to a homeomorphism (use observation 2 to see that for an ideal $J\subset k[x_1,\dots,x_n]/I$, the closed set $X(k)\cap V(J)$ of $X(k)$ is sent to the vanishing locus of $J$ inside $V(I)$). Finally, we tackle the structure sheaf of $X(k)$. Denote $\sO_{V(I)}$ to the structure sheaf of $V(I)$ (the sheaf of regular functions on this algebraic set). We want to show that $\sO_{X(k)}\cong\sO_{V(I)}$ via the previous homeomorphism $X(k)\cong V(I)$. Let $U\subset X$ be open. By definition, a section $s\in \sO_{X(k)}(U\cap X(k))$ is a continuous lifting of the inclusion $U\cap X(k)\to X$ to the projection $|\sO_X|\to X$ (where $|\sO_X|$ is the étale space of $\sO_X$) i.e., a continuous map $s:U\cap X(k)\to |\sO_X|$ that makes the triangle commute. Equivalently, $s$ is a lifting such that for all $f\in A$ and all sections $t=g/f^n\in\sO_X(D(f))=A[f^{-1}]$, the set $W=s^{-1}(\operatorname{Im}\dot{t})$ is open, where $\dot{t}:D(f)\to|\sO_X|$ is the induced continuous local section. That is, it holds $s(x)=(g/f^n)_x\in\sO_{X,x}$ for all $x\in W$. Since the sets $W$ cover $X(k)$, evaluation $\sO_{X,x}\to\kappa(x)\cong k$ provides us with a morphism $\sO_{X(k)}\to\sO_{V(I)}$.

Here's where I don't know how to continue. I don't know if $\sO_{X(k)}\to\sO_{V(I)}$ will be an isomorphism at all. I have been unable to find a counterexample nor a proof of the fact that it is an iso.

Denote $A=k[x_1,\dots,x_n]/I$, where $I$ is a radical ideal and suppose then $X=\Spec A$. The previous two observations imply that $X(k)\cong V(I)$ as sets. It is also not difficult to see that this bijection promotes to a homeomorphism (use observation 2 to see that for an ideal $J\subset k[x_1,\dots,x_n]/I$, the closed set $X(k)\cap V(J)$ of $X(k)$ is sent to the vanishing locus of $J$ inside $V(I)$). Finally, we tackle the structure sheaf of $X(k)$. Denote $\sO_{V(I)}$ to the structure sheaf of $V(I)$ (the sheaf of regular functions on this algebraic set). We want to show that $\sO_{X(k)}\cong\sO_{V(I)}$ via the previous homeomorphism $X(k)\cong V(I)$. Let $U\subset X$ be open. By definition, a section $s\in \sO_{X(k)}(U\cap X(k))$ is a continuous lifting of the inclusion $U\cap X(k)\to X$ to the projection $|\sO_X|\to X$ (where $|\sO_X|$ is the étale space of $\sO_X$) i.e., a continuous map $s:U\cap X(k)\to |\sO_X|$ that makes the triangle commute. Equivalently, $s$ is a lifting such that for all $f\in A$ and all sections $t=g/f^n\in\sO_X(D(f))=A[f^{-1}]$, the set $W=s^{-1}(\operatorname{Im}\dot{t})$ is open, where $\dot{t}:D(f)\to|\sO_X|$ is the induced continuous local section. That is, it holds $s(x)=(g/f^n)_x\in\sO_{X,x}$ for all $x\in W$. Since the sets $W$ cover $X(k)$, evaluation $\sO_{X,x}\to\kappa(x)\cong k$ provides us with a morphism $\sO_{X(k)}\to\sO_{V(I)}$. This morphism is onto, since it is onto on stalks.

Here's where I don't know how to continue. I don't know if $\sO_{X(k)}\to\sO_{V(I)}$ will be injective (hence, an isomorphism). I have been unable to find a counterexample nor a proof of the fact that it is an iso.

Somehow related to this issue, in MSE we find questions 1 and 2.

So the only reasonable alternative left is to define $G(X)=\Spec(\mathbb{R}[x])$. But this also brings us into trouble, because $\Gamma(X,\sO_X)$ identify with classical endomorphisms of the real affine line. Very concretely, if we consider the morphism \begin{align*} f:X&\to X\\ x&\mapsto \frac{1}{x^2+1},\tag{1}\label{1} \end{align*} then the map $f^*:\Gamma(X,\sO_X)\to \Gamma(X,\sO_X)$, obtained by precomposition by $f$, does not preserve $\mathbb{R}[x]$. It is not clear then what $G(f):\Spec(\mathbb{R}[x])\to \Spec(\mathbb{R}[x])$ should be.

The following is just trying to carry out the construction we know possible when $k$ is algebraically closed, but now performed for an arbitrary field.

Let $X$ be a schematic $k$-prevariety. We are going to define $F(X)$ in the following way: we declare its underlying set to be the subset of $X$ of $k$-rational points (recall that a point $x\in X$ is said to be $k$-rational if the canonical morphism $k\to\kappa(x)$ is an isomorphism); equivalently, via a natural identification, it is the set of $k$-valued points over $k$ on $X$, i.e., $\Hom_{\Spec k}(\Spec k,X)$. We denote it as $X(k)$. The topology on $X(k)$ is the subspace topology inherited from $X$. We turn $X(k)$ into a a locally ringed space by equipping it with the structure sheaf $\sO_{X(k)}:=i^{-1}\sO_X$, where $i:X(k)\to X$ is the inclusion. Thus, $i$ promotes to a morphism of of locally ringed spaces. In particular, $X(k)$ is a locally ringed space over $\Spec k$.

R. van Dobben de Bruyn provided a counterexample to question 1 in the comments. Also, I was looking at this and it seems that in order to define $F$ when $k=\mathbb{R}$ (or possibly $k$ a real-closed field) one has to work not with $\spvar_k$ but rather with the full subcategory $\spvar_k'$, whose objects are the schematic prevarieties $X$ such that $X(k)\subset X$ is dense (this doesn't happen in van Dobben de Bruyn's counterexample). However, density of $X(k)$ seems to be a too restrictive condition for more general $k$. For instance, for $k=\mathbb{F}_p$, the non-empty open set $D(x(x^{p-1}-1))$ of $\mathbb{A}_k^1=\Spec k[x]$ is disjoint from $\mathbb{A}_k^1(k)$. So I'm very lost about how we could construct $F$ for arbitrary $k$.

So the only reasonable alternative left is to define $G(X)=\Spec(\mathbb{R}[x])$. But this also brings us into trouble, because $\Gamma(X,\sO_X)$ identifies with classical endomorphisms of the real affine line. Very concretely, if we consider the morphism \begin{align*} f:X&\to X\\ x&\mapsto \frac{1}{x^2+1},\tag{1}\label{1} \end{align*} then the map $f^*:\Gamma(X,\sO_X)\to \Gamma(X,\sO_X)$, obtained by precomposition by $f$, does not preserve $\mathbb{R}[x]$. It is not clear then what $G(f):\Spec(\mathbb{R}[x])\to \Spec(\mathbb{R}[x])$ should be.

The following is just trying to carry out the construction we know is possible when $k$ is algebraically closed, but now performed for an arbitrary field.

Let $X$ be a schematic $k$-prevariety. We are going to define $F(X)$ in the following way: we declare its underlying set to be the subset of $X$ of $k$-rational points (recall that a point $x\in X$ is said to be $k$-rational if the canonical morphism $k\to\kappa(x)$ is an isomorphism); equivalently, via a natural identification, it is the set of $k$-valued points over $k$ on $X$, i.e., $\Hom_{\Spec k}(\Spec k,X)$. We denote it as $X(k)$. The topology on $X(k)$ is the subspace topology inherited from $X$. We turn $X(k)$ into a locally ringed space by equipping it with the structure sheaf $\sO_{X(k)}:=i^{-1}\sO_X$, where $i:X(k)\to X$ is the inclusion. Thus, $i$ promotes to a morphism of locally ringed spaces. In particular, $X(k)$ is a locally ringed space over $\Spec k$.

R. van Dobben de Bruyn provided a counterexample to question 1 in the comments. Also, I was looking at this and it seems that in order to define $F$ when $k=\mathbb{R}$ (or possibly $k$ a real-closed field) one has to work not with $\spvar_k$ but rather with the full subcategory $\spvar_k'$, whose objects are the schematic prevarieties $X$ such that $X(k)\subset X$ is dense (this doesn't happen in van Dobben de Bruyn's counterexample). However, the density of $X(k)$ seems to be a too restrictive condition for more general $k$. For instance, for $k=\mathbb{F}_p$, the non-empty open set $D(x(x^{p-1}-1))$ of $\mathbb{A}_k^1=\Spec k[x]$ is disjoint from $\mathbb{A}_k^1(k)$. So I'm very lost about how we could construct $F$ for arbitrary $k$.