$\def\cpvar{\mathsf{CPVar}} \def\spvar{\mathsf{SPVar}} \def\Spec{\operatorname{Spec}} \def\class{\mathrm{class}} \def\sO{\mathcal{O}} \def\Hom{\operatorname{Hom}}$This question is a follow-up to this other one. On my answer, I explain that “taking $k$-rational points” defines a functor $F$ from classicizable schematic $k$-prevarieties to classical $k$-prevarieties that is neither faithful nor full. So it seems natural to ask:
Is the functor $F$ essentially surjective?
To discuss this issue, we introduce the following terminology: if we have a classical prevariety $X\in\cpvar_k$ and a schematic prevariety $X'\in\spvar_k$ such that $X'(k)\cong X$, then we say that $X'$ is a scheme-theoretic thickening of $X$ (this terminology is taken from Huisman). So the question is rephrased: does every classical prevariety $X$ have a scheme-theoretic thickening?
The answer is yes if $X$ is affine: Suppose $X=V(I)\subset k^n$. Then $X'=\Spec(k[x_1,\dots,x_n]/I(X))$ is a scheme-theoretic thickening of $X$. This is because $X'(k)\cong V(I(X))=X$ (cf. the construction of $F$ on my original post) and because $X'(k)$ is dense in $X'$, so one can apply the lemma from my answer.
However, I do wonder: what happens for general $X\in\cpvar_k$? The first idea one probably thinks of is considering an open affine cover $X=\bigcup U_i$; then, one obtains a scheme-theoretic thickening $U_i'$ of $U_i$, using the method of the last paragraph. Nonetheless, after the following question emerges: how can one glue the schemes $U_i'$ to give rise to a scheme-theoretic thickening $X'$ of $X$? It happens that the so-called coordinate ring of a classical affine variety (in last paragraph, this is $k[x_1,\dots,x_n]/I(X)$) is not an invariant of the variety, but rather, it depends on the embedding into affine space (see this). It seems that we cannot rely on an arbitrary open affine cover $X=\bigcup U_i$ to perform the gluing. For instance, for $k=\mathbb{R}$, the cover could be $X=U_1\cup U_2$, with and $\mathbb{A}_k^1=X=U_1=U_2$, with choice of isomorphisms $U_1=\mathbb{A}_k^1$ and $U_2\cong Y$ (where $Y$ and the isomorphism is given in Wofsey's answer). As Wofsey points out, there is no isomorphism between $\mathcal{A}(X)=\mathbb{R}[x]$ and $\mathcal{A}(Y)=\mathbb{R}[x,(1+x^2)^{-1}]$, and so, we cannot glue $U_1'=\Spec\mathcal{A}(X)$ and $U_2'=\Spec\mathcal{A}(Y)$. (This supposes an example of two non-isomorphic scheme-theoretic thickenings of the same classical variety.)
Maybe there are classical prevarieties without a scheme-theoretic thickening? The previous example can be fixed by just changing to an appropriate cover and choice of isomorphisms, but can such a thing always be done? Or is there maybe a classical prevariety patched up from affines ones in a way that prevents s.-t. thickenings from happening?
The only place where I've found discussing anything similar is Huisman's answer. However, in the answer and the comments they only converse about a (separated) classical variety for which all s.-t. thickenings are non-separated. (They require their varieties to be always separated; nevertheless, I'm not interested on this condition.)
