3
$\begingroup$

$\def\cpvar{\mathsf{CPVar}} \def\spvar{\mathsf{SPVar}} \def\Spec{\operatorname{Spec}} \def\class{\mathrm{class}} \def\sO{\mathcal{O}} \def\Hom{\operatorname{Hom}}$Let $k$ be a field. By classical $k$-prevariety I mean a quasi-compact locally ringed space over $\Spec k$ that is locally isomorphic (over $\Spec k$) to an algebraic subset of $k^n$ with the sheaf of regular functions. This is the same as a “prevariety in the sense of FAC” (although Serre required $k$ to be algebraically closed). (A classical $k$-variety is a classical $k$-prevariety that is separated, see e.g. Milne's notes, 5.7, 5.8 and 5.26.) A morphism of classical $k$-prevarieties is a morphism of locally ringed spaces over $\Spec k$.

Now suppose $\overline{k}=k$. There is an equivalence of categories between classical $k$-prevarieties and reduced $k$-schemes of finite type. I wrote a very detailed account and proof of the result in The Classical-Schematic Equivalence. Alternatively, there also is Haiman's Varieties as Schemes.

However, I wonder: What happens when $\overline{k}\neq k$? Let us call a “schematic $k$-prevariety” to a reduced $k$-scheme of finite type. Can we relate somehow the categories of classical and schematic $k$-prevarieties? Is there some “canonical” functor in any direction between them? Denote $\cpvar_k$ and $\spvar_k$ to these categories.

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

Obstructions for the existence of a “canonical” functor $G:\cpvar_k\to\spvar_k$

consider the case $k=\mathbb{R}$.

Where does $G$ send the classical affine line to?

Denote $X$ to the classical affine line. Then one idea might be $G(X)=\Spec(\Gamma(X,\sO_X))$. Problem: this scheme is not of finite type over $\Spec \mathbb{R}$ (i.e., $\Gamma(X,\sO_X)$ is not a finitely generated $\mathbb{R}$-algebra): the real algebra $\Gamma(X,\sO_X)$ contains all regular functions $T=\left\{\frac{1}{(x-c)(x-\overline{c})}\mid c\in\mathbb{C}\setminus\mathbb{R}\right\}$. On the other hand, it is not difficult to see that $\Gamma(X,\sO_X)$ is generated by $\mathbb{R}[x]\cup T$. Moreover, given a finite subset $S\subset T$, one can check that the $\mathbb{R}$-algebra generated by $\mathbb{R}[x]\cup S$ is disjoint from $T\setminus S$. (In particular, a finite subset of $\Gamma(X,\sO_X)$ generates a real algebra that cannot contain all of $T$.)

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.

(Incomplete) attempt to construct a “canonical” functor $F:\spvar_k\to\cpvar_k$

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 $\Spec k$.

This defines an endofunctor in locally ringed spaces over $\Spec k$: Since morphisms of $k$-schemes preserve $k$-rational points, a morphism of schematic $k$-prevarieties $f:X\to Y$ induces a continuous map $g:X(k)\to Y(k)$. Since $f\circ i=j\circ g$, where $i:X(k)\to X$ and $j:Y(k)\to Y$ are the inclusions, we can pullback the morphism $f^{-1}\sO_Y\to\sO_X$ along $i$ to obtain a morphism $g^{-1}\sO_{Y(k)}\to\sO_{X(k)}$. This way $g$ becomes a morphism of locally ringed spaces over $\Spec k$.

It would be left to prove:

Conjecture. The locally ringed space $(X(k),\sO_{X(k)})$ is a classical $k$-prevariety.

Proof attempt. If $U\subset X$ is open, then $X(k)\cap U=U(k)$; thus, $\sO_{X(k)}|_{U\cap X(k)}=\sO_{U(k)}$. Hence, it suffices to show the claim for $X$ affine. (Quasi-compactness of $X(k)$ would also follow, since a finite affine covering $X=U_1\cup\dots\cup U_n$ would induce a finite affine covering $X(k)=U_1(k)\cup\dots\cup U_n(k)$, i.e., $X(k)$ is a finite union of Noetherian topological spaces; hence, Noetherian; hence, quasi-compact.)

First, I would like to make the following two observations: Let $I,\mathfrak{m}\subset k[x_1,\dots,x_n]$ be ideals.

  1. If the unique $k$-algebra morphism $\varphi:k\to k[x_1,\dots,x_n]/\mathfrak{m}$ is an isomorphism, then necessarily $\mathfrak{m}=(x_1-a_1,\dots,x_n-a_n)$, for some $(a_1,\dots,a_n)\in k^n$ (contemplate the composite map $\psi:k[x_1,\dots,x_n]\to k[x_1,\dots,x_n]/\mathfrak{m}\xrightarrow{\varphi^{-1}} k$, which is an onto $k$-algebra morphism, and consider $a_i=\psi(x_i)$).

  2. For $a=(a_1,\dots,a_n)\in k^n$, it holds $I\subset (x_1-a_1,\dots,x_n-a_n)$ if and only if $a\in V(I)$ (use that $I(a)=(x_1-a_1,\dots,x_n-a_n)$, which is always true).

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.

My questions are:

  1. Will the morphism $\sO_{X(k)}\to\sO_{V(I)}$ be an iso? If not, what's a counterexample?

  2. Is this the correct approach to define $F$? Can such an $F$ be defined at all? Or is there some obstruction I'm not taking into account?

  3. Are there any general ways to relate schematic and classical $k$-prevarieties? Or am I looking for something that doesn't exist? The construction I proposed for $F$ works for $\mathbb{A}_k^n$ and $\mathbb{P}_k^n$: for arbitrary $k$, “restricting the structure sheaf to the subset of $k$-rational points” maps $\Spec (k[x_1,\dots,x_n])$ and $\operatorname{Proj}(k[x_0,\dots,x_n])$, respectively, to the classical affine and projective spaces.


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 $k$.

$\endgroup$
11
  • $\begingroup$ I think that $I = (x^2+y^2) \subseteq \mathbf R[x,y]$ is a counterexample to what you're trying to do. It is a $1$-dimensional scheme consisting of two lines meeting at the origin, but the action of $\operatorname{Gal}(\mathbf C/\mathbf R)$ swaps the two lines so they are not defined over $\mathbf R$. But their point of intersection is, and $X(\mathbf R) = (0,0)$. If I understand your definitions correctly, then $\mathcal O_{X(k)}$ is the localisation of $\mathbf R[x,y]/(x^2+y^2)$ at the prime $(x,y)$, but $\mathcal O_{V(I)}$ is just $\mathbf R$. $\endgroup$ Commented Jul 4, 2023 at 19:57
  • 1
    $\begingroup$ As I'm sort of hinting in my comment, in the case of a perfect field $k$, there is a useful way to compare $k$-schemes to classical $\bar k$-varieties together with the natural $\operatorname{Gal}(\bar k/k)$-action on $\bar k^n$ given by $\sigma(x_1,\ldots,x_n) = (\sigma(x_1),\ldots,\sigma(x_n))$. $\endgroup$ Commented Jul 4, 2023 at 20:43
  • $\begingroup$ @R.vanDobbendeBruyn Thank you for your comments 😀. Yes, what you propose in your first comment is a counterexample. I just edited the post adding a new hypothesis not present in your example (denseness of $X(k)$). Regarding your second comment: do you know any literature for the idea you explain? $\endgroup$ Commented Jul 5, 2023 at 10:11
  • $\begingroup$ Once you have constructed the algebraic set $X(k) \subset k^n$, just take the sheaf of regular functions on $X(k)$. Restricting $\mathcal{O}_{X}$ doesn't work, as RvDB pointed out. $\endgroup$ Commented Jul 11, 2023 at 9:10
  • 1
    $\begingroup$ The action of the Galois group is given by simply composing a point $\operatorname{Spec}\bar{k} \to X$ with a Galois automorphism $\operatorname{Spec}\bar{k} \to \operatorname{Spec}\bar{k}$. Unfortunately, I do not have references for this approach. However, here is an instructive exercise: show that a point $x\in X(\bar{k})$ comes from a $k$-rational point if and only if it is Galois invariant. $\endgroup$ Commented Jul 12, 2023 at 16:42

1 Answer 1

1
$\begingroup$

$\def\cpvar{\mathsf{CPVar}} \def\spvar{\mathsf{SPVar}} \def\Spec{\operatorname{Spec}} \def\class{\mathrm{class}} \def\sO{\mathcal{O}} \def\Hom{\operatorname{Hom}}$Compiling all the phenomena different people have pointed at in the comments, I arrived to the following.

Definition. A schematic prevariety $X\in\spvar_k$ is said to be classicizable if $X(k)$ is an algebraic prevariety.

For instance, $\Spec (k[x_1,\dots,x_n])$ and $\operatorname{Proj}(k[x_0,\dots,x_n])$ are classicizable and their $k$-rational points are, respectively, the classical affine and projective spaces. An example of a non-classicizable schematic variety is, for $k=\mathbb{R}$, the scheme $\Spec(\mathbb{R}[x,y]/(x^2+y^2))$ (this is van Dobben de Bruyn's counterexample).

Lemma. Let $X\in\spvar_k$. If $X(k)$ is dense in $X$, then $X$ is classicizable.

This is just a sufficient condition, but not a necessary one, as the example with the affine space over a finite field shows. To prove the lemma, one needs to show that the map $\sO_{X(k)}\to\sO_{V(I)}$ constructed above is an isomorphim. Since it is onto, we need to show that it is injective. One does this by showing injectivity of the map on stalks; in turn, this follows from density of $X(k)$.

We denote $\spvar_k^+$ to the full subcategory of $\spvar_k$ of classicizable algebraic prevarieties. Thus, taking $k$-rational points is a functor $(-)(k)=F:\spvar_k^+\to\cpvar_k$.

The functor $F$ is not faithful: if $k=\mathbb{R}$, then the endomorphism of schemes $\Spec(\mathbb{R}[x]/(x^2+1))\to\Spec(\mathbb{R}[x]/(x^2+1))$ over $\Spec\mathbb{R}$ given by $x\mapsto -x$ is sent to the same morphism as the identity, namely, to $\varnothing\to\varnothing$.

The functor $F$ is not full: again, in the case $k=\mathbb{R}$, there is no morphism of schemes $\Spec \mathbb{R}[x]\to\Spec \mathbb{R}[x]$ over $\Spec\mathbb{R}$ that maps to the morphism \eqref{1}.

$\endgroup$
1
  • $\begingroup$ Now I wonder whether $F$ is essentially surjective or not. I wrote my insights here. $\endgroup$ Commented Aug 11, 2023 at 11:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .