# “Real algebraic varieties” vs finite type separated reduced $\mathbb{R}$-schemes with dense $\mathbb{R}$-points

This question is partly motivated by a few comments here. Let me denote by $$R$$ the (real-closed) field of real numbers $$\mathbb{R}$$; everything is probably the same over an arbitrary real-closed field.

When one has a polynomial subset $$V$$ of $$R^n$$, the following two are equally sensible ways of putting a structure sheaf on $$V$$:

1. One is by considering regular functions in the sense of usual scheme theory: in this case the global regular functions are $$R$$-polynomials in $$n$$ variables modulo the ideal of those polynomials vanishing on $$V$$. If, more precisely, we call $$X$$ the corresponding $$R$$-scheme (with all of its non-$$R$$-points too, which by the way are reconstructible from the set $$V\subseteq R^n$$) and $$O_X$$ its structure sheaf, then $$X(R)=V\subseteq R^n$$ and $$O_X(X)\simeq R[x_1,\ldots, x_n]/I_X$$.

2. The other way is by declaring that a regular function is a ratio of polynomials with nonvanishing denominator. We will call such functions $$R$$-regular, and $$R_V$$ the resulting structure sheaf. We call $$(V,R_V)$$ an $$R$$-algebraic variety. This definition seems to be standard in real algebraic geometry, see e.g. Bochnak-Coste-Roy - Real algebraic geometry (Section 3.2). I think it doesn't change much if we consider the topological space $$X$$ of the scheme in point 1) instead, endowed with the sheaf $$R_X$$ that sends an open set $$U$$ to the rational functions on $$U\subseteq X$$ that are regular at each point of $$U\cap X(R)$$.

The resulting structure sheaves are not the same. For example, consider the real line: the function $$\frac{1}{1+x^2}$$ is an $$R$$-regular function which is not (scheme-theoretically) regular.

Likewise, one can define abstract $$R$$-algebraic varieties, and $$R$$-regular maps thereof.

The curious thing is that every projective $$R$$-algebraic variety is $$R$$-biregularly isomorphic to an affine one. Indeed, the set theoretic map (example 1.5 in Ottaviani - Real algebraic geometry. A few basics or theorem 3.4.4 in BCR) $$\mathbb{P}^n(R)\to \operatorname{Sym}^2(R^{n+1})\;\;,\quad (x_0:\ldots : x_n)\mapsto \frac{x_ix_j}{\sum_{h=1}^n x_h^2}$$ is an $$R$$-regular embedding. This does not correspond to an everywhere-defined morphism of schemes, as is immediately seen by looking at any component of the map in a standard affine chart of $$\mathbb{P}^n$$.

Are there non-(quasi-)affine abstract $$R$$-algebraic varieties at all?

Edit: I think the "quasi" in "quasi-affine" may be pleonastic: I haven't checked the details but a quasi-affine $$R$$-algebraic variety should very often be affine. Indeed, if $$X=W\smallsetminus Y$$, $$Y\subset W \subseteq R^n$$ with $$W$$ affine and $$Y$$ closed (maybe with some assumptions on $$Y$$), the real blowup $$\operatorname{Bl}_Y W$$ is closed in some $$\mathbb{P}^{m}\times W$$ and the latter is affine; but now the "missing" set $$E$$ has become a divisor: $$X\simeq (\operatorname{Bl}_Y W)\smallsetminus E$$, and affine minus a divisor is still affine.

The above example (the one of projective space embedding in an affine space) shows that the category $$\text{R-Var}$$ of $$R$$-algebraic varieties is not a full subcategory of schemes over $$\operatorname{Spec}(R)$$. On the other hand, I think the category $$\operatorname{Sch}'_R$$ of finite type separated reduced schemes over $$\operatorname{Spec}(R)$$ is a full subcategory of $$\text{R-Var}$$. [Edit: following the comment of Julian Rosen, we probably also want to require the schemes in $$\operatorname{Sch}'_R$$ to have dense $$R$$-points]

Are there two non-isomorphic schemes in $$\operatorname{Sch}'_R$$ that become isomorphic in $$\text{R-Var}$$?

Edit: even before posting, I found example 3.2.8 in BCR. There is also proposition 3.5.2 in BCR, the $$R$$-biregular isomorphism between the circle $$x^2+y^2=1$$ and $$\mathbb{P}^1_R$$. And between the "quadric" sphere and the "Riemann" sphere (i.e. complex projective line thought of as a real algebraic variety).

In which other ways does $$\text{R-Var}$$ deviate from $$\operatorname{Sch}'_R$$?

Note: I'm not asking how real algebraic geometry deviates from complex algebraic geometry (which is surely addressed in a preexisting MO question).

For non real-closed fields, or fields of positive characteristic, do people consider varieties in the sense of 1) or in the sense of 2)?

For example, should $$1/(1+x^2)$$ be a regular function on the line over $$\mathbb{F}_7$$? (It's a well defined function on a finite field, so there will be a polynomial realizing its values set theoretically, but should it be enough?) -- Or, should 1/(x^2-3) be a regular function on the line over $$\mathbb{Q}(\sqrt{2})$$?

• For your second question, the scheme $\operatorname{Spec} R[x]/(x^2+1)$ becomes isomorphic to the empty scheme in $R$-Var. Maybe the definition of $\operatorname{Sch}'_R$ should include only those schemes whose $R$-points are Zariski-dense (this is true for every scheme arising in the construction (1)). – Julian Rosen Jun 11 '20 at 3:34
• Thanks. Edited accordingly. – Qfwfq Jun 11 '20 at 12:28

As for your first question, concerning nonaffine R-varieties as you call them, yes, there are nonaffine R-varieties. However, they are considered pathological. Example 12.1.5 on page 301 of Bochnak-Coste-Roy, Real algebraic geometry, constructs an R-line bundle over $$\mathbf R^2$$ whose total space is not affine. In fact, it is not affine since it does not have any separated complexification. Note that the R-variety itself, however, is separated!

The essential point here is that the set of real points of an irreducible affine scheme over $$\mathbf R$$ can be reducible. In the aforementioned example, the irreducible scheme in question is the one defined by the irreducible polynomial $$p=x^2(x-1)^2+y^2\in\mathbf R[x,y,z].$$ The set of real points in $$\mathbf R^3$$ defined by $$p$$ is the disjoint union of the affine lines $$L_0=\{(0,0)\}\times\mathbf R\ \mathrm{and}\ L_1=\{(1,0)\}\times\mathbf R.$$ This is clearly a reducible subset of $$\mathbf R^3$$. The separated R-variety that does not have a separated complexification is the one obtained by gluing the open subsets $$U_0=\mathbf R^3\setminus L_0\ \mathrm{and}\ U_1=\mathbf R^3\setminus L_1$$ along the open subsets $$U_{01}=U_0\cap U_1\subseteq U_0\ \mathrm{and}\ U_{10}=U_0\cap U_1\subseteq U_1$$ via the regular isomorphism $$\phi_{10}\colon U_{01}\rightarrow U_{10}$$ defined by $$\phi_{10}(x,y,z)=(x,y,pz).$$ Note that this is indeed a regular isomorphism since the map $$\phi_{01}=\phi_{10}^{-1}$$ is the regular map $$\phi_{01}\colon U_{10}\rightarrow U_{01}$$ defined by $$\phi_{01}(x,y,z)=(x,y,\tfrac{z}{p}).$$

Now, it is easy to see that the R-variety $$U$$ one obtains is separated, as defined in the founding paper of the whole theory: Faisceaux algébriques cohérents by Jean-Pierre Serre. Indeed, one easily checks that the diagonal in $$U\times U$$ is closed. However, if one wants to construct a real scheme $$X$$ whose set of real points coincides with $$U$$, then, inevitably, $$X$$ will not be separated. Indeed, the polynomial $$p$$ defines a nonclosed point $$x_0$$ in any scheme-wise thickening $$X_0$$ of $$U_0$$ since $$p$$ has zeros in $$U_0$$, and similarly it defines a non closed point $$x_1$$ of any scheme-wise thickening $$X_1$$ of $$U_1$$. The gluing morphisms $$\phi_{01}$$ and $$\phi_{10}$$ will extend to open subsets $$X_{01}$$ of $$X_0$$ and $$X_{10}$$ of $$X_1$$, but they won't contain $$x_0$$ and $$x_1$$, respectively. This is because the polynomial $$p$$ vanishes at $$x_0$$. As a result, any scheme-wise thickening of $$U$$ will be nonseparated!

As for your second question, if I understand correctly, you are asking whether the functor $$F\colon Sch_R'\rightarrow R-Var$$ defined by $$F(X)=X(\mathbf R)$$ is an equivalence onto a full subcategory, where $$Sch_R'$$ is the category of finite type separated reduced schemes over $$Spec(\mathbf R)$$ having dense sets of real points. This is an equivalence onto a full subcategory, its image category, if you localize $$Sch_R'$$ with respect to inlcusions of open subsets containing all real points: any morphism of $$R$$-varieties wil extend to a morphims defined on some open subset containing the real points. Uniqueness is implied by density of real points and separation.

As for your third question, I can't think of other differences between $$R$$-varieties and schemes over $$\mathbf R$$ that differ essentially from phenomena already present in the example above.

As for your final question about varieties in the sense of $$R$$-varieties over other fields, Serre certainly did define them in the paper I mentioned above. I'm not sure whether that has had a follow-up for other fields than real or algebraically closed fields.

• Thank you for the detailed answer. As for the second question, in the edit immediately following it I reported a couple of examples, that I found in Coste et al. just after having written that question, which show $Sch'_R\to R-Var$ cannot be full if you don't localize. If I get it correctly, your answer now adds the observation that localizing is the only thing left to do to obtain fullness. – Qfwfq Sep 28 '20 at 12:50
• As for essential surjectivity, your "non separated scheme theoretic thickening" example shows that $F$ cannot be ess. surjective (cause my $Sch_R'$ by definition only contained separated schemes). On the other hand we can't just extend $F$ from the cat. $Sch_R''$ of all possibly non separated schemes (with the rest of conditions unchanged) cause we would land outside $R-Var$. I'm wondering if the "thickification" is a functor $G:R-Var\to Sch_R$, and if we have an adjuction with the suitable extention $\tilde{F}:G(R-Var)\to R-Var$ where $G(R-Var)\subset Sch_R$ is the essential image in schemes. – Qfwfq Sep 28 '20 at 12:51
• Well, I think I said that the functor $F$ after localization is an equivalence onto a full subcategory. As you said, it cannot be essentially surjective. – Johannes Huisman Sep 28 '20 at 12:54
• Concerning a thickification functor, I think it can only be defined if at both sides you allow nonseparated objects. If you localize suitably, the functor of taking-real-points is then probably an equivalence of categories, so in particular, you would have an adjunction, yes. – Johannes Huisman Sep 28 '20 at 13:02
• Re your first comment (and my second one): you're right. (Btw, I wasn't making a correction, just recapping and asking an additional question) – Qfwfq Sep 28 '20 at 15:51