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).

Edit: (added following 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 $\mathbb{Q}(\sqrt{2})$?

  • 2
    $\begingroup$ 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)). $\endgroup$ Jun 11, 2020 at 3:34
  • $\begingroup$ Thanks. Edited accordingly. $\endgroup$
    – Qfwfq
    Jun 11, 2020 at 12:28

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Qfwfq
    Sep 28, 2020 at 12:50
  • $\begingroup$ 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. $\endgroup$
    – Qfwfq
    Sep 28, 2020 at 12:51
  • $\begingroup$ 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. $\endgroup$ Sep 28, 2020 at 12:54
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    $\begingroup$ 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. $\endgroup$ Sep 28, 2020 at 13:02
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    $\begingroup$ 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) $\endgroup$
    – Qfwfq
    Sep 28, 2020 at 15:51

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