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Let $V$ be a geometrically irreducible and reduced scheme defined over the real numbers, and let $K = K(V)$ be its function field.

  1. If $V$ does not have any real points, is it true that $K$ is not formally real? It seems this is a theorem due to (E.) Artin but I cannot find a modern reference and my German needs a little work.

  2. If $V$ does have real points, is $K$ necessarily formally real?

Thanks for the help.

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1 Answer 1

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The theorem you want is due to Serge Lang, from the following paper:

The theory of real places. Ann. of Math. (2) 57, (1953). 378–391.

The statement is almost, but not quite, what you suggest. To see the problem, a good example to consider is the affine plane curve over $\mathbb{R}$ defined by $\mathbb{R}[x,y]/(y^2+x^2+x^4)$. This defines a geometrically integral curve over $\mathbb{R}$ with non-formally real fraction field but possessing an $\mathbb{R}$-point, namely $(0,0)$. The key is that $(0,0)$ is the only $\mathbb{R}$-point on this curve and (thus!) it is a singular point.

So the correct result is that the function field of an integral affine variety $V_{/\mathbb{R}}$ is formally real iff $V$ admits a nonsingular $\mathbb{R}$-point. (Note that projective real algebraic varieties are also affine(!!).) Probably you could extend this to finite-type integral schemes without any trouble.

I also looked in Bochnak, Coste and Roy, following Thierry Zell's suggestion, but only found "half" of this result, namely the Artin-Lang Homomorphism Theorem. It seems likely though that I just didn't look hard enough: perhaps someone can enlighten me.

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Dear Pete, can you clarify what you mean by projective varieties over $\mathbf{R}$ being affine? The projective line over $\mathbf{R}$ is a counter-example if you take the usual, scheme-theoretic interpretations of these terms. –  JBorger Nov 30 '10 at 12:05
See Chapter 3 of Bochnak, Coste and Roy for more details, but briefly: one has the notion of a real algebraic variety, a locally ringed space locally isomorphic to the set of $\mathbb{R}$-points of an affine $\mathbb{R}$-variety with the analytic topology. Of course projective space $\mathbb{P}^n$ is a real algebraic variety. But as a real algebraic variety, it is affine. –  Pete L. Clark Nov 30 '10 at 13:00
For instance $\mathbb{P}^1$ is, as a real algebraic variety, isomorphic to $S^1= \operatorname{Spec}(\mathbb{R}[x,y]/(x^2+y^2−1))$. –  Pete L. Clark Nov 30 '10 at 13:04
Thanks. So, by "real algebraic variety" you don't mean an algebraic variety (in any of the standard scheme-theoretic senses) over the field of real numbers. Dangerous terminology! –  JBorger Dec 2 '10 at 9:04
@James: Yes, absolutely. In retrospect, I could have written this in a less confusing way. I find this quirk of real algebraic geometry amusing, but it's certainly not necessary to mention it in this context: it's pretty clear that the result in question extends to non-affine varieties. –  Pete L. Clark Dec 2 '10 at 10:29

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