Real spectrum of ring of continuous semialgebraic functions - MathOverflow

Real spectrum of ring of continuous semialgebraic functions

2009-12-19T07:36:15Z

Let R be a real closed field, and let U be a semialgebraic subset of $R^n$. Let $S^0(U)$ be the ring of continuous R-valued semialgebraic functions. Also let $\tilde{U}$ be the subset of Spec$_r (R[X_1, \ldots, X_n])$ corresponding to U.

What does the real spectrum of $S^0(U)$ look like? Is it related to $\tilde{U}$ in some way?

Answer by Jose Capco for Real spectrum of ring of continuous semialgebraic functions

2009-12-19T08:45:30Z

The real spectrum of ring of continuous functions (not necessarily semialgebraic) looks exactly like its prime spectrum (i.e. they are homeomorphic), this is also the same for ring of continuous semialgebraic functions (see "Semi-algebraic Function Rings via Reflectors of Partially Ordered Rings" by Schwartz and Madden, Corollary 7.8 ). And $U$ should be homeomorphic to $\text{Spec}_r (S^0(U))$ when we consider their constructible topology (see the book of Schwartz and Madden, Proposition 7.9).

Answer by Michel Coste for Real spectrum of ring of continuous semialgebraic functions

2009-12-19T09:44:48Z

I don't agree with the preceding answer.

When $U$ is a locally compact semialgebraic set, then $\widetilde{U}$ equipped with its sheaf of semi-algebraic continuous functions is isomorphic to the affine scheme $\mathrm{Spec}(S^0(U))$. This is proposition 6 in Carral, Coste : Normal spectral spaces and their dimensions, J. Pure Appl. Algebra 30 (1983) 227-235. In particular $\widetilde{U}$ is homeomorphic to the prime spectrum of $S^0(U)$, which is homeomorphic to its real spectrum. In case $U$ is not locally compact, the situation is different; there are more points in $\mathrm{Spec}(S^0(U))$.