an algebraically closed field definable in a real closed field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:50:42Z http://mathoverflow.net/feeds/question/63545 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63545/an-algebraically-closed-field-definable-in-a-real-closed-field an algebraically closed field definable in a real closed field Dima Sustretov 2011-04-30T17:22:59Z 2011-04-30T21:45:12Z <p>Is it true that an algebraically closed field $k$ definable (in the model theory sense) in a real closed field $\mathcal R$ is the algebraic closure $\mathcal{R}^{alg}=\mathcal{R}(\sqrt{-1})$?</p> <p>UPDATE: if we prove that the definable set $K \subset \mathcal{R}^n$ that defines $k$ in $\mathcal R$ has the same cardinality as $\mathcal R$ then by categoricity of $ACF_0$ we get that $K$ is isomorphic to $\mathcal{R}(\sqrt{-1})$ (which is definable in $\mathcal R$ in an obvious fashion). Would this isomorphism be definable?</p> http://mathoverflow.net/questions/63545/an-algebraically-closed-field-definable-in-a-real-closed-field/63556#63556 Answer by Simon Thomas for an algebraically closed field definable in a real closed field Simon Thomas 2011-04-30T19:44:48Z 2011-04-30T21:45:12Z <p>Suppose that $A$ is an infinite definable subset of a real closed field $R$ which is a zero-divisor-free ring under operations whose graphs are definable in $R$. Then $A$ is definably isomorphic to one of $R$, $R(\sqrt{-1})$ or the ring of quaternions over $R$. This is a special case of the main result of:</p> <p>Otero, Peterzil, and Pillay, On groups and rings definable in o-minimal expansions of real closed fields, Bull. London Math. Soc. 28 (1996), no. 1, 7–14.</p>