MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathcal{R}^{alg}=\mathcal{R}(\sqrt{-1})$?

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?

share|cite|improve this question
One can define the structure $(\mathbb{C}, +, \cdot)$ in $\mathbb R$, by interpreting $mathbb{C}$ as $\mathbb{R}^2$, $+$ as the vector addition, and $(a,b) \cdot (c,d):=(ac-bd,ad+bc)$. The question is whether this is the only algebraically closed field one can define in $\mathbb{R}$. – Dima Sustretov Apr 30 '11 at 18:28
Ansrew: Model theoretically, there are several ways to interpret algebraic structures within a given structure: Either by means of definable graphs (as in the example Dmitry mentions) or, more generally, as (definable) quotients of definable graphs (which is the modern approach). – Andrés E. Caicedo Apr 30 '11 at 19:19
Andres: as RCF has elimination of imaginaries, interpretable implies definable. – Dima Sustretov Apr 30 '11 at 19:22
Dmitry, it is certainly true that the field is the same cardinality as $\mathcal R$ seeing as RCF has no two cardinal models. – James Freitag Apr 30 '11 at 19:43
James: interesting, can you reproduce the argument or give a reference? – Dima Sustretov Apr 30 '11 at 19:47
up vote 11 down vote accepted

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:

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.

share|cite|improve this answer
Thank you! This reference is exactly what I needed. – Dima Sustretov May 1 '11 at 4:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.