# an algebraically closed field definable in a real closed field

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?

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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

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: