I have been looking at the discussion of real closed fields in Appendix B of Marker's Model Theory:an Introduction. I am baffled by what it says about the uniqueness of real closures. I have no problem with the statement just after Lemma B.13 that the field of rational functions $\Bbb{Q}(t)$ has $c$ non-isomorphic real closures.
However, I don't believe the statement that "because $\Bbb{Q}(\sqrt{2})$ has two distinct orderings, it has two non-isomorphic real closures". Surely the real closure of any subfield $F$ of the field $\cal R$ of real algebraic numbers is isomorphic to $\cal R$ (because each real algebraic $x$ is definable over $\Bbb{Q}$). The two possible orderings on $F = \Bbb{Q}(\sqrt{2})$ give rise to two distinct embeddings of $F$ in $\cal R$, not two non-isomorphic real closures. Am I wrong about this?
Just before these statements, there is a proof that $\Bbb{Q}(\sqrt{2}, \sqrt{-2})$ is a real field, which is surely impossible because $-2$ must be negative in any ordered field while non-zero squares must be positive. Guessing at fixes to some typos e.g., the appeal to a non-existent Corollary B.5, I think the argument actually just proves that $\Bbb{Q}(\sqrt{-\sqrt{2}})$ is a real field, which isn't surprising because that field is isomorphic to $\Bbb{Q}(\sqrt[4]{2})$. Can anyone make better sense of this argument or confirm my opinion that it doesn't prove that $\Bbb{Q}(\sqrt{2})$ has two non-isomorphic real closures?
I know that the wikipedia article on real closed fields singles out $\Bbb{Q}(\sqrt{2})$ as an example of a field whose real closure is not a field, but I think that holds for a somewhat different definition of the real closure that is intended to make sense for real closed rings, not just real closed fields.