# Algebraic numbers in real closed fields

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.

• It depends on what you mean by an isomorphism of real closures. The natural thing to ask for is an isomorphism of fields respecting the embedding of the field you started with; this is the sense in which algebraic closures are unique up to isomorphism. So the claim here is that there are two embeddings which are not isomorphic in this sense. – Qiaochu Yuan Mar 21 '16 at 16:23
• If that is what is intended, then it is very badly explained. Particularly as the proof offered that $\Bbb{Q}(\sqrt(2)$ has two non-isomorphic real closures incorrectly comes up with a real closed field containing $\sqrt{-2}$. – Rob Arthan Mar 21 '16 at 16:29
• I agree, by the way, with the way you've resolved the typos. They're annoying, aren't they... – Alex Kruckman Mar 21 '16 at 17:13

Qiaochu's comment is correct. When I read "$R_1$ and $R_2$ are isomorphic real closures of $F$", I automatically parse it as "there is an isomorphism $R_1\to R_2$ respecting the embeddings of $F$ in $R_1$ and $R_2$". A "real closure of $F$" implicitly comes with more structure than just a "real closed field", namely the embedding of $F$.
Marker could certainly afford to be clearer about this, but he is explicit at the end of the paragraph in question, when he writes "Then, $R$ is not isomorphic to the real algebraic numbers over $F$". An isomorphism "over $F$" is one which respects a given embedding of $F$, and this language is used throughout the book.
You write "The two possible orderings on $F = \mathbb{Q}(\sqrt{2})$ give rise to two distinct embeddings of $F$ in $R$, not two non-isomorphic real closures." The point is that there are two distinct embeddings of $F$ in $R$ which are not conjugate by an automorphism of $R$.
• I think that explains it, but the proof begins "for example there are non-isomorphic real closures of $F$" at a point when a real closure has just been defined to be a field, not an extension of one field by another, so the reader's expectations are pretty badly managed, particularly as it is followed fairly shortly by the $\Bbb{Q}(t)$ example where the real closures are not isomorphic as fields. – Rob Arthan Mar 21 '16 at 17:27