The real algebraic numbers form a maximal algebraic ordered field extension of $\mathbb{Q}$ in the sense that they are an algebraic ordered field extension of $\mathbb{Q}$, and no other ordered field extension of $\mathbb{Q}$ contained in $\overline{\mathbb{Q}}$ properly contains the real algebraic numbers. Are there any other maximal algebraic ordered field extensions of $\mathbb{Q}$? I believe that, by Zorn’s Lemma, this is equivalent to asking whether there is any ordered field extension of $\mathbb{Q}$ in $\overline{\mathbb{Q}}$ which contains a non-real number. Furthermore, assuming other such fields exist, does $\overline{\mathbb{Q}}$ necessarily have dimension $2$ as a vector space over a maximal algebraic ordered field extension of $\mathbb{Q}$?
2 Answers
A maximal algebraic ordered field extension of a field $K$ is called a real closure of $K$, and the fields arising in this way are called real closed.
Any ordered field $K$ admits a real closure, and this real closure is unique up to a unique isomorphism over $K$. But the real closure of $K$ can admit many different embeddings into the algebraic closure of $K$.
For example, to construct distinct real closures of $\mathbb{Q}$ inside $\overline{\mathbb{Q}}$, note that the field $\mathbb{Q}[\sqrt{2}]$ admits two orderings: the standard one, and the one obtained by applying the automorphism swapping $\sqrt{2}$ and $-\sqrt{2}$. Let $K$ be $\mathbb{Q}[\sqrt{2}]$ equipped with this nonstandard order, let $R$ be a real closure of $K$, and embed $R$ in $\overline{\mathbb{Q}}$. Then $R$ is not equal to the field of real algebraic numbers, since (for example) $-\sqrt{2}$ has a square root in $R$ (since it is positive in $K$).
One of the many characterizations of real closed fields is that a field $R$ is real closed if and only if it is not algebraically closed, but its algebraic closure is a finite extension: in particular, it is always the extension $R[\sqrt{-1}]$ of degree $2$. So the answer to your last question is yes. In fact, the real closed subfields of $\overline{\mathbb{Q}}$ are exactly the subfields $R$ such that $[\overline{\mathbb{Q}}:R] = 2$. With a little more work, one can show that there are continuum-many such subfields.
Well, you can take the image of the real algebraic numbers under any automorphism of $\overline{\mathbb{Q}}$ (or any isomorphism from $\overline{\mathbb{Q}}$ to any other algebraic closure of $\mathbb{Q}$). Since $\overline{\mathbb{Q}}$ is algebraically closed, it has lots of automorphisms; in particular, for any $a,b\in\overline{\mathbb{Q}}$ with the same minimal polynomial over $\mathbb{Q}$, there is an automorphism $f:\overline{\mathbb{Q}}\to\overline{\mathbb{Q}}$ such that $f(a)=b$. Taking $a$ to be real and $b$ to be non-real but have the same minimal polynomial, this gives lots of automorphisms that don't map the reals to themselves. (In fact, the only automorphisms of $\overline{\mathbb{Q}}$ that map the reals to themselves are the identity and complex conjugation, whereas $\overline{\mathbb{Q}}$ has uncountably many different automorphisms.)
However, this is the only kind of example. Indeed, suppose $K\subseteq\overline{\mathbb{Q}}$ is a maximal ordered field. Then $K$ must be archimedean, since any infinitely large element would be transcendental over $\mathbb{Q}$. It follows that $K$ embeds into $\mathbb{R}$ (each element of $K$ determines a Dedekind cut of rationals), and then by maximality the image of this embedding must be all of the real algebraic numbers. It follows that $K(i)$ is algebraically closed and so is all of $\overline{\mathbb{Q}}$. Moreover, our embedding $K\to\mathbb{R}$ extends to an embedding $\overline{\mathbb{Q}}=K(i)\to\mathbb{C}$ which then can be considered as an automorphism of $\overline{\mathbb{Q}}$ whose inverse maps the real algebraic numbers to $K$.
More generally, the following facts are true by theorems of Artin and Schreier. If $k$ is any ordered field, then it has a maximal ordered algebraic extension $K$ which is unique up to unique order-preserving isomorphism, known as the real closure of $k$. The extension $K(\sqrt{-1})$ is then algebraically closed. Conversely, if $L$ is an algebraically closed field that is a finite proper extension of a subfield $K$, then $K$ admits a unique ordering and has no proper algebraic extensions that can be ordered, and $L=K(\sqrt{-1})$.