In terms of the algebraic properties of $\mathbb{C}$, what (if anything) is so special about $\mathbb{R}$ as a subfield?
$\mathbb{R}$ is the subfield of $\mathbb{C}$ consisting of all 'conjugation-invariant' elements, yet complex conjugation is not a natural choice of field automorphism (although it is the unique non-trivial topological field automorphism of $\mathbb{C}$ with the Euclidean topology).
Then there is the order structure: $\mathbb{R}$ is maximal as an orderable/formally real subfield of $\mathbb{C}$. Yet as has been helpfully pointed out in response to my recent question regarding uniquely orderable subfields of $\mathbb{C}$, there are many subfields of $\mathbb{C}$ that are isomorphic to $\mathbb{R}$ that would therefore have an identical order structure.
Is there any intrinsic reason that $\mathbb{R}$ is, algebraically, a 'special' subfield of $\mathbb{C}$ or is it more to do with the historical development of the subject?
Any thoughts on this are most welcome.
Best Answer
Since you ignore the topological structure on $\mathbb{C}$, you need to realize that - as an abstract field - $\mathbb{C}$ is isomorphic to every other algebraically closed field of characteristic $0$ which has cardinality $c$ (the continuum cardinal). In particular, there is an isomorphism to $\overline{\mathbb{C}(x)}$, but also to $\mathbb{C}_p$ (see here) for example. Inside these abstract fields you do not have any natural copy of $\mathbb{R}$ over which the field has degree $2$.
The answer is therefore "no - when you ignore the topology".