What is so special about $\mathbb{R}$ as a subfield of $\mathbb{C}$? 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.
 A: 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".
A: Expanding on Martin Brandenburg's answer and Eric Wofsey's commands to same:
Consider $\mathbb C$ as a structure over the signature $(0,1,{+},{\times},R)$ where $R$ is a unary predicate with the intuitive meaning "is a real". Since the positive reals are exactly the squares of nonzero reals, this makes the ordering of reals first-order definable in this structure. Complex conjugation and therefore $z\mapsto |z|$ is also first-order definable.
Using standard tools from model theory (compactness and the Löwenheim-Skolem theorem) we can find an extension of $(\mathbb C,0,1,{+},{\times},\mathbb R)$ which is elementarily equivalent to it (meaning it has the same first-order properties), still has cardinality $2^{\aleph_0}$, but contains a real that is larger than $n$ ($=1+1+\cdots+1$) for every natural number $n$. The "reals" in this structure are still a field, but a non-archimedean one, and therefore manifestly not isomorphic to our usual $\mathbb R$.
However, our extension is isomorphic to $\mathbb C$ as a field -- that is, ignoring the $R$ predicate in the signature -- because it is algebraically closed (a collection of first-order properties that it inherits from $\mathbb C$ due to being an elementary extension), and two algebraically closed fields of the same characteristic and cardinality are necessarily isomorphic.
On the other hand, the non-standard reals of the extended model sit inside their "$\mathbb C$" in the same way as $\mathbb R$ sits inside the actual $\mathbb C$ -- at least as long as "the same way" means according to any first-order description we can possibly write down.
So it's not possible to characterize $\mathbb R$ even up to isomorphism by a first-order statement of how it behaves as a subset of $\mathbb C$.
Can we get around this by using a higher-order description, such as one that quantifies over subfields, symmetries, subgroups of the symmetry group, or generally sets of sets of sets of subsets of the field, etcetera? It seems that no matter what we do, if it is to retain any arguable "algebraic" flavor, it would be something we can express in a first-order form if we first add finite number of new sorts and symbols to our signature to denote subsets, etc. And then the same construction as above will make us a non-archimedean subfield of $\mathbb C$ that the proposed criterion cannot distinguish from $\mathbb R$. (Hmm, this won't work -- there are so many subfields of $\mathbb C$ that if we want to quantify over them, Löwenheim-Skolem will not be able to guarantee that the "numbers" in the extended structure are few enough that they can be isomorphic to our original $\mathbb C$).

You might conclude from this that "$\mathbb R$ is not all that special, once we know $\mathbb C$". However, this feels rather backwards to me.
In most cases where we use $\mathbb C$ in particular (rather than immediately generalize to a wider class of fields), the reason we care about $\mathbb C$ in the first place is that it is the algebraic closure of the $\mathbb R$ we already know and love.
So it's hard to see in which context one would start from $\mathbb C$ and then look for the reals.
