TITLE: Is $\mathbf{C}$ the algebraic closure of any field other than $\mathbf{R}$?
QUESTION [29 upvotes]: It seems to me (intuitively) that there should be no other fields whose algebraic closure is $\mathbf{C}$, even though I have no reason to believe it. The facts I've been using to formulate an argument are $[\mathbf{C}\mathbin{:}\mathbf{R}]=2$ and $\mathbf{R}$ is the only field with the usual analytic properties. I mean, it seems that for the complexes to be even defined we need to reference the analytic properties of the reals. Also, we know that such a field would have to be uncountable, right? 
This question might end up being trivial, but any information or references would be appreciated.

REPLY [7 votes]: Let me be clear from the outset that I am assuming the Axiom of Choice.
There are $2^{c} = 2^{2^{\aleph_0}}$ isomorphism classes of subfields $R$ of $\mathbb{C}$ with $[\mathbb{C}:R] = 2$.  It follows that $\operatorname{Aut} \mathbb{C}$ has $2^c$ orbits on the set of index $2$ subfields of $\mathbb{C}$.  [Note: this contradicts the last line of JSchlather's answer and David Speyer's comment on it.]
It follows from the answers to this old MO question of mine that there are 
$2^c$ isomorphism classes of real-closed fields of continuum cardinality.  By Artin-Schreier, each of these fields $R$ is a degree $2$ subfield of its algebraic closure $C$.  Since $C$ is an algebraically closed field of characteristic $0$ and continuum cardinality, it is isomorphic to the complex field $\mathbb{C}$.  Composing $R \hookrightarrow C \cong \mathbb{C}$ realizes $R$ as an index $2$ subfield of $\mathbb{C}$.
Although the argument that there are the largest conceivable number of real-closed fields of continuum cardinality is rather technical, it is easier to see that there must be one other than $\mathbb{R}$ and thus that $\mathbb{R}$ is not unique up to isomorphism among index $2$ subfields of $\mathbb{C}$.  Namely, we can take an ordering on $\mathbb{R}(t)$ which extends the usual ordering on $\mathbb{R}$ and makes $t$ larger than any real number.  This is a non-Archimedean ordered field of continuum cardinality; its real-closure is thus a non-Archimedean ordered field of continuum cardinality.  Since two real-closed fields are isomorphic as ordered fields iff they are abstractly isomorphic, this gives a second real-closed field of continuum cardinality.