TITLE: Do maximal proper subfields of the real numbers exist?
QUESTION [19 upvotes]: To clarify the problem, consider the field ${\Bbb R}$ as a field extension of ${\Bbb Q}$ using some sort of Hamel basis. Does there exist a field $F\subsetneq{\Bbb R}$ such that $F(\sqrt{2})={\Bbb R}$ (or $F(x)={\Bbb R}$ for any other irrational $x$)? If $(z_\alpha)_{\alpha<{\frak c}}$ is a basis for ${\Bbb R}$ as a ${\Bbb Q}$-vector space such that $z_0=\sqrt 2$, then $F={\Bbb Q}(z_1,z_2,\dots,z_\alpha,\dots)$ (where all $z_\alpha$ are adjoined for $0\ne \alpha<{\frak c}$) should be a field not containing $\sqrt2$, even though $F(\sqrt2)={\Bbb R}$. (I am using that $\sqrt{2}$ is algebraic of degree $2$ in this argument to prove that $\sqrt{2}\notin F$, although it is probably not necessary.)
My question relates to the process used to find this field. As you can see, I've used a nonconstructive proof using a Hamel basis (and the axiom of choice, implicitly), and this is not philosophically satisfactory for me. Is there an "explicit" proof of existence of such a field $F$, and do its elements have any nice characterization? Is $F$ uniquely determined by the properties $F\subsetneq{\Bbb R}$ and $F(\sqrt{2})={\Bbb R}$? What are the closure properties of $G:={\Bbb R}\setminus F$?
Edit: it seems the construction above doesn't work, and there are good reasons why the $F$ described above doesn't exist. I would like to patch my construction instead by specifying only that $F\subseteq{\Bbb R}$ and $\sqrt2\notin F$, and $F$ is maximal in the sense that there is no proper field extension $F(\alpha)$ satisfying the same properties. Surely this definition will work by Zorn's lemma, although I doubt that such an $F$ is unique. Am I right?

REPLY [2 votes]: Not only is your construction not philosophically satisfactory, I'm sorry to say it is also just wrong. Though $z_0=\sqrt 2$ is $\Bbb Q$-linearly independent of the other $z_i$, it will actually be in the field $F$, in other words it is a quotient of polynomial expressions in the other $z_i$.
In fact there are no fields such that $\Bbb R$ is a finite degree (other than$~1$) algebraic extension of it. As Jyrki Lahtonen explained the existence of such a field would imply that $\Bbb R$ as a field has non-trivial automorphisms, but it hasn't. This is quite easy to prove directly: since the real numbers that are a square (of a real number) are precisely the non-negative ones, and field automorphism must stabilise the set of positive numbers, hence be an automorphism of ordered fields, and since the rational numbers are (point-wise) fixed under any automorphism, the identity automorphism of $\Bbb R$ is the only possibility.
I think there is an even more sweeping statement (but since I have no idea about its proof I hope that somebody will either correct me or provide a reference) which says:

If any algebraically closed field is a finite degree extension of any other field, then that degree is either $1$ or $2$.

Or equivalently (I guess), the automorphism group of an algebraically closed field can have elements of finite order $1$ or $2$ only. By the way this does not say that complex conjugation is the only automorphism of $\Bbb C$ of order$~2$, there are many of them (giving rise to "strange copies" of $\Bbb R$ inside $\Bbb C$). However any two such elements generate an infinite subgroup of $\operatorname{Aut}(\Bbb C)$.
Since the degree of $\Bbb C$ over your hypothetic field $F$ would be twice the degree of $\Bbb R$ over $F$, it should be clear that the above statement rules out the existence of such$~F\subsetneq\Bbb R$.