Do maximal proper subfields of the real numbers exist?

Question

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?

Answer 1

Such a field $F$ does not exist.

Assume contrariwise that $F$ is a field with the properties $\sqrt2\notin F$, $F(\sqrt2)=\mathbb{R}$. In that case there exists a non-trivial $F$-automorphism $\sigma$ of $F(\sqrt2)$ with the property $\sigma(z)=z$ for all $z\in F$ and $\sigma(\sqrt2)=-\sqrt2$.

This contradicts the known fact that the field $\mathbb{R}$ has no non-trivial automorphisms. I outline the steps in the argument in case you have not seen them before. Below $\tau$ is an arbitrary automorphism of $\mathbb{R}$.

1. We have $\tau(q)=q$ for all the rational numbers $q$.
2. The automorphism $\tau$ maps any square in the field $\mathbb{R}$ to (possibly another) a square.
3. A real number is a square, iff it is non-negative, so by Step 2 $\tau$ maps any positive real number to a positive real number.
4. In the field $\mathbb{R}$ we have $x\le y\Leftrightarrow y-x\ge0$. Therefore Step 3 implies that the automorphism $\tau$ is strictly increasing as a real function.
5. Step 4 implies that $\tau$ is continuous on all of $\mathbb{R}$. Therefore its fixed points form a closed set.
6. The (topological) closure of $\mathbb{Q}$ is all of $\mathbb{R}$, so combining Steps 1 and 5 shows that $\tau(x)=x$ for all real numbers $x$.

Answer 2

There is a theorem that if $[\bar{F} : F]$ is finite (where $\bar{F}$ is the algebraic closure of $F$), then $\bar{F} = F(\mathbf{i})$. e.g. this means $[\bar{F} : F] \in \{1, 2, \infty\}$.

If there were a field $F$ such that $F(\sqrt{2}) = \mathbb{R}$, then

$$ [\bar{F} : F] = [\mathbf{C} : F] = [\mathbf{C}:\mathbf{R}][\mathbf{R}:F] = 2 [\mathbf{R}:F] $$

All of the above information implies that $[\mathbf{R}:F]=1$, and so $F = \mathbb{R}$.

Answer 3

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$~F\subsetneq\Bbb R$.