Deduce the degree of the extension $[\mathbb C:K]$ is countable and not finite

This problem has been posted more than once on the site, but I think none of these gave a satisfactory answer.

Let $K$ be a subfield of $\mathbb C$ maximal with respect to the property $\sqrt 2\notin K$. Deduce the degree of the extension $[\mathbb C:K]$ is countable and not finite.

The full problem asks one to show
(a) such $K$ exists.
(b) $\mathbb C$ is algebraic over $K$.
(c) Every finite extension of $K$ in $\mathbb C$ is cyclic with Galois group some $2$-group.

And I am able to solve these three ones.

But for the one described in the title, I am not sure how to show it can't be finite, any suggestions ?

Solution 1:

I assume the axiom of choice, as without it the claim may be unprovable.

Some maximal $K$ exists with $Aut(\Bbb{C}/K) \cong \Bbb{Z}_2$ (the $2$-adic integers)

Take a $\Bbb{Q}$-transcendental basis $X$, that you can identify with an uncountable set of variables, so that $\Bbb{C}=\overline{\Bbb{Q}(X)}$. Take $\sigma\in Aut(\Bbb{C}/\Bbb{Q}(X))$ such that $\phi (\sqrt2)=-\sqrt2$. Then the orbit of any element under the action of $\phi^\Bbb{Z}$ is finite.

For all $n$, $\lim_{m\to \infty}(m! / 2^{v_2(m!)})^{2^m}$ converges to $1$ modulo $2^n$ and to $0 $ modulo $(2n+1)$.

Whence $$\sigma= \lim_{m\to \infty} \phi ^{(m! / 2^{v_2(m!)})^{2^m}} \text{ is well-defined and } \sigma\in Aut(\Bbb{C}/\Bbb{Q}(X))$$

Let $K=\Bbb{C}^\sigma$ (the subfield fixed by $\sigma$). You'll have that $\Bbb{C}^{\sigma^{2^n}}/K$ is a degree $2^n$ cyclic extension and $\Bbb{C}=\bigcup_{n\ge 1}\Bbb{C}^{\sigma^{2^n}}$. This implies that $\Bbb{C}/K$ is algebraic of countable degree. Any finite extension $F/K$ is contained in some $\Bbb{C}^{\sigma^{2^n}}$, so $F$ will be the subfield of $\Bbb{C}^{\sigma^{2^n}}$ fixed by some subgroup of $Aut(\Bbb{C}^{\sigma^{2^n}}/K)$ ie. $F=\Bbb{C}^{\sigma^{2^m}}$ for some $m\le n$. Since $\sqrt{2} \in \Bbb{C}^{\sigma^2}$ we are done.

Any $K$ is of this kind

Any $K$ maximal for the property $\sqrt{2}\not \in K$ will contain a $\Bbb{Q}$-transcendental basis $X$.

That is we have the tower $\Bbb{C}/K/\Bbb{Q}(X)$. Take an automorphism $\phi\in Aut(\Bbb{C}/K)$ such that $\phi(\sqrt2)=-\sqrt2$. Let $\sigma= \lim_{m\to \infty} \phi ^{(m! / 2^{v_2(m!)})^{2^m}} \in Aut(\Bbb{C}/K)$, it must be that $K=\Bbb{C}^\sigma$.