Degree of extension $\mathbb{C}/K$, where $K$ is maximal with the property $\sqrt{2} \notin K$

Question

This question has been asked before but not really answered, but my query is a bit separate. To summarise the details: $K$ is a field maximal with respect to the property $\sqrt{2}\notin K$, any finite extension $L$ of $K$ is cyclic with the Galois group as a $2$-group, and $\mathbb{C}/K$ is an algebraic extension.

We can show that $[\mathbb{C}:K]$ is not finite, but the question asks us to show that the degree is in fact countable. Now, here's where I have a doubt: I think we can argue that the, $\overline{K}$, the algebraic closure of $K$ is just $\mathbb{C}$, right? Because since $\mathbb{C}/K$ is algebraic, so is $\mathbb{C}/\overline{K}$ is algebraic, giving us $\overline{K} = \mathbb{C}$.

So, talking about the degree $[\mathbb{C}:K]$ is the same as talking about the order of $\text{Gal}(\mathbb{C}/K)$, and since the degree is infinite, the Galois group is a profinite group and necessarily has to be uncountable, right (topological arguments apply here)? But the question (in Dummit and Foote) is asking us to prove that degree is countable. What is going wrong in this argument of mine?

Answer 1

For infinite algebraic extensions we don't have $[L:K]=|Gal(L/K)|$ (in the cardinality sense, ie. injection of one in the other) but $[L:K] =\inf \{ \ |G|, G \ is \ dense \ in \ Gal(L/K)\}$.

With such infimum $G$ then $K$ is the subfield of $L$ fixed by $G$ and $$Gal(L/K)=Gal(L/K)=\varprojlim_{H\ normal and \ open \ in\ Gal(L/K)} G/(H\cap G)$$

Let $\sigma\in Aut(\Bbb{C})$ such that $\sigma(\sqrt{2})=-\sqrt{2}$ and $F$ the subfield of $\Bbb{C}$ fixed by $\sigma$.

The question is (from a transcendental basis of $\Bbb{C/Q}$) to find $\sigma$ such that $\Bbb{C}/F$ is algebraic. Then $Gal(\Bbb{C}/F)$ is a quotient of $\varprojlim \sigma^\Bbb{Z}/\sigma^\Bbb{nZ}=\sigma^\widehat{\Bbb{Z}}$ and $K= F(\{ a, 2\nmid [F(a):F]\}$ satisfies the requirement.

$\phi=\lim_{k\to \infty} \sigma^{\prod_{m\le 2^k} (2m+1)}$ converges in $Gal(\Bbb{C}/F)$ and $K$ is the subfield of $\Bbb{C}$ fixed by $\phi$ and $Gal(\Bbb{C}/K)=\varprojlim \phi^\Bbb{Z/2^nZ}$.