# Does a maximal subfield of $\mathbb C$ not containing $\sqrt{2}$ have index $2$?

Let $K$ be a maximal subfield of $\mathbb C$ which doesn't contain $\sqrt{2}$ (one exists by Zorn's Lemma). Then $\mathbb C$ is algebraic over $K$, since if a complex number $\alpha$ is a transcendental over $K$, then adjoining $\alpha$ cannot result in $\sqrt{2}$ appearing, since squaring any expression of $\sqrt{2}$ as a rational function in $K(\alpha)$ gives a polynomial with $\alpha$ as a root.

The next part of this problem asks to prove that any finite extension of $K$ is Galois and cyclic. We can assume that the extension $L$ is Galois, since once we prove that the Galois group is cyclic, every subextension is automatically Galois since all subgroups of cyclic groups are normal. So, $Gal(L/K)$ contains an element $\sigma$ that doesn't fix $\sqrt{2}$, but then $\langle\sigma\rangle$'s fixed field is an extension of $K$ which doesn't contain $\sqrt{2}$, and thus must be $K$ itself. Thus, the Galois group is cyclic, and it's of order $2^n$, since there is no nontrivial odd extension of $K$.

Now, the final part of the problem is to prove that the degree $[\mathbb C:K]$ is countable and not finite. However, I just realized that it might in fact be $2$: shouldn't there be an automorphism $\sigma$ of $K(\sqrt{2})$ that fixes $K$ and maps $\sqrt{2}\mapsto -\sqrt{2}$, which extends to an automorphism of $\mathbb C$? I guess it's possible that this automorphism doesn't have order $2$ because of unforseen relations between complex numbers.

I assume I have to prove that the degree is not finite by assuming $\mathbb C$ is cyclic over $K$ and deriving a contradiction, but I'm not sure how to do that, nor how to show that the degree is not uncountable.

• See item 4 in the list of equivalent conditions for a real closed field. I don't remember how the proof goes. It is in a paper by Artin & Schreier and contained in M. Rosen's book of collected papers by Artin. My copy is in my office, so I cannot check right now. – Jyrki Lahtonen Jun 17 '14 at 7:01
• Hmm, so since $2$ must be positive in any ordering, if the extension were finite, then it would have to contain $\sqrt{2}$. How can I show that the degree is countable? – Nishant Jun 17 '14 at 7:12
• @Nishant You said "....and thus must be K itself. Thus, the Galois group is cyclic, " Could you give the reason? Thanks. – nicksohn Jan 3 '16 at 16:23
• Since the fixed field of $\langle\sigma\rangle$ is $K$. this group must be the full Galois group. – Nishant Jan 3 '16 at 18:19
• Please see my answer to the related question, I decided to post there since it is newer. – Rene Schipperus Jan 3 '16 at 18:36