# Field of characteristic 0 such that every finite extension is cyclic

I am trying to construct a field $F$ of characteristic 0 such that every finite extension of $F$ is cyclic. I think that I have an idea as to what $F$ should be, but I am not sure how to complete the proof that it has this property. Namely, let $a\in \mathbb Z$ be an element which is not a perfect square and let $F$ be the a maximal subfield of $\bar{\mathbb Q}$ which does not contain $\sqrt{a}$ (such a field exists by Zorn's lemma). Intuitively, a finite extension of $F$ should be generated by $a^{1/2^n}$ for some $n$, in which case its Galois group will be cyclic since $F$ contains the $2^n$th roots of unity. However, I cannot find a nice way to prove this. Any suggestions?

-

Let $F$ be a maximal subfield of $\bar{\mathbb Q}$ with respect to not containing $\sqrt{a}$. Let $F \subset E$ be a Galois extension. Show that $F(\sqrt{a})$ is the unique subfield of $E$ of degree $2$. Deduce that $\mathrm{Gal}(E/F)$ contains a maximal normal subgroup of index $2$. Conclude that $\mathrm{Gal}(E/F)$ is cyclic.
Simple examples of such fields include $\Bbb{C}$ (which has no nontrivial finite extensions) and $\Bbb{R}$ (which has one, namely $\Bbb{C}$, cyclic of degree $2$).