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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?

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2 Answers 2

up vote 2 down vote accepted

This is a fairly common question in algebra texts. Here's a series of hints taken from a prelim exam.

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.

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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 $2$).

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More generally any real closed field works. –  Martin Brandenburg Feb 6 '13 at 23:37
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I would like to try and complete the proof for the field I listed, though. Can anyone offer any suggestions for that? –  user15464 Feb 7 '13 at 0:07

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