Is there an algebraic extension $ F \neq \bar{\mathbf Q} $ of $ \mathbf Q $ such that every irreducible polynomial in $ F[X] $ has odd degree?

The algebraic closure of $ \mathbf Q $ is excluded as a "trivial example" - I am interested in the proper subfields of $ \bar{\mathbf Q} $.

I would very much like to provide some effort on my part to answer this question, but unfortunately all of my ideas seem to be completely ineffective. Some obvious observations include that the constructible numbers are a subfield of any such $ F $, and any polynomial $ f \in \mathbf Q[X] $ whose Galois group over $ \mathbf Q $ is a $ 2 $-group splits completely over $ F $. Furthermore, no polynomial of degree $ > 1 $ in $ F[X] $ can have full symmetric Galois group over $ F $.

The question is not taken from any source, so I am unsure if there is an easy answer or not - approach cautiously.

  • $\begingroup$ A less trivial solution is $\Bbb Q(A)$, where $A$ is the set of the roots of all irreducible polynomials of even degree. But $\Bbb Q(A)$ might be $\bar{\Bbb Q}$, or not. I'd bet that $\sqrt[3]2\notin\Bbb Q(A)$, but my field theory is very rusty... $\endgroup$ – ajotatxe Jan 9 '17 at 16:48
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    $\begingroup$ $x^6-2$ could be trouble! $\endgroup$ – gobucksmath Jan 9 '17 at 16:54
  • $\begingroup$ $ \mathbf Q(A) = \bar{\mathbf Q} $, so that doesn't work. (Just let $ \alpha $ be any algebraic integer, replacing by $ 2\alpha $ if its minpoly has constant term $ \pm 1 $, and repeatedly take square roots until you land outside of $ \mathbf Q(\alpha) $. Then, what you get has even degree over $ \mathbf Q $, and repeated squaring gives $ 2 \alpha $.) $\endgroup$ – Ege Erdil Jan 9 '17 at 16:55
  • $\begingroup$ @Starfall The title should be only informative. The question should be typed into the question box. $\endgroup$ – user26857 Jan 11 '17 at 11:30
  • $\begingroup$ Related on MO: mathoverflow.net/questions/16778/… $\endgroup$ – Watson Jan 11 '17 at 14:14

This reminds me of Lang's Algebra question 26 from chapter 6: Let $\mathbb{Q}^a$ be a fixed algebraic closure of $\mathbb{Q}$. Let $E$ be a maximal subfield of $\mathbb{Q}^a$ not containing $\sqrt{2}$. Show that every finite extension of $E$ is cyclic.

As it turns out, every extension is cyclic of order $2^k$! We can do something similar with your problem. Let $E$ be the maximal extension of $\mathbb{Q}$ that does not contain $\zeta_7+\zeta_7^{-1}$ (the seventh root of unity and its conjugate). In other words, $x^3+x^2-2x-1$ is still irreducible in $E[X]$.

We want to show that any extension of $E$ is a power of $3$. Let $\alpha=\zeta_7+\zeta_7^{-1}$. Let $\beta\in\bar{\mathbb{Q}}\cap E[\alpha]^c$. By the maximality of $E$, we know that $E[\alpha]\subset E[\beta]$. Let $K$ be the splitting field of the minimal polynomial of $\beta$ over $E$. Let $G=Gal(K/E)$. Suppose there is a prime $q\not=3$ such that $q\mid |G|$. Then let $P$ be the Sylow $3$-subgroup of $G$. By Galois correspondence, there is a field extension $L$ of $E$ that corresponds to the group $P$, but by our maximality $E[\alpha]\subset L$. Thus $3\mid[L:E]$. This is our contradiction, there is no such prime $q$.

In fact, it can be shown that all extensions should be cyclic of order $3^k$!

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    $\begingroup$ What about $ E(\zeta_3) $? $\endgroup$ – Ege Erdil Jan 9 '17 at 16:46
  • $\begingroup$ @Starfall - I edited the post to reflect your comment. I think this should work out now. $\endgroup$ – gobucksmath Jan 9 '17 at 16:57
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    $\begingroup$ Very nice! You should write out the argument in greater detail for posterity. $\endgroup$ – Ege Erdil Jan 9 '17 at 17:01
  • $\begingroup$ @Starfall, I proved the $3$ power part, as this answers your original question. I'm going to leave the cyclic claim to others! $\endgroup$ – gobucksmath Jan 9 '17 at 17:23
  • $\begingroup$ I think cyclicity follows the same way it did here (in the $\sqrt2$ variant of the question). There is probably an easier argument than what I cooked up back then. $\endgroup$ – Jyrki Lahtonen Jan 12 '17 at 22:21

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