For a field $L$, let $\widetilde L$ be the splitting field of all irreducible polynomials over $L$ having prime-power degree.

Question: Do we have $\widetilde{\mathbf Q}=\overline{\mathbf Q}$?

My money is on "no", because I see no obvious reason why it should be true. If the answer is indeed negative, can one say what degree occurs as the smallest degree of an $f\in \mathbf Q[X]$ which does not split in $M$?

In any case, it seems quite difficult (to me)...

A variation: We have a chain $$L \subset \widetilde L \subset \widetilde {\widetilde L} \subset \dots$$ Let $\widehat L$ be the limit of this chain. Is it even true that $\widehat{\mathbf Q}=\overline{\mathbf Q}$? Does the chain stabilize? The field $\widehat{\mathbf Q}$ has the strange property of having no finite extensions of prime-power degree. Correspondingly, the Galois group $\text{Gal }(\overline{\mathbf Q}/\widehat{\mathbf Q})$ has the strange property of having no open subgroups of prime-power index...

For $L$ a finite field, it is easy to see that $\widetilde{L}=\overline{L}$. We obviously have $\widetilde{\mathbf R}=\overline{\mathbf R}$. I do not know if $\widetilde{\mathbf Q_p}=\overline{\mathbf Q_p}$, or if $\widehat{\mathbf Q_p}=\overline{\mathbf Q_p}$. (Edit: Every finite Galois extension of $\mathbf Q_p$ is solvable, and I believe it follows from this that $\widehat{\mathbf Q_p}=\overline{\mathbf Q_p}$.)

(I have asked this on MSE: please see the discussion there.)

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    $\begingroup$ Obviously $\tilde{L}$ contains all abelian (or even nilpotent) Galois extensions of $L$, and therefore $\hat{L}$ contains all solvable extensions of $L$. $\endgroup$ – YCor Jun 3 '14 at 20:25
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    $\begingroup$ The question is tricker than one could expect (see the discussion on MSE!). For instance it's possible that $L$ contains a Galois extension of $L$ with Galois group the alternating group $A_6$. Indeed, let $A_6$ act on $F_p^5$ in a standard way. Then in the semidirect product $G=A_6\ltimes F_p^5$, the intersection of conjugates of $A_6$ is trivial. Thus if one has a Galois extension $L\subset L'$ with Galois group $G$, then this Galois extension is generated by splitting fields of polynomial of $p$-power degree. (...) $\endgroup$ – YCor Jun 3 '14 at 20:32
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    $\begingroup$ (...) because $A_6$ has $p$-power index in $G$ and the intersection of conjugates of $A_6$ in $G$ is trivial. On the other hand, $L'$ admits a smaller Galois extension $L''$ with Galois group $A_6$ over $L$, which contains no subextension $\neq L$ of prime power degree (because $A_6$ has no proper subgroup of prime power index), still we have $L''\subset L'\subset\tilde{L}$. Such extensions can be found on a suitable field of characteristic zero, and probably over $\mathbf{Q}$. $\endgroup$ – YCor Jun 3 '14 at 20:35

$\def\QQ{\mathbb{Q}}$Building on YCor's construction, your field is $\overline{\mathbb{Q}}$. Let $L$ be any finite Galois extension of $\QQ$ with Galois group $G$; I will show that $L$ is contained in your field.

We can find an element $x$ in $L$ so that, for any nonempty subset $S$ of $G$, the product $\prod_{\sigma \in S} x^{\sigma}$ is not square. For example, choose $x$ to generate a principal prime ideal of $L$ lying over a prime of $\mathbb{Q}$ which is completely split in $L$. Let $M$ be the result of adjoining to $L$ all square roots $\sqrt{x^{\sigma}}$, for all $\sigma \in G$. So $Gal(M/\mathbb{Q}) \cong G \ltimes (\mathbb{Z}/2)^G$, where the first $G$ acts by permuting the factors in the second $G$.

Let $F \subset M$ be the fixed field of $G \ltimes \{ 0 \}$. This group has index $2^{|G|}$ in $Gal(M/\mathbb{Q})$, so $[F:\QQ]=2^{|G|}$. Let $f$ be the minimal polynomial of a primitive element of $F$. So $f$ splits in your field. Thus, your field contains $F$ and all Galois conjugates of $F$. But the intersection $\bigcap_{h \in G \ltimes (\mathbb{Z}/2)^G} h (G \ltimes \{ 0 \}) h^{-1}$ is trivial, so the collection of Galois conjugates of $F$ generates $M$. We have shown that your field contains $M$, and hence contains $L$.

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    $\begingroup$ Nice. In other words, this shows that every Galois extension of $\mathbf{Q}$ of degree $d$ is contained in the splitting field of a polynomial of degree $2^d$ (the splitting field having degree $d2^d$). $\endgroup$ – YCor Jun 4 '14 at 12:32
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    $\begingroup$ Very nice (and, for me, unexpected!) $\endgroup$ – Bruno Joyal Jun 4 '14 at 12:44
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    $\begingroup$ @DavidSpeyer: Why is the extension $1\to(\mathbf{Z}/2)^G\to Gal(M/\mathbf{Q})\to G\to 1$ split? $\endgroup$ – YCor Jun 4 '14 at 13:05
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    $\begingroup$ Right, but probably working harder than is necessary: The field was just being used as a convenient thing for the groups to act on. I think the following is a restatement in the language of groups: Given $1 \to A^G \to E \to G \to 1$, choose a set-theoretic section $s: G \to E$. Define $\Gamma \subset E$ to be the group sending the set $s(G)$ to itself. I think you should get that $\Gamma \to G$ is an isomorphism. (Maybe I'll assign this next time I teach group theory :).) $\endgroup$ – David E Speyer Jun 4 '14 at 16:51
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    $\begingroup$ Dear David: if you'd like to post your beautiful answer at MSE also, I'll happily accept it there as well. Cheers! $\endgroup$ – Bruno Joyal Jun 9 '14 at 2:56

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