What is $\mathbb Q^{\mathrm{hypoab}}$?

Question

$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Gal{Gal}\newcommand{\ab}{\mathrm{ab}}$Let $G(\mathbb Q) = \Gal(\overline{\mathbb Q} / \mathbb Q)$ be the absolute Galois group. It's well-known that the abelianization $G(\mathbb Q)^{\ab}$ of $G(\mathbb Q)$ is isomorphic to $\Aut(\mathbb Q / \mathbb Z) = \widehat {\mathbb Z}^\times$, and that the fixed field $\mathbb Q^{\ab}$ of the commutator subgroup of $G(\mathbb Q)$ may be constructed by adjoining all roots of unity to $\mathbb Q$.

Abelian Galois groups are generalized by solvable Galois groups, or more generally hypoabelian Galois groups (recall that a group is hypoabelian if its derived series stabilizes at the trivial group, possibly after transfinitely many steps).

Question 1: What is the hypoabelianization of $G(\mathbb Q)$?

Question 2: What is the fixed field of the maximal perfect normal subgroup of $G(\mathbb Q)$?

(Recall that in general, the maximal perfect normal subgroup is the subgroup at which the derived series stabilizes, possibly after transfinitely many steps; the hypoabelianization of a group is its quotient by its maximal perfect normal subgroup.)

Answer 1

My comment, Wojowu's answer, YCor's comment, and Z. M's comment already contain everything we need. Let me provide a little more detail here. I will shift the indices by $1$ for reasons that will become apparent:

Definition. Set $K_0 = \mathbf Q$ and let $K_1 = \mathbf Q(\boldsymbol \mu)$ be the field obtained by adjoining the roots of unity $\boldsymbol \mu \subseteq \bar{\mathbf Q}$. Inductively define $K_{i+1}=K_i\big(\sqrt[\infty\ \ ]{K_i^\times}\big)$, and set $K_\infty = \underset{\substack{\longrightarrow \\ i}}{\operatorname{colim}} K_i$.

We claim that this is the extension we're after. We first introduce some notation:

Definition. Given a profinite group $G$, its (profinte) derived series is the transfinite chain of closed subgroups $$G = G^{(0)} \trianglerighteq G^{(1)} \trianglerighteq \cdots \trianglerighteq G^{(\alpha)} \trianglerighteq \cdots$$ defined by $G^{(\alpha+1)} = \overline{[G^{(\alpha)},G^{(\alpha)}]}$ and $G^{(\beta)} = \bigcap_{\alpha < \beta} G^{(\alpha)}$ for any limit ordinal $\beta$ (which is already closed as each $G^{(\alpha)}$ is closed). One could alter the notation to distinguish it from the abstract derived series, but I will never use the latter (the same goes for the Kronecker–Weber theorem: it computes the topological abelianisation, not the abstract one!). Note that for any continuous surjective homomorphism $G \to H$ of profinite groups, the image of $G^{(\alpha)}$ is $H^{(\alpha)}$.

Lemma. Let $G$ be a profinite group. Then $G^{(\omega + 1)} = G^{(\omega)}$, and this group is trivial if and only if $G$ is pro-soluble¹.

Proof. For any finite group $G$, the descending chain $G^{(i)}$ stabilises after finitely many steps, so $G^{(\omega + 1)} = G^{(\omega)}$. The same statement for profinite groups follows since any closed normal subgroup $H \trianglelefteq G$ is the intersection of the open normal subgroups $U \trianglelefteq G$ containing it. Similarly, $G^{(\omega)} = 1$ if and only if the same holds in every finite quotient $G/U$, i.e. if and only if all $G/U$ are soluble. $\square$

Let's denote $G^{(\omega)}$ by $G^{(\infty)}$. For $n \in \mathbf N \cup \{\infty\}$, we will say that $G$ is $n$-soluble if $G^{(n)} = 1$, and we write $G^{n\text{-}\!\operatorname{sol}} = G/G^{(n)}$ for its maximal $n$-soluble quotient (in which we omit $n$ if $n = \infty$). For instance, $G$ is $1$-soluble if and only if it is abelian, and $\infty$-soluble if and only if it is pro-soluble (equivalently, hypoabelian as profinite group).

Theorem. Let $\Gamma = \Gamma_{\mathbf Q}$ be the absolute Galois group of $\mathbf Q$.

1. For $n \in \mathbf N \cup \{\infty\}$, the fixed field of $\Gamma^{(n)}$ is $K_n$ (i.e. $K_n$ is the maximal pro-soluble extension of derived length $\leq n$);
2. For $n \in \mathbf N \setminus \{0\}$, the Galois group $\operatorname{Gal}(K_{n+1}/K_n) = \Gamma^{(n)}/\Gamma^{(n+1)}$ is isomorphic to $$\operatorname{Hom}_{\operatorname{cont}}\!\big(K_n^\times,\hat{\mathbf Z}(1)\big),$$ where $K^\times$ has the discrete topology and $\hat{\mathbf Z}(1) = \lim_m \boldsymbol \mu_m$ is the Tate module of $\bar{\mathbf Q}^\times$.

Proof. Statement (1) is trivial for $n=0$, and is the Kronecker–Weber theorem for $n=1$. Statements (1) and (2) for finite $n \geq 2$ follow inductively by Kummer theory (see the corollary below). Finally, statement (1) for $n = \infty$ follows from the statement at finite levels, since $K_\infty = \bigcup_n K_n$ and $G^{(\infty)} = \bigcap_n G^{(n)}$. $\square$

Note also that the Galois group $\operatorname{Gal}(K_1/K_0)$ is isomorphic to $\operatorname{Aut}(\boldsymbol \mu) = \hat{\mathbf Z}^\times$. However, explicitly computing $\operatorname{Gal}(K_{n+1}/K_n)$ in a meaninful way is pretty hard, let alone saying anything about how the various pieces fit together.

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Edit: After writing this answer, I became aware of the following two striking results:

Theorem (Iwasawa). The Galois group $\operatorname{Gal}(K_\infty/K_1) = \Gamma^{(1)}/\Gamma^{(\infty)}$ is a free pro-soluble group $\widehat{F_\omega}^{\operatorname{sol}}$ on countably infinitely many generators.

So we know that $\Gamma^{\operatorname{sol}}$ sits in a short exact sequence $$1 \to \widehat{F_\omega}^{\operatorname{sol}} \to \Gamma^{\operatorname{sol}} \to \hat{\mathbf Z}^\times \to 1.$$ I find it hard to imagine that this sequence splits as a semi-direct product (but I am more optimistic about the derived length $\leq 2$ situation).

Theorem (Shafarevich). Any finite soluble group $G$ occurs as a quotient of $\operatorname{Gal}(K_\infty/\mathbf Q) = \Gamma^{\operatorname{sol}}$.

A modern reference is Neukirch–Schmidt–Wingberg's Cohomology of number fields, Corollary 9.5.4 (Iwasawa) and Theorem 9.6.1 (Shafarevich). (This is a truly great book, but even at $>800$ pages it can be a bit terse at times.)

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We used the following general result:

Lemma (Kummer theory). Let $m \in \mathbf Z_{>0}$, and $K$ be a field of characteristic not dividing $m$ that contains $\boldsymbol \mu_m$.

1. The maximal abelian extension of exponent $m$ of $K$ is $L=K\big(\sqrt[m\ \ ]{K^\times}\big)$;
2. The map \begin{align*} \operatorname{Gal}(L/K) = \Gamma_K^{\operatorname{ab}}/m &\to \operatorname{Hom}_{\operatorname{cont}}\!\big(K^\times,\boldsymbol \mu_m\big) = \left(K^\times/(K^\times)^m\right)^\vee \\ \sigma &\mapsto \left(a \mapsto \frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right) \end{align*} is an isomorphism of profinite groups, where $K^\times/(K^\times)^m$ has the discrete topology and $A^\vee$ denotes the Pontryagin dual of a locally compact abelian group $A$.

We avoid the notation $\widehat A$ for Pontryagin duals, since it clashes with the notation for profinite completions. (Note that Z. M's comment uses $(-)^\vee$ for a linear dual, which differs from my notation by a Tate twist.)

Because it's not very hard, let's include a proof.

Proof. For (2), by Pontryagin duality it suffices to show that the dual map \begin{align*} K^\times/(K^\times)^m &\to \operatorname{Hom}\!\big(\Gamma_K,\boldsymbol \mu_m\big) = \left(\Gamma_K^{\operatorname{ab}}/m\right)^\vee \\ a &\mapsto \left(\sigma \mapsto \frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right) \end{align*} is an isomorphism. Note that it is well-defined since any two $m$-th roots of $a$ differ (multiplicatively) by an element of $\boldsymbol \mu_m \subseteq K$, on which $\sigma$ acts as the identity. Since $\boldsymbol \mu_m \subseteq K$, the $\Gamma_K$-module $\boldsymbol \mu_m$ has trivial action, so $\operatorname{Hom}_{\operatorname{cont}}(\Gamma_K,\boldsymbol \mu_m) = H^1(K,\boldsymbol \mu_m)$. The Kummer sequence $$1 \to \boldsymbol \mu_m \to \mathbf G_m \stackrel{(-)^m}\to \mathbf G_m \to 1$$ and Hilbert's theorem 90 compute $K^\times/(K^\times)^m \stackrel\sim\to H^1(K,\boldsymbol \mu_m)$ via the map above. Now (1) follows since $\sigma \in \Gamma_K$ is in the kernel of $\Gamma_K \to \big(K^\times/(K^\times)^m\big)^\vee$ if and only if $\sigma$ fixes all $m$-th roots of elements in $K$. $\square$

Corollay. Let $K$ be a field of characteristic $0$ containing $\boldsymbol \mu$.

1. The maximal abelian extension of $K$ is $L=K\big(\sqrt[\infty\ \ ]{K^\times}\big)$.
2. The map \begin{align*} \operatorname{Gal}(L/K) &\to \operatorname{Hom}_{\operatorname{cont}}\!\big(K^\times,\hat{\mathbf Z}(1)\big) \\ \sigma &\to \left(a \mapsto \left(\frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right)_{m \in \mathbf Z_{>0}}\right) \end{align*} is an isomorphism of profinite groups.

Proof. Take inverse limits over all $m \in \mathbf Z_{>0}$ in the lemma above, noting that the inverse limit pulls out of $\operatorname{Hom}(K^\times,-)$. $\square$

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¹Linguistic footnote: soluble and solvable mean the same thing. I used to think that this is one of those BrE vs AmE things (for instance, my Oxford Advanced Learner's dictionary does not contain the word solvable at all). But I think some folks in the UK also use solvable, so it's not entirely clear to me.

Answer 2

EDIT: This answer is incorrect, see comments. I've assumed hypoabelianness is preserved under quotients, which is not the case.

An algebraic extension $L/K$ is hypoabelian iff it is prosolvable, i.e. $L$ is a union of finite solvable extensions of $K$. Indeed, since Galois groups are residually finite, it is easy to see $Gal(L/K)$ is hypoabelian iff all its finite quotients are (this is easy to see using the perfect core characterization), and a finite group is hypoabelian group iff it is solvable.

Therefore, the maximal hypoabelian extension of $\mathbb Q$ (or any field) is the maximal (pro)solvable extension, which as R. van Dobben de Bruyn mentions in the comments, is the maximal radical extension, i.e. it is formed by repeatedly adjoining all roots of all elements of the field.