$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Gal{Gal}\newcommand{\ab}{\mathrm{ab}}$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.)



What is $\mathbb Q^{\mathrm{hypoab}}$?
My [comment](https://mathoverflow.net/questions/421107/what-is-mathbb-q-mathrmhypoab?noredirect=1#comment1081832_421107), Wojowu's [answer](https://mathoverflow.net/a/421125/82179), YCor's [comment](https://mathoverflow.net/questions/421107/what-is-mathbb-q-mathrmhypoab?noredirect=1#comment1081902_421125), and Z. M's [comment](https://mathoverflow.net/questions/421107/what-is-mathbb-q-mathrmhypoab?noredirect=1#comment1081944_421107) 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](https://mathoverflow.net/questions/421107/what-is-mathbb-q-mathrmhypoab?noredirect=1#comment1081944_421107) 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.


