Galois group of constructible numbers

Let $\mathcal{C}$ be the field of constructible numbers, that is, the complex numbers constructible by compass and straightedge. It can be shown that it consists of all the numbers obtainable by adding square roots iteratively to the rational numbers, and it's easy to see that the extension $\mathcal{C}/\mathbb{Q}$ is Galois.

Can anything be said about the group $\mathrm{Gal}(\mathcal{C}/\mathbb{Q})$?

For example, it has a quotient isomorphic to the multiplicative group of invertible $2$-adic numbers since it contains the $2^n$ roots of unity.

Edit: The biggest quotient I can think of explicitly is the one given by the subextension $\mathbb{Q}(\sqrt[2^\infty]{\mathbb{Q^*}})$ (add all the $2$-power roots of every non-zero rational) which by Kummer theory has Galois group isomorphic to $\mathbb{Z}_2^\infty\rtimes\mathbb{Z}_2^*$.

The question is analogous if one changes the prime $2$ by any prime $p$. I just liked the relationship with compass and straightedge constructions in the case $p=2$.

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Start with $K_0={\bf Q}$ and let $K_1$ be the maximal abelian extension of $K_0$ of exponent $2$, so $K_1=K_0(\sqrt{K_0^\times})$. Repeating the process $K_{n+1}=K_n(\sqrt{K_n^\times})$ indefinitely gives your field $\mathscr{C}$. Clearly the group $\mathrm{Gal}(K_{1}|K_0)$ can be explicitly described. It would be nice to be able to describe $\mathrm{Gal}(K_{n}|K_0)$ for all $n$. –  Chandan Singh Dalawat May 6 '13 at 7:14
It follows from a well known result of Galois theory that $\operatorname{Gal}(\mathcal{C}/\mathbf{Q})$ is the maximal pro-2-quotient of the absolute Galois group of $\mathbf{Q}$. –  François Brunault May 6 '13 at 8:42
@Chandan Singh Dalawat Yes, in fact $\mathrm{Gal}(K_{n+1}/K_n)$ is isomorphic to a countable product of copies of the group of order two. But already $\mathrm{Gal}(K_2/K_0)$ seems difficult to me, it properly contains the field $K_0(\sqrt[4]{K_0})$ which has Galois group over $K_0$ isomorphic to the semidirect product of $\mathbb{Z}/2\mathbb{Z}$ by a countable product of copies of $\mathbb{Z}/4\mathbb{Z}$ –  Cristos A. Ruiz May 6 '13 at 8:46
@Cristos A. Ruiz : I don't think so, since the latter group is not (pro)solvable : it has a linear group in the composition series. On the other hand, the Galois group you mention is only a subgroup of $\operatorname{GSp}_{2n}$ in general, so it might be possible that it is pro-2 in some cases, especially when $n=1$ i.e. $A$ is an elliptic curve. I don't know explicit examples though. –  François Brunault May 6 '13 at 9:20
To get the exact analogue of the question for primes $p\neq2$, perhaps it would be a good idea to start with $\mathbf{Q}(\zeta_p)$ instead of $\mathbf{Q}$. –  Chandan Singh Dalawat May 6 '13 at 9:20

It is always a good idea to look at the local question first. What is the maximal pro-$p$ extension of $\mathbf{Q}_p$ ? There is a vast literature on the subject, starting with Demushkin whose results are exposed in an old Bourbaki talk by Serre on the Structure de certains pro-p-groupes (d'après Demuškin), (http://www.numdam.org/numdam-bin/fitem?id=SB_1962-1964__8__145_0). They turn out to be extremely interesting groups which satisfy a kind "Poincaré duality" and have a striking presentation mysteriously similar to the presentation of the fundamental group of a compact orientable surface. Now I have to go for my yoga class; I hope you can find more on the web by yourself.

Addendum. Another good idea which I record here although I'm sure you don't need to be reminded of it is that in the local situation one should first ask what happens at primes $l\neq p$. What is the maximal pro-$l$ extension of $\mathbf{Q}_p$ ? This is much easier to answer because the only possible ramification is tame. You have the maximal unramified $l$-extension, which is a $\mathbf{Z}_l$-extension, and then a totally ramified extension of group $\mathbf{Z}_l(1)$, the projective limit of the roots of $1$ of $l$-power order. As a pro-$l$ group it admits the presentation

$\langle\tau,\sigma\mid\sigma\tau\sigma^{-1}=\tau^p\rangle$.

All this can be found in Chapter 16 of Hasse's Zahlentherie and in a paper by Albert in the Annals in the late 30s.

Addendum 2. A third good idea is to look at the function field analogues, wherein you replace $\mathbf{Q}$ by a function field $k(X)$ over a finite field $k$ of characterisitc $p$, where $X$ is a smooth projective absolutely irreducible curve over $k$. There would again be two cases, according as you are looking at the maximal pro-$l$ extension for a prime $l\neq p$ or for the prime $l=p$; the former should be much easier. You can simplify the problem by replacing $k$ by an algebraic closure. We are then in the setting of Abhyankar's conjecture (1957) which was very useful to Grothendieck as a first test for his theory of schemes. In any case, the conjecture was settled by Raynaud and Harbatter in the early 90s.

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... or in "Cohomology of number fields". –  Timo Keller May 6 '13 at 17:20
... or in Cohomology of number fields by Jürgen Neukirch, Alexander Schmidt and Kay Wingberg. Thanks, Timo. –  Chandan Singh Dalawat May 6 '13 at 18:02
Thanks for pointing me to this bibliography, I have a lot to read now. –  Cristos A. Ruiz May 6 '13 at 20:56