# Galois group of $\Bbb Q(\{\sqrt[2^n]{2}, \zeta_{2^n} \;:\; n \geq 1 \})$ over $\Bbb Q$

Let $K_n = \Bbb Q(\sqrt[2^n]{2}, \zeta_{2^n})$ be a Galois extension of $\Bbb Q$ (where $\zeta_{2^n}=e^{2\pi i / 2^n}$), and let $K$ be the compositum of all the $K_n$'s. It is a Galois extension of $\Bbb Q$. Notice that $m ≤ n \implies K_m \subset K_n$.

What does the Galois group of $K/\Bbb Q$ look like?

According to the proposition $1.1.$ of this document, $\text{Gal}(K/\Bbb Q)$ is the inverse limit of $\text{Gal}(L/\Bbb Q)$ where $L/\Bbb Q$ is a finite Galois extension such that $L \subseteq K$. In particular, we have to consider $L=K_n$, which has Galois group isomorphic to the "affine group" $\Bbb Z/2^n\Bbb Z \rtimes (\Bbb Z/2^n\Bbb Z)^{\times}$ (the holomorph of $\Bbb Z/2^n\Bbb Z$). What other subextensions should I consider? Moreover, I have some trouble as for understanding the inverse limit of all these Galois groups...

Some related questions are: (1), (2), (3). Here is a question with a similar infinite extension.

Any help would be highly appreciated. Thank you in advance!

First, it does suffice to just take the inverse limit of just your groups $\mathrm{Gal}(K_n/\mathbb Q)$, instead of including all finite Galois subextensions. This is a general fact, only requiring the $K_n$'s to be finite Galois extensions of the base field whose union is all of $K$. The proof is the same as in the notes you linked: an element of $\mathrm{Gal}(K/\mathbb Q)$ restricts to elements of $\mathrm{Gal}(K_n/\mathbb Q)$, giving a map $\mathrm{Gal}(K/\mathbb Q) \to \lim\limits_{\leftarrow n}\mathrm{Gal} (K_n/\mathbb Q)$. Then an element of $\mathrm{Gal}(K/\mathbb Q)$ is determined by its restrictions (giving injectivity), and compatible automorphisms of the $K_n$'s glue to an automorphism of $K$ (giving surjectivity).
Second, the inverse limit of your groups is the affine group over the $2$-adic integers, $\mathbb Z_2 \rtimes \mathbb Z_2^{\times}$. Proof. Regard the various affine groups as the groups of linear functions $mx + b$, where $m$ is invertible. (The group law is composition. Explicitly, for $K_n$, $mx + b$ is the automorphism sending $\zeta_{2^n}^x 2^{1/2^n}$ to $\zeta_{2^n}^{mx+b} 2^{1/2^n}$.) Notice that the restriction maps from $\mathrm{Gal}(K_{n+1}/\mathbb Q)$ to $\mathrm{Gal}(K_n/\mathbb Q)$ are the obvious quotient maps, given by modding $m$ and $b$ out by $2^n$.) Then $\mathbb Z_2 \rtimes \mathbb Z_2^{\times}$ maps (compatibly) to each $\mathbb Z/2^n \mathbb Z \rtimes \mathbb (Z/2^n \mathbb Z)^{\times}$ by taking $m$ and $b$ mod $2^n$, so it maps to the inverse limit. But this map is an isomorphism, because $\mathbb Z_2$ is (by definition) the inverse limit of $\mathbb Z/2^n \mathbb Z$ under the quotient maps, and $(\mathbb Z_2)^{\times}$ is the inverse limit of $(\mathbb Z/2^n \mathbb Z)^{\times}$. (The latter is because an element in either $\mathbb Z/2^n \mathbb Z$ or $\mathbb Z_2$ is invertible if and only if it is $1 \bmod 2$.)