Galois group of $\mathbb{Q}(\underset{n\geq 1}{\bigcup}\mu_n)/\mathbb{Q}$ I was trying to formalise the fact that $G_{\mu_\infty}:=Gal(\mathbb{Q}(\underset{n\geq 1}{\bigcup}\mu_n)/\mathbb{Q})\simeq \underset{n}{\varprojlim}\ (\mathbb{Z}/n\mathbb{Z})^{\times}=\hat{\mathbb{Z}}^{\times}$. When I was done I tried finding something on this problem online to check my argument. The only things I was able to find are these two ancient posts here and here and the last page of these notes. The two posts only deal with the isomorphism $\underset{n}\varprojlim\ Gal(\mathbb{Q}(\mu_n)/\mathbb{Q})\simeq\underset{n}\varprojlim\ (\mathbb{Z}/n\mathbb{Z})^{\times}$ which is in my opinion the most trivial part of this problem whereas the short notes on infinite Galois theory do mention a subtlety involved but fail to make it precise. The subtelty here is that the isomorphism $G_{\mu_\infty}\simeq\underset{n}\varprojlim\ Gal(\mathbb{Q}(\mu_n)/\mathbb{Q})$ is not immediately obvious. Here is my attempt to formalise this
Claim 1: The extenstion $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)/\mathbb{Q}$ is normal.
Proof: Let $p(x)\in\mathbb{Q}[x]$ be an irreducible polynomial with a root $a\in\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)$ Then $a$ is some finite expression in $\mu_{n_1},\dots,\mu_{n_i}$ for some $i\geq 1$, involving field operations. Then letting $m:=lcm(n_1,\dots,n_i)$ we have that $a\in\mathbb{Q}(\mu_m)$ which is normal over $\mathbb{Q}$. Hence $p(x)$ splits in $a\in\mathbb{Q}(\mu_m)$ and hence in $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)\ \ \ \blacksquare$
Remark: We immediately get that $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)/\mathbb{Q}$ is Galois since seperability is automatic. Hence we have that $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)=\underset{K}\bigcup K$ where $K$ runs over all Galois number fields contained in $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)$. But we can actually improve this as follows.
Claim 2: It suffices to have $K$ run over all abelian, Galois number fields contained in $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)$.
Proof: One containment is trivial. To see the other one, if $K$ is a Galois number field contained in $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)$, write $K=\mathbb{Q}(\alpha)$. Then by the above remark and the proof of the first claim we see that $\alpha\in\mathbb{Q}(\mu_m)$ for some $m$. Hence $K\subset\mathbb{Q}(\mu_m)$ and since $K$ is Galois, we have by Galois theory that $Gal(K/\mathbb{Q})$ is isomorphic to a quotient of the abelian group $Gal(\mathbb{Q}(\mu_m)/\mathbb{Q})$ and hence is itself abelian. $\blacksquare$
Because of this we get an isomorphism $$G_{\mu_\infty}\simeq\underset{K}\varprojlim\ Gal(K/\mathbb{Q})$$ where now $K$ runs over all abelian Galois number fields contained in $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)$ and the inverse limit is with respect to inclusions of number fields. We now define the group homomorphism $$\phi:\underset{K}\varprojlim\ Gal(K/\mathbb{Q})\longrightarrow\underset{n}\varprojlim\ Gal(\mathbb{Q}(\mu_n)/\mathbb{Q})$$ where we just forget all non cyclotomic number fields and the inverse limit on the right is taken with respect to projections, for $m|n$, $Gal(\mathbb{Q}(\mu_m)/\mathbb{Q})\longrightarrow Gal(\mathbb{Q}(\mu_n)/\mathbb{Q})$.
Claim 3: $\phi$ is an isomorphism.
Proof: Let $(\sigma_K)\in\ker(\phi)$, then $\sigma_{\mathbb{Q}(\mu_n)}=id, \forall n$. Now from claim number 2 and its proof we see that given any abelian Galois number field $K$ contained in $\mathbb{Q}(\underset{n\geq 1}\bigcup\mu_n)$, we have that $K\subset\mathbb{Q}(\mu_m)$ for some $m$. Hence by the inverse system of $\underset{K}\varprojlim\ Gal(K/\mathbb{Q})$ we have that $\sigma_K=id$, for all such $K$. Hence $\phi$ injects.
Now let $(\sigma_n)$ be any element of $\underset{n}\varprojlim\ Gal(\mathbb{Q}(\mu_n)/\mathbb{Q})$. We define the element $(\sigma_K)\in\underset{K}\prod Gal(K/\mathbb{Q})$ as follows $$\sigma_K:=\sigma_m|_K, \text{if}\ K\subset\mathbb{Q}(\mu_m)\ \text{for some}\ m$$ Where again we've implicitly used claim number 2 for this definition. To see that this is well defined, suppose we have $K\subset\mathbb{Q}(\mu_n)\bigcap\mathbb{Q}(\mu_m)=\mathbb{Q}(\mu_d)$, where $d=gcd(n,m)$. Then since $d$ divides both $n$ and $m$, we have by the inverse system of $\underset{n}\varprojlim\ Gal(\mathbb{Q}(\mu_n)/\mathbb{Q})$ that $\sigma_n|_K=\sigma_d|_K=\sigma_m|_K$ as required. We also have by construction that $(\sigma_K)\in\underset{K}\varprojlim\ Gal(K/\mathbb{Q})$ and maps to $(\sigma_n)$ under $\phi$. Hence $\phi$ surjects $\blacksquare$
Hence indeed, in order to compute $G_{\mu_\infty}$, it suffices to compute $\underset{n}\varprojlim\ Gal(\mathbb{Q}(\mu_n)/\mathbb{Q})$.
Also note that in place of claim number 2 we could have also used Kronecker-Webber but that would be a bit of an overkill?
What i'd like to know is whether everthing here makes sense? Is it necessary? Can it be made shorter and/or more precise?
 A: For a simpler proof, first note that $\{\Bbb Q(\mu_n):n\geq 1\}$ is a set of Galois $\Bbb Q$-subfields of $\Bbb{\bar Q}$ directed respect to the inclusion because for every $n,m\geq 1$ we have $\Bbb Q(\mu_n)\cup\Bbb Q(\mu_m)\subseteq\Bbb Q(\mu_{nm})$.
Let $\mu=\bigcup_n\mu_n$.

$\Bbb Q(\mu)=\bigcup_n\Bbb Q(\mu_n)$ is Galois over $\Bbb Q$.

Proof. Let $E=\bigcup_n\Bbb Q(\mu_n)$.
Clearly, $E$ is a field and $\mu\subseteq E$, hence $\Bbb Q(\mu)\subseteq E$.
Conversely, from $\Bbb Q(\mu_n)\subseteq\Bbb Q(\mu)$ for every $n\geq 1$ follows $E\subseteq\Bbb Q(\mu)$.
Finally, for every $\Bbb Q$-homomorphism $\sigma:\Bbb Q\to\Bbb{\bar Q}$, we have $\sigma[\Bbb Q(\mu_n)]=\Bbb Q(\mu_n)$, hence $\sigma[E]=E$ and then $E$ is Galois over $\Bbb Q$.
$\newcommand\Gal{\operatorname{Gal}}\require{AMScd}$
Then we can prove directly the group isomorphism
$$\Gal_{\Bbb Q}(\Bbb Q(\mu))\cong\varprojlim\nolimits_n\Gal_{\Bbb Q}(\Bbb Q(\mu_n))$$
Let $\varrho_n:\Gal_{\Bbb Q}(\Bbb Q(\mu))\to\Gal_{\Bbb Q}(\Bbb Q(\mu_n))$ for $n\geq 1$ be group homomorphisms induced by the restriction so that we have a commutative diagram
\begin{CD}
\Gal_{\Bbb Q}(\Bbb Q(\mu))@>\varrho_m>>\Gal_{\Bbb Q}(\Bbb Q(\mu_m))\\
@|@VVV\\
\Gal_{\Bbb Q}(\Bbb Q(\mu))@>>\varrho_n>\Gal_{\Bbb Q}(\Bbb Q(\mu_n))
\end{CD}
for every $n,m\geq 1$ such that $n\mid m$, hence induces a group homomorphism
$$\varrho:\Gal_{\Bbb Q}(\Bbb Q(\mu))\to\varprojlim_n\Gal_{\Bbb Q}(\Bbb Q(\mu_n))$$
Since $\Bbb Q(\mu)=\bigcup_n\Bbb Q(\mu_n)$, this homomorphism is injective, for if $\sigma\in\ker\varrho$ then $\sigma\in\bigcap_n\ker(\varrho_n)$ hence the restriction $\sigma|_{\Bbb Q(\mu_n)}$ is the identity for every $n\geq 1$, hence $\sigma$ is the identity over $\Bbb Q(\mu)$.
To prove that $\varrho$ is surjective, let $\sigma_n\in\Gal_{\Bbb Q}(\Bbb Q(\mu_n))$ for $n\geq 1$ be a sequence of automorphisms such that $\sigma_m|_{\Bbb Q(\mu_n)}=\sigma_m$ for every $n,m\geq 1$ such that $n\mid m$.
For every $n\geq 1$ we have a $\Bbb Q$-field homomorphism
$$\Bbb Q(\mu_n)\xrightarrow{\sigma_n}\Bbb Q(\mu_n)\hookrightarrow\Bbb Q(\mu)$$
By taking the union of the graphs of these functions for every $n\geq 1$, we get a function $\sigma:\Bbb Q(\mu)\to\Bbb Q(\mu)$.
This function is, in fact, an automorphism of $\Gal_{\Bbb Q}(\Bbb Q(\mu))$ and $\sigma|_{\Bbb Q(\mu_n)}=\sigma_n$ for every $n\geq 1$.
