Automorphisms of infinite cyclotomic extension field Denote $s_n$ be the $n^{th}$ root of unity. Let $K=\Bbb Q(s_1, s_2, s_3,\dots)$.
Then how to prove that $\operatorname{Aut}(K/\Bbb Q)$ is abelian?
 A: That's really the definition. Your field is $K = \lim_{n \to \infty} \mathbb{Q}(\zeta_{n!})$ the limit of a tower of fields extensions $\mathbb{Q}(\zeta_{(n+1)!})/\mathbb{Q}(\zeta_{n!})$


*

*if $\sigma,\sigma_2  \in Gal(K/\mathbb{Q})$ then $\sigma(\zeta_{n!}) = \zeta_{n!}^{k(n)},\sigma_2(\zeta_{n!}) = \zeta_{n!}^{l(n)}$ where $gcd(n!,k(n))=gcd(n!,l(n))=1$ and so $$\sigma \circ \sigma_2(\zeta_{n!}) = \sigma(\zeta_{n!}^{l(n)})=\zeta_{n!}^{l(n)k(n)}=\sigma_2 \circ \sigma(\zeta_{n!})$$

*any element of $K$ can be written as $\sum_{m=0}^{n!} c_m \zeta_{n!}^m$ for some $n \in \mathbb{N}, c_m\in \mathbb{Q}$ 

*by definition of the Galois group $\sigma \circ \sigma_2$ and $\sigma_2 \circ \sigma$ are elements of $Gal(K/\mathbb{Q})$ 
i.e. 
$$\sigma \circ \sigma_2(\sum_{m=0}^{n!} c_m \zeta_{n!}^m) = \sum_{m=0}^{n!} c_m\sigma \circ \sigma_2(\zeta_{n!}^m)=\sum_{m=0}^{n!}c_m\sigma_2 \circ \sigma(\zeta_{n!}^m)=\sigma_2 \circ \sigma(\sum_{m=0}^{ n!} c_m \zeta_{n!}^m)$$
$\implies$ $Gal(K/\mathbb{Q})$ is abelian.
