Brauer group of maximal abelian extension of rationals I am trying to show that if $K$ is the compositum of all cyclotomic fields (so by Kronecker-Weber, consists of all abelian extensions of $\mathbb{Q}$), then its Brauer group is trivial.
My attempt goes as follows: firstly, it can be shown that $H^2(G_K,\mu_n)\simeq Br(K)[n]$, where $G_K=\mathrm{Gal}(K_s/K)$ by looking at the long exact sequence associated to $1\rightarrow \mu_n\rightarrow K_s^\times\rightarrow K_s^\times\rightarrow 1$ and using Hilbert 90. Furthermore, there is a fundamental exact sequence from class field theory $$0 \to Br(K) \to \bigoplus_v Br(K_v) \to \mathbb Q/\mathbb Z \to 0 $$ which implies that $Br(K)$ splits as a sum copies of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Q}/\mathbb{Z}$. If I could show that the $n$-torsion $H^2(G_K,\mu_n)\simeq Br(K)[n]$ is trivial for all $n$, this would imply that the group is trivial. I know $G_K$ acts trivially on $\mu_n$ so we can also interpret $H^2(G_K,\mu_n)$ as equivalence classes of central extensions. Another thing that we could use is that $H^2(G_K,\mu_n)=\varinjlim H^2(\mathrm{Gal}(F/K),\mu_n),$ where $F/K$ range over the finite extensions of $K$. However I am stuck here and don't know how to proceed, or even if this would work - any help would be appreciated.
EDIT: In Local Fields, page 162, Serre says the following: "An algebraic extension of Q containing all the roots of unity [has trivial Brauer group]. This follows, by passage to the limit, from the determination of the Brauer group of a number field. " As far as I know, $Br(K)=\varinjlim H^2(F/K)$, but am not sure how this would be of any help and how Serre's argument goes.
 A: Since $K$ is a union of finite extensions of $\mathbf{Q}$, any elemeny $\mathrm{Br}(K)$ is certainly defined over some finite cyclotomic extension $E = \mathbf{Q}(\zeta_n)$. For a number field, there is an injection
$$\mathrm{Br}(E) \rightarrow \bigoplus \mathrm{Br}(E_v).$$
If $c \in \mathrm{Br}(E)$, the image of $c$ in $\mathrm{Br}(E_v)$ is non-trivial for only a finite set of primes $S$. Assume that $c$ has order $d$. Now choose a finite extension $F/E$ contained in $K$ such that $F_w/E_v$ contains a cyclic extension of degree dividing $d$ for all $v \in S$. Note that this is easy to do, since there are unramified cyclic extensions of any local field of any degree, and they are all given by adjoining roots of unity. But now the class $c_v \in \mathrm{Br}(E_v)$ becomes trivial in $\mathrm{Br}(F_w)$, and then the restriction of $c$ to $\mathrm{Br}(F)$ is everywhere locally trivial and thus trivial.
A: The  Merkurjev-Suslin theoremsays that $Br_n(K)$ is generalted by the class of cyclic algebras over $K$ for any field $K$ of characteristic $0$ containing $\mu_n$.
In particular, if $K=\mathbb{Q}^{ab}$, it is enough to prove that any cyclic algebra over $K$ splits. But a cyclic algebra of degree $d$ contains a cyclic extension of $K$ of degree $d$. Since $K$ does not have any nontrivial abelian extension, $d=1$ and the result follows.
I'm not sure you can find an elementary argument avoiding this theorem.
