Brauer group of global fields Is the Brauer group $\text{Br}(K)$ of a global field $K$ 


*

*an $\ell$-divisible group for some prime $\ell$? If so, what $\ell$?

*Is $\text{Br}(K)[n]$ finite, for $n$ integer?
I know from class field theory that it fits into an exact sequence
$$0\to \text{Br}(K)\to\bigoplus_v\text{Br}(K_v)\xrightarrow{\sum_v \text{inv}_v} \mathbf{Q}/\mathbf{Z}\to 0$$
with $v$ running over all places of $K$, and $K_v$ the completion of $K$ at $v$.
but I can't conclude. 
Thanks very much.
 A: It is $\ell$-divisible for every odd number $\ell$.
To see this, let $\alpha \in Br(K)$, and look at its image in $(\alpha_\nu)_\nu  \in\oplus_\nu \mathbb{Q}/ \mathbb{Z}$. You know that each component is divisible by $\ell$, so you can form, in several ways, the element $(\alpha_\nu /\ell)_\nu$. The problem is that now it might be that $inv((\alpha_\nu /\ell)_\nu)$ is not $0$.
denote this number by $d \in \mathbb{Q} / \mathbb{Z}$.  This is actually an element of $(1/\ell) \mathbb{Z} / \mathbb{Z}$. We can choose some $\nu_0$ and modify 
$(\alpha_\nu / \ell)_\nu$ by subtracting $d|_{\nu_0}$ (i.e. the element that has $d$ in the $\nu_0$ summand and 0 elsewhere) and get a new element 
$(\beta_\nu)_\nu$. The difference between $(\alpha_\nu / \ell)_\nu$ and 
$(\beta_\nu)_\nu$ is $\ell$-torsion, so clearly we still have 
$\ell (\beta_\nu)_\nu = (\alpha_\nu)_\nu$, but on the other hand 
$inv(\beta_\nu) = 0$ so it comes from an element $\beta$ of $Br(K)$. 
Since the map $Br(K) \to \oplus_\nu Br(K_\nu)$ is injective, this implies 
$\ell \beta = \alpha$. 
Regarding the second question, $Br(K)[n]$ is not finite. For example, choose $\nu_0$ and consider all the elements of the form $(1/n)|_{\nu_0} - (1/n)|_{\nu}$ for $\nu$ a non-Archimedean place. Then it is an infinite sequence of different elements in it.  
