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.



Brauer group of global fields
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. 


