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.
The exact sequence splits (non-canonically) to prove that $Br(K)$ is a direct sum of a finite number of copies of $\Bbb Z/2\Bbb Z$ (coming from the real places) and countably many copies of $\Bbb Q/\Bbb Z$. So the $n$-torsion is always infinite (for $n\ge2$) and the group is $\ell$-divisible for all odd $\ell$. For $\ell=2$, $Br(K)$ is $2$-divisible iff $K$ is totally complex.
Galois cohomology provides perhaps the "best" proof for the Brauer-Hasse-Noether theorem (= the global-local exact sequence that you recalled in the number field case; in the case of a function field of positive characteristic, the exact sequence is slightly different), so it's natural to apply the cohomological approach to your problem.
1) Over any field $K$, the starting point is the canonical isomorphism $Br(K)\cong H^2(G_K,{K^{sep}}^*)$, where $G_K=Gal(K^{sep}/K)$. For any integer $n$ prime to char($K)$, raising to the $n$-th power yields an exact sequence $1\to \mu_n \to {K^{sep}}^* \to {K^{sep}}^* \to 1$, where $\mu_n$ denotes the group of $n$-th roots of unity. The resulting cohomology exact sequence reads ...$H^1(G_K, {K^{sep}}^*)\to H^2(G_K, \mu_n) \to H^2(G_K, {K^{sep}}^*) \to H^2(G_K, {K^{sep}}^*)$... But the first $H^1$ group is trivial (Hilbert's thm. 90), so that actually $H^2(G_K, \mu_n)\cong Br(K)[n]$, and for a prime $l\neq$ char($K$), taking inductive limit over the powers of $l$ shows that $H^2(G_K, \mu(l))\cong Br(K)(l)$ (the notation $(.)(l)$ means the $l$-primary part).
2) For simplicity, consider only the case where $K$ is a number field. Suppose $l$ odd or $K$ totally imaginary (@), and let us show that $H^2(G_K, \mu(l))$ is divisible. For any $m$, consider the exact sequence $1\to \mu_{l^m}\to \mu(l) \to \mu(l)\to 1$, and the corresponding cohomology exact sequence ... $H^2(G_K, \mu(l))\to H^2(G_K, \mu(l))\to H^3(G_K,\mu_{l^m})$... Under our assumptions (@), it's known that $cd_l(G_K)\le 2$ (see e.g. Serre, "Galois Cohomology", 4.4, prop. 16), which implies the vanishing of the $H^3$ group and shows the desired $l$-divisibility of $Br(K)(l)$. If $l=2$ and $K$ is not totally imaginary, the BHN exact sequence shows at once that $Br(K)(2)$ is not $2$-divisible.
