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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.

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    $\begingroup$ The Brauer group of $\mathbb{Q}$ is not 2-divisible (because $\operatorname{Br}(\mathbb{R})=\mathbb{Z}/2 $). $\endgroup$
    – abx
    Commented Feb 20, 2018 at 6:32
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    $\begingroup$ @abx Yes, but what about $\ell$-divisibility for primes $\ell > 2$? $\endgroup$
    – user120812
    Commented Feb 20, 2018 at 6:49
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    $\begingroup$ Do you know what the Brauer group of $K_v$ is for nonarchimedean $v$? $\endgroup$
    – KConrad
    Commented Feb 20, 2018 at 7:56
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    $\begingroup$ I see the question has been asked also at math.stackexchange.com/questions/2658210/…, essentially at the same time. The OP generally should ask just at one of these sites and wait a little while before posting it at the other site, to avoid duplicate efforts. $\endgroup$
    – KConrad
    Commented Feb 20, 2018 at 8:00
  • $\begingroup$ Sure, it's $\mathbf{Q}/\mathbf{Z}$, but if for some $x\in\text{Br}(K)$, then we have $\sum_v\text{inv}_v(x_v)=0$, for $(x_v)\in\bigoplus_v(\mathbf{Q}/\mathbf{Z})$, then there is some $(y_v)$ such that $\ell(y_v) = (x_v)$. Who tells us that, at least for $\ell >2$, $\sum_v\text{inv}_v(y_v)=0$? $\endgroup$
    – user120812
    Commented Feb 20, 2018 at 8:01

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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.

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