Let $k$ be a number field. We have the well-known Kummer exact sequence of etale sheaves on $\mathrm{Spec}\, k$: $$1 \rightarrow \mu\_n \rightarrow \mathbb{G}\_m \rightarrow \mathbb{G}\_m \rightarrow 1.$$


**Question 0.** Applying the etale cohomology functor $H\_{et}^i(k,-)$, I know that $H\_{et}^i(k,\mathbb{G}\_m) = H^i(k,\bar{k}^\*)$, where the latter is a Galois cohomology group. What is the Galois cohomological equivalent for $H^i\_{et}(k,\mu\_n)$?


Let $\mu\_\infty := \mathrm{colim}\_n\mu\_n$, this group can be interpreted as the set of all $n$-th roots of unity, i.e., $\mu\_\infty \cong \mathbb{Q}/\mathbb{Z}$. We work only in Galois cohomology.


**Question 1.** I would like to know if we could obtain a similar Kummer sequence involving $\mu\_\infty$ instead of $\mu\_n$ for a fixed $n$. The reason for this is that I came across an exact sequence $$0 \rightarrow \mathrm{Br}(k) \otimes \_\mathbb{Z} \mathbb{Q}/\mathbb{Z} \rightarrow H^3(k,\mu\_\infty) \rightarrow H^3(k,\bar{k}^\*) \rightarrow 0$$ but I am unable to derive it from the usual Kummer sequence.


**Question 2.** The idea for this exact sequence is to prove that the middle term is 0. It is known that we have $H^3(k,\bar{k}^\*)=0$ in our setting and for any non-archimedean place $v$, we have $\mathrm{Br}(k\_v) \subset \mathbb{Q}/\mathbb{Z}$. But I'm also unsure how does it imply that $\mathrm{Br}(k) \otimes \_\mathbb{Z} \mathbb{Q}/\mathbb{Z}= 0$.



A Kummer exact sequence involving $\mu_\infty$
Regarding you question 1, it suffices to take the short exact sequence
$$1\to \mu\_\infty\to \mathbb{G}\_m\to \mathbb{G}\_m\otimes\_{\mathbb{Z}}\mathbb{Q}\to 1\,.$$
This sequence is exact in the étale topos of any scheme over $\mathbb{Q}$. Indeed to show surjectivity it is enough to show that for every ring $R$, any $x\in R^\times=\mathbb{G}\_m(R)$ and any $n\ge1$ integer, there is an étale extension $R\to R'$ and $y\in \mathbb{G}\_m(R')$ such that $y\otimes 1=x\otimes \frac{1}{n}$. But it is enough to take $y=\sqrt[n]{x}$ in $R'=R[y]/(y^n-x)$. Moreover $\mu\_\infty$ is just the kernel of the second map by definition.


Another way of thinking about the above short exact sequence is that it is the colimit of the short exact sequences
$$ 1\to \mu\_n \to \mathbb{G}\_m\xrightarrow{n}\mathbb{G}\_m\to 1$$
as $n$ goes to infinity along the divisibility poset (where the arrows on the central term are all the identities, while the arrows on the second term are multiplications by suitable integers).


Once you have this short exact sequence, you can look at the long exact sequence in cohomology


$$ H^2(k;\mathbb{G}\_m)\to H^2(k;\mathbb{G}\_m\otimes\mathbb{Q})\simeq H^2(k;\mathbb{G}\_m)\otimes\mathbb{Q}\to H^3(k;\mu\_\infty)\to H^3(k;\mathbb{G}\_m) \to H^3(k;\mathbb{G}\_m)\otimes\mathbb{Q}$$


and you can deduce the vanishing of $H^3(k;\mu\_3)$ from the vanishing of $H^2(k;\mathbb{G}\_m)\otimes\mathbb{Q}$ and $H^3(k;\mathbb{G}\_m)$.




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Regarding your second question, the simplest reason I can think of is that $\operatorname{Br}(k)\otimes\_{\mathbb{Z}}\mathbb{Q}/\mathbb{Z}=0$ because the Brauer group of a field is torsion (e.g. Example II.2.22 in Milne's *Étale cohomology*) and $\mathbb{Q}/\mathbb{Z}$ is divisible. I don't see much point in passing through the completions of $k$.


