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 $\mathrm{Br}(k) \otimes _\mathbb{Z} \mathbb{Q}/\mathbb{Z}= 0$.

  • $\begingroup$ Can't you just take $1\to \mu_\infty\to \mathbb{G}_m\to \mathbb{G}_m\otimes \mathbb{Q}\to 1$? Moreover $\operatorname{Br}(k)\otimes_{\mathbb{Z}}\mathbb{Q}/\mathbb{Z}=0$ because $\operatorname{Br}(k)$ is torsion. $\endgroup$ Jul 12, 2021 at 9:39
  • $\begingroup$ @DenisNardin I'm not sure the map you defined is exact. For instance, how is $\mathbb{G}_m \rightarrow \mathbb{G}_m \otimes \mathbb{Q}$ surjective? As for the part on the Brauer group, I agree with you, but I was just wondering how do we deduce that it's zero solely from the fact the each $\mathrm{Br}(k_v) \otimes \mathbb{Q}/\mathbb{Z}$ is zero. $\endgroup$
    – oleout
    Jul 12, 2021 at 10:56
  • $\begingroup$ The sequence is exact. For example, to show that the map $\mathbb{G}_m\to \mathbb{G}_m\otimes \mathbb{Q}$ is surjective it is enough to show that every class of the form $x\otimes \frac{1}{n}$ with $x\in\overline{k}$ and $n\ge1$, is in the image, since those are generators as an abelian group. But $x\otimes \frac{1}{n}=y\otimes 1$ where $y$ is an $n$-th root of $x$. Regarding your second question, I don't see why you expect there to be a proof passing through $\operatorname{Br}(k_v)$. $\endgroup$ Jul 12, 2021 at 11:04

1 Answer 1


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 $H^3(k;\mathbb{G}_m)$.

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 $k$.

  • $\begingroup$ Thank you for your detailed answer. So, say we take $x \otimes \frac{1}{n} \in \mathbb{G}_m \otimes \mathbb{Q}$, even though both are rings over $\mathbb{Z}$, they exhibit different 'behaviours', like this $$x \otimes 1/n = (x^{1/n} \cdot \ldots \cdot x^{1/n}) \otimes 1/n = x^{1/n} \otimes (1/n + \ldots + 1/n = x^{1/n} \otimes 1.$$ But what about the identity of the right hand side, i.e., what does $1 \in \mathbb{G}_m$ map to? It seems we have both $1 \otimes 1$ and $1 \otimes 0$ in $\mathbb{G}_m \otimes \mathbb{Q}$. $\endgroup$
    – oleout
    Jul 12, 2021 at 11:53
  • $\begingroup$ Regarding your edit, it seems you are doing it in étale cohomology, this is okay, but I have two questions. Do you have an explanation or any references for the isomorphism $$H^2(k,\mathbb{G}_m \otimes \mathbb{Q}) \cong H^2(k,\mathbb{G}_m) \otimes \mathbb{Q}?$$ Also, the exact sequence in my question is tensored with $\mathbb{Q}/\mathbb{Z}$, how do we account for that? $\endgroup$
    – oleout
    Jul 12, 2021 at 11:57
  • 1
    $\begingroup$ @KelvinLian Sorry, maybe I should have been more precise. The map $\mathbb{G}_m\to \mathbb{G}_m\otimes \mathbb{Q}$ sends $x$ to $x\otimes 1$ ($x\otimes 0=0$ for every $x$). We can take $-\otimes\mathbb{Q}$ out because $\mathbb{Q}$ is a flat $\mathbb{Z}$-module, and so it commutes with taking (co)homology of a complex. Finally, I'm just using that $M\otimes -$ is right exact for every $M$, so that $M\otimes\mathbb{Q}/\mathbb{Z}$ is the cokernel of $M\to M\otimes \mathbb{Q}$. $\endgroup$ Jul 12, 2021 at 12:08
  • $\begingroup$ Thanks again. So I wonder, what will be mapped to $0 \in \mathbb{G}_m \otimes \mathbb{Q}$? $\endgroup$
    – oleout
    Jul 12, 2021 at 13:21
  • 1
    $\begingroup$ @KelvinLian Perhaps you're confusing $\mathbb{G}_m(R)\otimes\mathbb{Q}$ with $(R\otimes\mathbb{Q})^\times=\mathbb{G}_m(R\otimes\mathbb{Q})$. These are not the same thing! For example when $R=\mathbb{Z}$ we have $\mathbb{G}_m(\mathbb{Z})\otimes \mathbb{Q}=\mathbb{Z}/2\otimes\mathbb{Q}=0$, but $\mathbb{G}_m(\mathbb{Z}\otimes\mathbb{Q})=\mathbb{Q}^\times\neq 0$. $\endgroup$ Jul 12, 2021 at 14:31

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