Let $X$ be a smooth geometrically integral variety over a number field $k$, we have an exact sequence of complexes of $\mathrm{Gal}(\bar{k}/k)$-modules $$0 \rightarrow [\bar{k}^* \rightarrow 0] \rightarrow [\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X})] \rightarrow [\bar{k}(X)^*/\bar{k}^* \rightarrow \mathrm{Div}(\bar{X})] \rightarrow 0.$$

As usual, the complexes in the middle (called 1-motives) are placed in degrees $-1$ and $0$. Taking hypercohomology groups, Lemma 2.1 of Local-global principles for 1-motives by Harari and Szamuely gives us the isomorphism $$\mathbf{H}^1(k,[\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X})]) \cong \mathrm{Br}_1(X),$$ this is the algebraic Brauer group of $X$. I have two questions regarding its proof:

(1) It says that we have a distinguished triangle $$\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X}) \rightarrow [\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X})] \rightarrow \bar{k}(X)^*[1],$$ but how do we check which exact triangle is this isomorphic to? This triangle itself is clearly not exact.

(2) The result seem to follow directly from the fact that the permutation module $\mathrm{Div}(\bar{X})$ has trivial $H^1$ (Galois cohomology), but I can't seem to see why.

  • $\begingroup$ There seems to be a misprint in the first term of (1). $\endgroup$ Jun 26, 2021 at 17:34
  • $\begingroup$ @FernandoMuro Indeed, edited, thanks. $\endgroup$
    – oleout
    Jun 27, 2021 at 0:02
  • $\begingroup$ The distinguished triangle (1) is a particular instance of the one of the form $A\to B\to C \to A[1]$ associated to any map of chain complexes $f:A\to B$, with $C$ the mapping cone of $f$, thus fitting in a canonical exact sequence $0\to B\to C\to A[1]\to 0$. This has nothing to do with any permutation module structure of any sort. In many textbooks, distinguished triangles are defined as those isomorphic to those as above in the derived category. $\endgroup$ Jun 27, 2021 at 8:42
  • $\begingroup$ @Denis-CharlesCisinski Ok in this case we have $C=\bar{k}(X)^\times[1] \oplus \mathrm{Div}(\bar{X})$, so the distinguished triangle is $$\bar{k}(X)^\times \rightarrow \mathrm{Div}(\bar{X}) \rightarrow \bar{k}(X)^\times[1] \oplus \mathrm{Div}(\bar{X}) \rightarrow \bar{k}(X)^\times[1].$$ Do we have $$[\bar{k}(X)^\times \rightarrow \mathrm{Div}(\bar{X})] = \bar{k}(X)^\times[1] \oplus \mathrm{Div}(\bar{X})?$$ $\endgroup$
    – oleout
    Jun 28, 2021 at 1:41
  • $\begingroup$ @Denis-CharlesCisinski My apologies regarding Question (2). What I meant was by applying the functor $H^i(k,-)$ to the distinguished triangle, the proof used the fact that $\mathrm{Div}(\bar{X})$ is a permutation module to deduce that its $H^1$ is trivial. Thus the isomorphism in the question follows. What I don't understand is how being a permutation module leads to a trivial $H^1$. $\endgroup$
    – oleout
    Jun 28, 2021 at 1:45

1 Answer 1


After referring to some sources, I shall try to provide an answer my own question.

With reference to the last paragraph of the answer by Dan Petersen to this question What is a triangle?, we apply the (hyper)cohomology functor $H^i(k,-)$ to the distinguished triangle given in question (1). This gives us the long exact sequence $$H^1(k,\mathrm{Div}(\bar{X})) \rightarrow \mathbf{H}^1(k,[\bar{k}(X)^\times \rightarrow \mathrm{Div}(\bar{X})]) \rightarrow H^2(k,\bar{k}(X)^\times) \rightarrow H^2(k,\mathrm{Div}(\bar{X})).$$

Since $\mathrm{Div}(\bar{X})$ is a permutation module (indeed $\mathrm{Div}(\bar{X})$ is a free abelian group and it has a $\mathbb{Z}$-basis permuted by $\Gamma_k :=\mathrm{Gal}(\bar{k}/k)$), it is the sum of modules of the form $\mathbb{Z}[\Gamma_k/H]$ where $H$ is a subgroup of $\Gamma_k$. By Shapiro's lemma, for $i\geq 1$, we have $$H^i(\Gamma_k,\mathbb{Z}[\Gamma_k/H]) \cong H^i(H,\mathbb{Z}) = 0.$$ Hence $H^1(k,\mathrm{Div}(\bar{X}))=0$ and so the middle map above is injective. The kernel of the last map is identified as $\mathrm{Br}_1(X)$ as seen in Diagram (4.17) on page 72 of Skorobogatov's book Torsors and rational points. Hence we obtain the desired isomorphism.


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