Let $k$ be a field of characteristic 0 with a fixed algebraic closure $\bar k$ and absolute Galois group $\Gamma={\rm Gal}(\bar k/k)$. Let $M$ be a finite $\Gamma$-module, that is, a finite abelian group endowed with a continuous action of $\Gamma$. Write $$M^D={\rm Hom}(M,\bar k^\times).$$ We have a canonical pairing $$ M\times M^D\to \bar k^\times,\quad (m,d)\mapsto d(m),$$ which induces a cup product pairing $$ H^0(\Gamma, M)\times H^2(\Gamma, M^D)\to H^2(\Gamma,\bar k^\times)=:{\rm Br}(k), \quad (m, [d_{s,t}])\mapsto [d_{s,t}(m)],$$ where ${\rm Br}(k)$ denotes the Brauer group of $k$, and $(d_{s,t})\in Z^2(\Gamma,M^D)$. Note that $ H^0(\Gamma, M)=M^\Gamma$, the group of fixed points of $\Gamma$ in $M$.

Let $x\in H^2(\Gamma, M^D)$. Using the cup product pairing, we obtain a homomorphism $$ x^0\colon M^\Gamma\to {\rm Br}(k),\quad m\mapsto m\,.\, x,$$ where $ m\,.\, x\in {\rm Br}(k)$ denotes the cup product of $m$ and $x$.

If $k$ is a $p$-adic field, then by the Tate duality theorem (see Serre, "Galois cohomology", or Milne, "Arithmetic duality theorems", or Harari, "Cohomologie galoisienne et théorie du corps de classes") the cup product pairing is perfect, that is, the homomorphism $$ H^2(\Gamma,M^D)\to {\rm Hom}(M^\Gamma, {\rm Br}(k)), \quad x\mapsto x^0$$ is an isomorphism. It follows that $x^0=0$ if and only if $x=0$. Moreover, the assertion "$x^0=0$ if and only if $x=0$" is true also for any finitely generated $\Gamma$-module $M$.

If $k=\bf R$, then using the Tate duality over $\bf R$ (see Milne, "Arithmetic duality theorems") one can show that again $x^0=0$ if and only if $x=0$ for any finitely generated $\Gamma$-module $M$.

Question: What is an example of a field $k$ of characteristic 0, a finitely generated $\Gamma$-module $M$, and a cohomology class $x\in H^2(\Gamma, M^D)$ such that $x\neq 0$, but $x^0=0$?


1 Answer 1


Let us take $k=\mathbf{Q}$ and $M=\mu_3$. Then $H^0(\Gamma,M)=0$. On the other hand, $M^D\simeq \mathbf{Z}/3\mathbf{Z}$, and one can show that $H^2(\Gamma,M^D)\neq 0$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because $H^0(\Gamma,M)=0$. (We thank Ofer Gabber for this example.)

Alternatively, there exists a field $k$ of characteristic 0 such that ${\rm Br}(k)=0$, but $k$ is not of dimension $\le 1$; see Serre, Galois Cohomology, II.3.1, Exercise 1. For such $k$, there exists a finite $\Gamma$-module $A$ such that $H^2(\Gamma,A)\neq 0$. Take $M=A^D$; then $A=M^D$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because ${\rm Br}(k)=0$. (We thank David Harari for this example.)


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