Vanishing question for self-products in Galois cohomology

Question

Suppose that $k$ is a number field. Let $G$ be the absolute Galois group of $k$ , let $M$ be a torsion $G$-module and $\alpha \in H^{1} (G, M)$. Is it true that $$\alpha \cup \alpha \cup \ldots \cup \alpha \in H^{N} (G, M \otimes M \otimes \ldots \otimes M)$$ vanishes for $N \gg 0$? Apologies if this question is a repeat.

Answer 1

There are two cases to be considered:

Case: $k$ has no real embeddings

It's a classical result that the cohomological dimension of a number field without real embeddings is at most 2 (Neukirch, et. al. Cohomology of Number Fields, 8.3.18). So $H^n(G,N)=0$ for all $n \ge 3$ and all torsion $G$-modules $N$. In particular, the cup products in the question vanish for $n \ge 3$.

Case: $k$ has a real embedding

In this case there are counter-examples for each field $k$. We'll construct a cohomology class $\alpha \in H^1(G,\mathbb{F}_2)$ such that the $n$-fold cup product $0 \neq \alpha^n \in H^n(G,\mathbb{F}_2)$ for each $n \ge 1$.

Let $C$ be the subgroup of $G$ of order $2$ generated by complex conjugation $c$. Note that $H^\ast(C,\mathbb{F}_2)=\mathbb{F}_2[\beta]$ is a polynomial ring in a generator $\beta$ of degree $1$. Hence it suffices to find a class $\alpha \in H^1(G,\mathbb{F}_2) = Hom(G, \mathbb{F}_2)$ that restricts to $\beta$.

$\beta$ can be represented by the group isomorphism $\beta: C \xrightarrow{\sim} \mathbb{F}_2$. Define $$\alpha: G = G(\bar{k}/k) \xrightarrow{\text{res}} G(k(\sqrt{-1})/k)\cong C \xrightarrow{\beta} \mathbb{F_2}$$ If $c'$ denotes complex conjugation on $k(\sqrt{-1})$ than the isomorphism above is just given by $c' \leftrightarrow c$. Now, by definition, $\alpha(c)=\beta(c)$ and hence $\text{res}^G_C(\alpha)=\beta$ and we are done.

Remark: Since $cd_p(k) \le 2$ for all $p\neq 2$, the cup product in the question also vanishes in the latter case if $M$ has no $2$-torsion.