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Let $A$ be a $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$-module which is a finitely generated free $\mathbb{Z}$-module. I'm interested in the behaviour of cohomology classes in $$\mathrm{H}^1(\mathbb{Q}, A)$$ upon extension to $\mathbb{Q}_p$. In particular, the set of primes $p$ for which a cohomology class $c$ trivialises upon base-change to $\mathbb{Q}_p$ (I denote the base-change of $c$ to $\mathbb{Q}_p$ by $c \otimes \mathbb{Q}_p \in \mathrm{H}^1(\mathbb{Q}_p, A)$).

Let $c \in \mathrm{H}^1(\mathbb{Q}, A)$. Does the set of primes $p$ for which $$c \otimes \mathbb{Q}_p = 0$$ have a density?

Remarks:

  • Each $c$ trivialises after some finite extension $k/\mathbb{Q}$. Thus $c \otimes \mathbb{Q}_p = 0$ for any prime $p$ which is completely split in $k$. This gives a positive density of primes $p$ which trivialise $c$; I want to understand all of them.

I understand the problem well in other, similar looking, situations (i.e. for different choices of $A$). For example:

  • The answer to the analogous question is true when $A$ is finite; this follows from the Chebotarev density theorem.
  • If $A$ is instead the set of algebraic points on an abelian variety the answer is again yes; in fact one has $c \otimes \mathbb{Q}_p = 0$ for all but finitely many primes $p$ by Lang-Weil and Hensel's lemma, as follows by interpreting $c$ as the class of a torsor for an abelian variety.
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  • $\begingroup$ In your last example ($A$ is the set of algebraic points on an abelian variety), the group $A$ is neither a finitely generated $\mathbb Z$-module nor a free $\mathbb Z$-module. So it doesn't seem to fit the parameters of your question. (It is, of course, still an interesting situation.) $\endgroup$ – Joe Silverman Nov 2 '16 at 17:16
  • $\begingroup$ I know, that is why I said "instead". I will rewrite my question to make it clearer; I'm just trying to make clear that in these different cases, I understand the problem completely. $\endgroup$ – Daniel Loughran Nov 2 '16 at 17:28
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    $\begingroup$ Doesn't Inf-Res just say that $H^1(Q,A)=H^1(Gal(K/Q),A)$ and similarly for the local guys, so Cebotarev gives you an answer? $\endgroup$ – znt Nov 2 '16 at 18:48
  • $\begingroup$ @znt: This is a nice observation! Please feel free to post this as an answer. $\endgroup$ – Daniel Loughran Nov 2 '16 at 21:53
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I think it follows again from Chebotarev. Represent the cohomogy class as an extension of the trivial representation by $A$. Such an extension is itself a Galois action on a finitely-generated $\mathbb Z$-module and thus has finite image. For unramified primes, whether this extension splits over $\mathbb Q_p$ depends only on the conjugacy class of Frobenius in this finite image group.

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  • $\begingroup$ This works even when $A$ is not free. $\endgroup$ – Will Sawin Nov 2 '16 at 19:04

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