5
$\begingroup$

Let $k$ be a number field. Let $M$ be a (continuous) $\text{Gal}(\overline{k}/k)$-module.

One can define two subgroups of the Galois cohomology group $H^i(k,M)$:

  • the group of elements of $H^i(k,M)$ mapping to zero in $H^i(\langle g \rangle, M)$ for all $g \in \text{Gal}(\overline{k}/k)$,

  • the group of elements of $H^i(k,M)$ mapping to zero in $H^i(k_v,M)$ for almost all places $v$ of $k$.

Is it true that these groups coincide? (The answer should be yes.) And how does one prove this?

$\endgroup$
3
  • $\begingroup$ A bit of context perhaps? For instance, I'm guessing you wouldn't mind assuming that $M$ is a continuous $\operatorname{Gal}(\bar{k}/k)$-module but you don't really say. $\endgroup$
    – Olivier
    Oct 18, 2011 at 5:30
  • $\begingroup$ Yes, of course I assume continuity, thanks. $\endgroup$
    – Wanderer
    Oct 18, 2011 at 10:41
  • $\begingroup$ The context: various texts about arithmetic duality theorems... Also, this might have something to do with Chebotarev's theorem. $\endgroup$
    – Wanderer
    Oct 18, 2011 at 10:43

1 Answer 1

1
$\begingroup$

Some ideas: I it holds when $M$ is a trivial Galois module (and $i=1$, $M$ finite) . For, if $\phi \in H^1(G_k,M) = Hom(G_k,M)$ restricts in almost all decomposition groups to $0$, it is $0$ by Chebotarev (it even suffices for the densitiy of the set of primes to be greater than $1/p$, $p$ the smallest prime divisor of $|M|$. On the other hand, if it maps all generators (all Frobenii suffice) to $0$, it must be $0$. So the groups coincide in this case and are equal to $0$.

Now for a counterexample: Take $M = \mu_n$, $i=2$. Then $H^2(k,M) = Br(k)[n] \hookrightarrow \oplus_v Br(k_v)[n] = \oplus_v \frac{1}{n}\mathbf{Z}/\mathbf{Z}$. Take $x = (\frac{1}{n}, \frac{1}{n}, 0, 0, 0, \ldots)$. We have $0 \neq x_v = \frac{1}{n} \in Br(k_v)[n] = \frac{1}{n} \mathbf{Z}/\mathbf{Z}$ $= Br(k_v^{nr}/k_v)[n] = Hom(<Frob_v>, \frac{1}{n} \mathbf{Z}/\mathbf{Z})$. Can someone finish from here (perhaps http://www.univ-valenciennes.fr/lamav/preprints/lamav-07.10.pdf) might help.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.