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If $K/k$ is a finite Galois extension of number fields and $\kappa\in H^2(K/k,\mathbb{G}_m)=H^2(\mathrm{Gal}(K/k),K^\times)$, then for each prime $v$ of $k$, upon choosing a prime $w$ of $K$ above $v$, get a localization map $H^2(K/k,\mathbb{G}_m)\rightarrow H^2(K_w/k_v,\mathbb{G}_m)$. Doing this for each place of $k$ gives a map $\beta:H^2(K/k,\mathbb{G}_m)\rightarrow\prod_vH^2(K_w/k_v,\mathbb{G}_m)$ (this requires choosing a place above each $v$).

My question: is there an obvious reason that the image of $\beta$ is contained in the direct sum of the local cohomology groups? Phrased differently: is a global class with coefficients in the multiplicative group locally trivial almost everywhere?

I know that the image of $\beta$ is contained in the direct sum because if $I_K$ is the idele group of $K$, then by considerations with Shapiro's lemma and some cohomological results from local class field theory (specifically the vanishing of the cohomology of units for an unramified extension), one gets $H^2(K/k,I_K)\cong\sum_vH^2(K_w/k_v,\mathbb{G}_m)$, and the composite $H^2(K/k,\mathbb{G}_m)\rightarrow H^2(K/k,I_K)\rightarrow\sum_vH^2(K_w/k_v,\mathbb{G}_m)$, where the first map comes by embedding $K$ into $I_K$ diagonally, is $\beta$. I'm wondering if there is an easier way.

By passing to the limit over all finite Galois extensions, one gets the corresponding locally trivially almost everywhere result for $H^2(\bar{k}/k,\mathbf{G}_m)$, which is what I really care about. I know that this result can be proved by identifying $H^2$ with the Brauer group of $k$ and invoking the fact that a central simple algebra over $k$ is split almost everywhere (admittedly I don't know much about this point of view, so I don't know how hard this fact is). I'm not sure how to describe what kind of argument I'm looking for, and it's entirely possible I've missed an easy one. I'm hoping to use just basic facts about Galois cohomology. I guess I have in mind something like proving that a class in $H^1(\bar{k}/k,A)$ is locally unramified almost everywhere for a discrete $\mathrm{Gal}(\bar{k}/k)$-module $A$, which just uses the fact that any class comes from $H^1$ of a finite Galois extension, which is unramified outside finitely many places.

Thanks!

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I wouldn't call it an obvious reason, but this can be seen by interpreting the elments of $H^2(K/k,K^\times)$ as elements of a Brauer group, and using the fact that central simple algebras split off a finite set. This is done in the Proposition of §18.5 in Pierce's Associative Algebras, for example.

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Thanks Mariano. I mentioned the identification with the Brauer group in the last paragraph, and maybe this is the most straightforward way. It certainly makes the statement $\mathrm{Br}(K_v^{unr}/K)=\mathrm{Br}(K_v)$ easier (existence of an unramified splitting field), but I've never been completely comfortable with the identification...perhaps it's time I fixed that :) –  Keenan Kidwell Nov 19 '11 at 0:24
    
I confess to not having read the last paragraph :) I don't know why, though! –  Mariano Suárez-Alvarez Nov 19 '11 at 0:26

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