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Feb 3, 2016 at 18:16 vote accept Sasha
Feb 3, 2016 at 18:16 comment added Sasha Thank you for your answer! I think that it is basically what I outlined, and I just reproved Shapiro's lemma, or something similar.
Feb 3, 2016 at 12:52 comment added R.P. You're totally right. I edited the answer.
Feb 3, 2016 at 12:52 history undeleted R.P.
Feb 3, 2016 at 12:52 history edited R.P. CC BY-SA 3.0
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Feb 3, 2016 at 12:51 history deleted R.P. via Vote
Feb 3, 2016 at 6:42 comment added nfdc23 @René: For the $K$-group $R$, the $L/K$-twists of $R$ are classified by ${\rm{H}}^1(G, {\rm{Aut}}(R_L))$, not by ${\rm{H}}^1(G, R(L))$; i.e., the coefficient group is the automorphism scheme of $R$ rather than $R$ itself. Also, twisted forms of $\mathbf{G}_m^n$ are $n$-dimensional tori over $K$ split by $L$, of which there are many non-split ones (no link to GL$_n(L)$). Rather, $R$ is the Aut-scheme of the underlying $K$-vector space of an $L$-line, so $H^1(G,R(L))=1$ since any such $L$-line has a basis.
Feb 3, 2016 at 5:45 history edited R.P. CC BY-SA 3.0
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Feb 3, 2016 at 5:39 comment added R. van Dobben de Bruyn I guess $R$ is a twist of $\mathbb G_m^{[L:K]}$ as opposed to $\mathbb G_m^2$. (I think I can tell which field extension you were thinking of...)
Feb 3, 2016 at 5:29 history edited R.P. CC BY-SA 3.0
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Feb 3, 2016 at 5:23 history undeleted R.P.
Feb 3, 2016 at 5:17 history edited R.P. CC BY-SA 3.0
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Feb 2, 2016 at 23:21 history deleted R.P. via Vote
Feb 2, 2016 at 23:18 history edited R.P. CC BY-SA 3.0
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Feb 2, 2016 at 22:53 history answered R.P. CC BY-SA 3.0