Timeline for First Galois cohomology of Weil restriction of $\mathbb{G}_m$
Current License: CC BY-SA 3.0
15 events
when toggle format | what | by | license | comment | |
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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 |
deleted 487 characters in body
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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 |
edited body
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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 |
added 85 characters in body
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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 |
added 677 characters in body
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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 |
made iso more explicit
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Feb 2, 2016 at 22:53 | history | answered | R.P. | CC BY-SA 3.0 |