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Aug 19, 2022 at 5:28 vote accept Desunkid
Aug 18, 2022 at 12:59 answer added SashaP timeline score: 2
Aug 18, 2022 at 8:41 comment added Damian Rössler The cokernel of $G[p^k](L)\to A[p^k](L)$ is a subgroup of $H^1(T[p^k],L))$. So, supposing $T=G_m^t$, the sequence is exact if $H^1(T[p^k],L)=(L^{*}/L^{*p^k})^t=0$, so eg if $L$ is algebraically closed.
Aug 17, 2022 at 6:21 history edited Desunkid CC BY-SA 4.0
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Aug 16, 2022 at 21:52 history asked Desunkid CC BY-SA 4.0