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May 25, 2010 at 18:54 comment added Emerton The standard technique for computing $H^n(G,A)$ when $n = 0$ is just the definition: one looks for $G$-fixed elements in $A$. For $n = 1$, here is a common method: the $G$-action on $A$ gives a homo. $G \to Aut(A)$; let $H$ be the kernel. (So $H$-acts trivially on $A$.) Then one has the inflation-restriction sequence $0 \to H^1(G/H,A) \to H^1(G,A) \to Hom_{G/H}(H^{ab},A)$ (where the last term is written as $Hom$ rather than $H^1$ precisely because $H$ acts trivially on $A$). Now one hopes that the outer two terms are easier to compute, and can be pieced together to understand $H^1(G,A)$.
May 25, 2010 at 4:04 answer added Torsten Ekedahl timeline score: 2
May 25, 2010 at 3:33 answer added Cam McLeman timeline score: 2
May 25, 2010 at 3:27 history asked user1832 CC BY-SA 2.5