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Apr 23, 2011 at 6:54 vote accept David Loeffler
Apr 22, 2011 at 21:32 comment added Kevin Buzzard @Keerthi: I think that Serre in his book on Galois cohomology talks about how one can do Galois cohomology "intrinsically"---probably all that it boils down to though is that he's doing etale cohomology really :-). Serre defines $H^i(k,M)$ without making a specific choice of alg closure of $k$.
Apr 22, 2011 at 19:54 answer added Olivier timeline score: 16
Apr 22, 2011 at 16:29 comment added Ramsey There's an old paper of Mazur that discusses these kinds of groups, though it's been a long time since I looked at it so I don't recall what the specific goals of the paper are. I think it's called "Notes on the etale cohomology of number fields." It might be worth a look.
Apr 22, 2011 at 14:46 comment added mephisto There is quite a bit on this in the second chapter of Milne's Arithmetic Duality Theorems --- see for example Proposition 2.9, the discussion on pages 195--197 (of the second edition), and Lemma 5.5.
Apr 22, 2011 at 13:44 comment added Keerthi Madapusi One thing is that Galois cohomology, by its very definition, requires the choice of a base-point (the `biggest' extension in which the primes outside $S$ are unramified), while etale cohomology is defined intrinsically in terms of the etale site over $\text{Spec}\mathcal{O}_{K,S}$. Usually, when you actually compute things, you end up returning to group cohomology, but it seems nicer to set up the theory without it.
Apr 22, 2011 at 10:57 comment added user19475 I asked the same question here: mathoverflow.net/questions/60310/…
Apr 22, 2011 at 9:29 history asked David Loeffler CC BY-SA 3.0