Skip to main content
MathOverflow

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

5
  • 2
    $\begingroup$ I asked the same question here: mathoverflow.net/questions/60310/… $\endgroup$ Commented Apr 22, 2011 at 10:57
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Apr 22, 2011 at 13:44
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Apr 22, 2011 at 14:46
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Apr 22, 2011 at 16:29
  • $\begingroup$ @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 $k$. $\endgroup$ Commented Apr 22, 2011 at 21:32