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This question may be too detailed but perhaps somebody knows the answer: Neukirch proofs in his algebraic number theory book in Chapter IV, Proposition 6.2, that his class field axiom implies that the Tate cohomology groups Hn(G(L|K),UL) for n=0,1 vanish for finite unramified extensions L|K, where UL is the group of units. He mentions in the proof that every element aAL can be written as a=ϵπmK, where ϵUL and πK is a prime element in AK. Why does this work? I absolutely understand this argument when the image of the valuation just lies in Z! But how does this work for a valuation whose image is ˆZ? Unless A is not a profinite module, I don't know what πmK is for some general mˆZ. Unfortunately, this must work in this generality for global class field theory.

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You don't need to make sense out of πmK for a general m in ˆZ. All you really need to know for his argument is that vK(AK)=vL(AL) as subgroups of ˆZ. I didn't think this through but I think it should be pretty easy to establish from the fact that πK is prime for both valuations.

All he really uses is that the Galois group fixes πK.

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  • The problem is somehow that if m \in \ZZ, then v<sub>L</sub>(a) = v<sub>K</sub>(\epsilon * \pi<sub>K</sub>^m) = m and therefore a would be a very special element of A<sub>L</sub>. I don't know why the valuation of a should lie in \ZZ and not somewhere in \widehat{Z}...
    – user717
    Commented Oct 26, 2009 at 16:48
  • Ah, no HTML available in comments. Sorry.
    – user717
    Commented Oct 26, 2009 at 16:50
  • Again, you don't need to show that you can take m in Z. Forget about pi as well for the moment. It would suffice to show that a=epsilon.x where v(epsilon)=0 and x is fixed by the Galois group. Show that the two valuations take the same values and you're done.
    – Joel Dodge
    Commented Oct 26, 2009 at 19:20
  • Oh, that works of course! But then this looks like a mistake there!? Anyway.
    – user717
    Commented Oct 27, 2009 at 9:13

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