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 a∈AL can be written as a=ϵπmK, where ϵ∈UL 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 m∈ˆZ. Unfortunately, this must work in this generality for global class field theory.
1 Answer
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}...– user717Commented Oct 26, 2009 at 16:48
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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. Commented Oct 26, 2009 at 19:20
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Oh, that works of course! But then this looks like a mistake there!? Anyway.– user717Commented Oct 27, 2009 at 9:13