# Neukirch's class field axiom and cohomology of units for unramified extension

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 H^n(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 \in AL can be written as a = \epsilon * \piK^m, where \epsilon \in UL and \piK is a prime element in AK. Why does this work? I absolutely understand this argument, when the image of the valuation just lies in \ZZ! But how does this work for a valuation whose image is \widehat{\ZZ}? Unless A is not a profinite module, I don't know what \piK^m is for some general m \in \widehat{\ZZ}. Unfortunately this has to work in this generality for global class field theory.

(\ZZ denotes the integers of course, sorry for my personal notation.)

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

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

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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 Oct 26 '09 at 16:48
Ah, no HTML available in comments. Sorry. –  user717 Oct 26 '09 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 Oct 26 '09 at 19:20
Oh, that works of course! But then this looks like a mistake there!? Anyway. –  user717 Oct 27 '09 at 9:13