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.)


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.

  • $\begingroup$ 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}... $\endgroup$ – user717 Oct 26 '09 at 16:48
  • $\begingroup$ Ah, no HTML available in comments. Sorry. $\endgroup$ – user717 Oct 26 '09 at 16:50
  • $\begingroup$ 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. $\endgroup$ – Joel Dodge Oct 26 '09 at 19:20
  • $\begingroup$ Oh, that works of course! But then this looks like a mistake there!? Anyway. $\endgroup$ – user717 Oct 27 '09 at 9:13

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