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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 $H^n(G(L|K),U_L)$ for $n=0,-1$ vanish for finite unramified extensions $L|K$, where $U_L$ is the group of units. He mentions in the proof that every element $a \in A_L$ can be written as $a = \epsilon \pi_K^m$, where $\epsilon \in U_L$ and $\pi_K$ is a prime element in $A_K$. Why does this work? I absolutely understand this argument when the image of the valuation just lies in $\mathbb{Z}$! But how does this work for a valuation whose image is $\hat{\mathbb{Z}}$? Unless $A$ is not a profinite module, I don't know what $\pi_K^m$ is for some general $m \in \hat{\mathbb{Z}}$. 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 $\pi_K^m$ for a general $m$ in $\hat{\mathbb{Z}}$. All you really need to know for his argument is that $v_K(A_K) = v_L(A_L)$ as subgroups of $\hat{\mathbb{Z}}$. I didn't think this through but I think it should be pretty easy to establish from the fact that $\pi_K$ is prime for both valuations.

All he really uses is that the Galois group fixes $\pi_K$.

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  • $\begingroup$ 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, 2009 at 16:48
  • $\begingroup$ Ah, no HTML available in comments. Sorry. $\endgroup$
    – user717
    Oct 26, 2009 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, 2009 at 19:20
  • $\begingroup$ Oh, that works of course! But then this looks like a mistake there!? Anyway. $\endgroup$
    – user717
    Oct 27, 2009 at 9:13

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