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Example text

12) Proof. 10) gives G χ(g)k dλ = × p Z xk d(χ(λ)). ˜ Λ(Z× p) Also, via our identification of with a subset of Λ(Zp ), the integral above on the right has the same value if we integrate over the whole ˜ ˜ of Zp . Now take λ = L(u), so that, by definition, we have χ( ˜ L(u)) = Υ(L(fu )), where we recall that L(fu )(T ) = fu (T )p 1 . 12) is equal to Dk−1 (hu (T ) − ϕ(hu )(T )) T =0 , where hu (T ) = (1 + T ) fu (T ) . 12), and the proof of the proposition is complete. 12). 4 of the previous chapter determines the image of δk for k = 1, · · · , p − 1.

If Y is any subset of R, then we denote by Y its image in Ω under the reduction map. 7. If ∆(W ) = Rψ=1 , then ∆(W ) = Rψ=1 . 4 The Logarithmic Derivative 23 Proof. Assume that the reductions of ∆(W ) and Rψ=1 do coincide, and take any g in Rψ=1 . Hence there exists h1 in W such that ∆(h1 ) = g. This implies that ∆(h1 ) − g = pg2 for some g2 in R, and again we have that ψ(g2 ) = g2 . Repeating this argument, we conclude that there exists h2 in W such that ∆(h2 ) − g2 = pg3 , with g3 in W . Note that since ∆(a) = 0 for all a in Z× p , it can be assumed, by multiplying by an appropriate (p − 1)-th root of unity, that h1 , h2 , · · ·, all have constant term which is congruent to 1 modulo p.

5) above. Of course, the series on the right converges because an tends to zero as n −→ ∞. Since the cn lie in Zp , it is clear that |L(f )|p ≤ f for all f . Hence there exists λ in Λ(Zp ) such that L = Mλ , and we define Υ(g(T )) = λ. It is plain that Υ is an inverse of M. In fact, it can also be shown that M preserves products, although we omit the proof here. 4. We have M(1Zp ) = 1 + T , and thus M : Λ(Zp ) −→ R is the unique isomorphism of topological Zp -algebras which sends the topological generator 1Zp of Zp to (1 + T ).

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