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1 . By the uniqueness of the expansions β = i αi π i for β ∈ D, conjugation by π shows that β is in k if and only if the expansion is actually of the form αi (π ν )i β= i and with all αi ∈ (M ∪ {0}) ∩ k. In particular, this shows that ∆(π ν ) = ∆(π)ν is largest value of ∆ on k less than 1, so π ν is a local parameter in k. From this (and from the ultrametric property) it follows that 1, π, π 2 , . . , π ν−1 are linearly independent over k. Thus, k(π) is a subfield of D, with [k(π) : k] = ν.

Since the field P/P is finite, there is an integer m such that for λ( ) x = Then define ordα α−1 = m = xp mod P mod P2 αα−1 = (q − 1) =− mod P2 µ∈M So π is also a local parameter, λ(π) = λ( ), and π normalizes M . /// Corollary: With M and π as in the theorem, every x ∈ D× has a unique expression of the form αi π i x= i≥m with αi ∈ M , and where m is the uniquely determined integer so that x ∈ πm · O× Proof: If x ∈ O× , then αo = ω(x) ∈ M satisfies α = x mod P, and by the theorem is uniquely determined in M by this property.

Necessarily m is prime to the residue characteristic of k, so ϕ factors modulo p as i ϕi where each ϕi is of degree n. By Hensel’s lemma, ϕ factors in such manner over k. That is, a Galois unramified extension of k is cyclic and is generated over k by a root of unity. Thus, every unramified extension of k is in fact cyclic. And then recapitulation of this argument shows that K is generated by a root of unity. To show that the norm from K to k is surjective when restricted to a map O× → o× on the local units, where O and o are the local rings of integers, We first reprove the even more elementary fact that norms O/P → o/p are surjective on finite fields.

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Algebras and Involutions(en)(40s) by Garrett P.

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