By Garrett P.

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**Example text**

D−1 } is a K-basis for D, and since π i ∈ k if and only if i = 0 mod d, for 1 ≤ i < d we have (full trace of left multiplication by απ i on D = 0 Thus, the reduced trace is also 0 on such elements. On the other hand, the restriction of the reduced trace on D to K is equal to the Galois trace from K to k. For x ∈ D, we may write αi π −i x= 0≤i

If D = k we are done. This leaves the unique quaternion division algebra D to be considered. The case that the involution θ on the quaternion division algebra D is of first kind is easy, since we already know that D has a main involution, so by Skolem-Noether any other involution of first kind differs by a conjugation. Now suppose that θ is of second kind. Let \alf → α be the main involution. Then α → (αθ ) is an automorphism of order 2 of D and gives a non-trivial automorphism τ on k. The set Do = {x ∈ D : xθ } is a subring of D containing the subfield ko of k fixed by τ .

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.

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

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