By Peter Schenzel

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37 The natural maps yield an isomorphism of the form as required. 34 simplifies to become Proof. Let 1,XI, ~ To be precise, if 2 ~ , 1 x denote 2 the four one-dimensional representations of Q8. Q8 = {x, y 1 x2 = y2, y4 = 1,xyx = y) set xl(x) = -1 = x2(y) and xl(y) = 1 = x2(x). Suppose that f : R(Q8) --+ Za is a homomorphism. By inflation f induces a homomorphism on R(Qgb). Since Since determinantal functions on R(Q8) take values in Z; there is a short exact sequence of the form Ram this sequence we see that there is an isomorphism of the form When p is an odd prime then every function in Hom(R(Q8), Z;) is a determinantal function because in this case Zp[Q8] is a maximal order in QP[Q8]Hence, if p is odd, then TorsKO(Z, [Q8];Qp) = 0.

Q8 = {x, y 1 x2 = y2, y4 = 1,xyx = y) set xl(x) = -1 = x2(y) and xl(y) = 1 = x2(x). Suppose that f : R(Q8) --+ Za is a homomorphism. By inflation f induces a homomorphism on R(Qgb). Since Since determinantal functions on R(Q8) take values in Z; there is a short exact sequence of the form Ram this sequence we see that there is an isomorphism of the form When p is an odd prime then every function in Hom(R(Q8), Z;) is a determinantal function because in this case Zp[Q8] is a maximal order in QP[Q8]Hence, if p is odd, then TorsKO(Z, [Q8];Qp) = 0.

In this case, recalling the isomorphism of Galois modules G(L/K) = (a, g I gd = ac, a' = 1, gag-1 = a") where v = IKI, the order of the residue field, K,of K . Here, if W/K is the maximal -unramified subextension then G(L/W) = (a) and the image of g in G(L/K) is the Frobenius automorphism. Note that, as in ([34] p. 369), we may arrange that c is a divisor of r. When char(K) = p > 0 we may arrange that c = r, since K Fv((X)) and L is a Kummer extension of L(") = FVd((X)). 22. Set q = v2 and let w be an integer satisfying the congruence vw = 1 (modulo (qd - I ) ~ )Here .

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