By D. Arnold, R. Hunter, E. Walker

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We need to calculate We need the trace of s: M Q N - ~ M ~ N . s(mQn) sm~sn. If m i i~ a base for M and nj a base for N then M @ N mi@ If sm i nj. ~ aiKm K and snj E bjenj, then = has - 28 - s ( m i ~ n j) = s m i ~ s n j = k , ~ a i k b j ~ ( m k ~ n ~ ) . So the trace is XM(S)XN(S). Next we check that the map preserves the identity. the unit of Go(K~). sl = 1 for all s ~ ~. [K] is Thus XK(S) = 1. Done. Thus X identifies Go(K~)with a subring of K valued functions on ~. Now let F(~, K) be the ring of K valued functions on ~.

F~ II (R~)~ and F/~ = II(R~/~). Hence we only need to prove the lemma for M = R~. N. Now the = R to N ~ (R~) ~ and is an isomorphism. 8° If 11"is a finite solvable group of order n, R is a Dedekind ring of characteristic 0 such that no prime dividing n is a unit in R, and P is a finitely generated projective R ~ m o d u l e , then rankRP = n rankRP~. Proof. 7 allows us to use P / ~ for P~. We proceed by in- duction on I ~ l l) If ~ i s of ~o not of prime order, there is a proper normal subgroup N rankRP -- INlrankR(P/N) by inductive hypothesis.

IT,K) ....... i" f ~- ~ F(T~ ,K) fi If i is a monomorphism (which we can assume is an inclusion), then we have ~ = U s i ~ where the s i are coset representatives. Then K ~ is free as a right K ~ K~K~N = ~ ( s i ~ M). That is, every element is uniquely expres- sible in the form ~ s i ~ m t(~si@m Ubsi~ i) = ~ t s i @ m module with s i as a base and i mi~ N. If t ~ ~, then i is not in the correct form. ~ t ~ = ~ and the union is disjoint. However, Hence there is a permu- tation of the cosets induced by t such that t s i ~ = st(i)~.