SKl(B[z,z-l])--~SKl(~Z,Z'~) --~coker 0o--~0 is exact. 5: -l Define Dn by DO = $ and Dn+l = Dn[Zn+l, Zn+l]. xS l is the n-torus. 8] and the fact that KO(sI) = 0 and K-I(sI) = Z we have~O(x x Sl) = E-l(x) ~ O ( x ) and K-I(x x SI) = ~-2(X)~B ~-l(x)~) Z =~O(x)~K-I(x)q) Z where the last equality follows from the periodicity theorem. Thus starting with Tl = Sl we find by induction that KO(Tn) is free of rank 2n-l-l for n ~ I. 7 and the above for each n we have an exact sequence (6) 0 ~ SKl(Dn[Zn+l,Z~11]) ~ SKl(Dn+l) ~ Now ~n is an integral domain so D n [ Z , z - l ] ' ~ KO(Tn) ~0 x Z from which i t follows that that SKl(~n[Z,z'l]) ~ SKI(Dn) using the fundamental theorem Kl(Dn[z,z-l])~ Ko(Dn)~) KI(Dn) and the previous computation KO(~n) = 0 which 40 implies Ko(Dn) = Z.

# Algebraic K-Theory: Proceedings of the Conference Held at by Michaeol J. Stein

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