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Corollary: 19 For every multiplicative set S of the ring R NSK1{S-lR) ~ [S]-lNSKl{R) ~ W{S-lR) ® NSKl{R); NU(S-lR) ~ [S]-lNU(R) ~ W(S-lR) ® NU(R). 3. these groups also equal S-lNSK 1(R) respectively. S-lNU(R), Remark: S ~ Z, The result for Theorem: S ~ Z and NU is classical. ) S NSKO(S-lR) ~ [S]-lNSKO{R) ~ W(S-lR) ® NSKO(R); NPic(S-lR} ~ [S]-lNPic(R) ~ W(S-lR) ® NPic(R). If R is a ~algebra, or -1 and S NPic(R), Remark: S ~ Z, these groups also equal S-1 NSKO (R) respectively. 1]. 3 supplies the answer to Swan's problem of formulating that result in greater generality.
And let its ring of global functions. I. OX* ). SP (X) SL " of is the famous isomorphism Similarly. the set set n which are trivial on each of the n n = 1 Pic (X) P (X) corresponds to (See [Weil]. [Hirz. b]. [Milne. 134] ...... ) ~. This gives a GL . n denote of global units acts on SP (X)/A* n vector bundles P on is isomorphic X with trivial. In particular. if P is a rank n vector bundle with trivial B. DAYTON AND C. WEIBEL determinant. then open cover {U} P 25 comes from of X such that 0U-modules together via matrices Proof: For each a € A*.
Is an R-module. or if S ~ Z. this group also equals S-INK (R). n Proof: See [Vorst. 4]. [vdK. 6] and [WNK. 8]. If M is any continuous W(R)-module. then [S]-IM is the same as W(S-IR) ~ M by o [WNK. 2]. 5 with A = R[t]. 1. Consider the following diagram of W{S-1R)-modules. whose rows are exact: o ----+ [S]-I NSK1 (R) ------+1 [S] -1 NKI (R) - 1~ 1 o ---+ NSK 1(S-I R) II [S] -----+1 -1 ~ -1 NK1 (S R) 1 1 I -1 NSK 1 (S Rred ) ------+1 NKI (S Rred ) I [S]-INSK1(Rred) Since NSK 1 (R) ~ NK 1(R red ). ~ -1 I~ [S]-1NK1(Rred) a diagram chase proves: NU(R) -1 NU(S R) - ---+ 0 0 B.