A geometric introduction to K-theory [Lecture notes] by Daniel Dugger

By Daniel Dugger

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4). ). The mystery factor is often called SK1 (R). We will not calculate K1 for many rings, but let us at least do the easiest examples. 12. If F is a field then K1 (F ) = F ∗ . Proof. One must show that if A ∈ GL(F ) satisfies det(A) = 1 then A ∈ [GL(F ), GL(F )] = E(F ). We first observe that if A is a diagonal matrix of determinant 1 then A lies in E(F ). This can be proven by matrix manipulation, but the following argument is a bit easier to write. We use that GL(F )/E(F ) ∼ = K1 (F ). Let d1 , .

For the relative K-groups one can’t quite decompose that far, but one can always get down to complexes of length 1. To state a theorem along these lines, consider maps f : P → Q where P and Q are finitely-generated R-projectives and S −1 f is an isomorphism (it is convenient to regard such maps as chain complexes concentrated in degrees 0 and 1). Let K(R, S)≤1 be the quotient of the free abelian group on such maps by the following relations: (1) [f ] = 0 if f is an isomorphism; (2) [f ] = [f ] + [f ] if there is a commutative diagram GP GP GP G0 0 f 0  GQ f  GQ  GQ f G0 where the rows are exact.

It will be convenient to prove this at the same time that we give other descriptions for K1 (R). In particular, we make the following definitions: (1) K1f r (R) is the group defined similarly to K1 (R) but changing all occurrences of ‘projective’ to ‘free’. 5). The “sp” stands for “split”. (3) K1sp,f r (R) is the group defined by making both the changes indicated in (1) and (2). 7) GL(R)ab colimP Aut(P )ab y G G K sp (R) 1 y G G K1 (R) y colimn GLn (R)ab G G K sp,f r (R) 1 G G K f r (R). 1 The maps labelled as surjections are obviously so.

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