By Daniel Dugger

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**Extra resources for A geometric introduction to K-theory [Lecture notes]**

**Example text**

48 DANIEL DUGGER We know K0 (D, S) ∼ = G(M | S −1 M = 0). The condition S −1 M = 0 just says that M is a torsion module. Consider the evident map G(D/P ) → G(M | S −1 M = 0) j: P =0 where the direct sum is over all nonzero prime ideals and where the map just forgets that a module is defined over D/P and instead regards it as a D-module. 15 and its proof). Note that each D/P is a field, and so G(D/P ) ∼ = Z. If M is a torsion D-module then MP is a torsion DP -module. Since DP is a discrete valuation ring, this means that MP has finite length.

This function is additive, and in fact it is clearly the universal additive invariant for bounded complexes of finitely-generated projectives. The usual Euler characteristic is χ(P• ) = χt (P• )|t=−1 . 5) χt (ΣP• ) = t · χt (P• ). If we differentiate χt with respect to t then we obtain χt (P• ) = jtj−1 [Pj ]. Clearly this is also an additive invariant of complexes. The invariant we called χ is just χt (P• )|t=−1 . 6) χt (ΣP• ) = χt (P• ) + t · χt (P• ), and consequently χ (ΣP• ) = χ(P• ) − χ (P• ).

Given a map β : (S −1 R)n → (S −1 R)n , choose a u ∈ S such that the standard matrix representing uβ has entries in R. Consider the assignment uβ u β → F (β, u) = [Rn −→ Rn ] − n[R −→ R] ∈ K0 (R, S). 16). It remains to show that the above formula does indeed define a homomorphism. We first show that F (β, u) does not depend on the choice of u. It suffices to prove that F (β, tu) = F (β, u) for any t ∈ S; for if u is another choice for u then 46 DANIEL DUGGER we would have F (β, u) = F (β, u u) = F (β, u ).