Lemma 15.119.2. Let $R$ be a ring. Let $D_{perf}(R)$ be the derived category of perfect objects, see Lemma 15.78.1. The map $c$ of Lemma 15.119.1 gives an isomorphism $K_0(D_{perf}(R)) = K_0(R)$.

**Proof.**
It follows from the definition of $K_0(D_{perf}(R))$ (Derived Categories, Definition 13.28.1) that $c$ induces a homomorphism $K_0(D_{perf}(R)) \to K_0(R)$.

Given a finite projective module $M$ over $R$ let us denote $M[0]$ the perfect complex over $R$ which has $M$ sitting in degree $0$ and zero in other degrees. Given a short exact sequence $0 \to M \to M' \to M'' \to 0$ of finite projective modules we obtain a distinguished triangle $M[0] \to M'[0] \to M''[0] \to M[1]$, see Derived Categories, Section 13.12. This shows that we obtain a map $K_0(R) \to K_0(D_{perf}(R))$ by sending $[M]$ to $[M[0]]$ with apologies for the horrendous notation.

It is clear that $K_0(R) \to K_0(D_{perf}(R)) \to K_0(R)$ is the identity. On the other hand, if $M^\bullet $ is a bounded complex of finite projective $R$-modules, then the the existence of the distinguished triangles of “stupid truncations” (see Homology, Section 12.15)

and induction show that

in $K_0(D_{perf}(R))$ (with again apologies for the notation). Hence the map $K_0(R) \to K_0(D_{perf}(R))$ is surjective which finishes the proof. $\square$

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