We quickly show that the zeroth K-group of the derived category of perfect complexes of a ring $R$ is the same as $K_0(R)$ defined in Algebra, Section 10.55.

**Proof.**
Let $K$ be a perfect object of $D(R)$. By definition we can represent $K$ by a finite complex $M^\bullet $ of finite projective $R$-modules. We define $c$ by setting

\[ c(K) = \sum (-1)^ n[M^ n] \]

in $K_0(R)$. Of course we have to show that this is well defined, but once it is well defined, then (1) and (3) are immediate. For the moment we view the map $c$ as defined on complexes of finite projective $R$-modules.

Suppose that $L^\bullet \to M^\bullet $ is a surjective map of finite complexes of finite projective $R$-modules. Let $K^\bullet $ be the kernel. Then we obtain short exact sequences of $R$-modules

\[ 0 \to K^ n \to L^ n \to M^ n \to 0 \]

which are split because $M^ n$ is projective. Hence $K^\bullet $ is also a finite complex of finite projective $R$-modules and $c(L^\bullet ) = c(K^\bullet ) + c(M^\bullet )$ in $K_0(R)$.

Suppose given finite complex $M^\bullet $ of finite projective $R$-modules which is acyclic. Say $M^ n = 0$ for $n \not\in [a, b]$. Then we can break $M^\bullet $ into short exact sequences

\[ \begin{matrix} 0 \to M^ a \to M^{a + 1} \to N^{a + 1} \to 0,
\\ 0 \to N^{a + 1} \to M^{a + 2} \to N^{a + 3} \to 0,
\\ \ldots
\\ 0 \to N^{b - 3} \to M^{b - 2} \to N^{b - 2} \to 0,
\\ 0 \to N^{b - 2} \to M^{b - 1} \to M^ b \to 0
\end{matrix} \]

Arguing by descending induction we see that $N^{b - 2}, \ldots , N^{a + 1}$ are finite projective $R$-modules, the sequences are split exact, and

\[ c(M^\bullet ) = \sum (-1)[M^ n] = \sum (-1)^ n([N^{n - 1}] + [N^ n]) = 0 \]

Thus our construction gives zero on acyclic complexes.

It follows formally from the results of the preceding two paragraphs that $c$ is well defined and satisfies (2). Namely, suppose the finite complexes $M^\bullet $ and $L^\bullet $ of finite projective $R$-modules represent the same object of $D(R)$. Then we can represent the isomorphism by a map $f : M^\bullet \to L^\bullet $ of complexes, see Derived Categories, Lemma 13.19.8. We obtain a short exact sequence of complexes

\[ 0 \to L^\bullet \to C(f)^\bullet \to K^\bullet [1] \to 0 \]

see Derived Categories, Definition 13.9.1. Since $f$ is a quasi-isomorphism, the cone $C(f)^\bullet $ is acyclic (this follows for example from the discussion in Derived Categories, Section 13.12). Hence

\[ 0 = c(C(f)^\bullet ) = c(L^\bullet ) + c(K^\bullet [1]) = c(L^\bullet ) - c(K^\bullet ) \]

as desired. We omit the proof of (2) which is similar.
$\square$

The following lemma shows that $K_0(R)$ is equal to $K_0(D_{perf}(R))$.

**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)

\[ \sigma _{\geq n}M^\bullet \to \sigma _{\geq n - 1}M^\bullet \to M^{n - 1}[-n + 1] \to (\sigma _{\geq n}M^\bullet )[1] \]

and induction show that

\[ [M^\bullet ] = \sum (-1)^ i[M^ i[0]] \]

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.
$\square$

## Comments (0)