## 15.109 Perfect complexes and K-groups

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 10.54.

Lemma 15.109.1. Let $R$ be a ring. There is a map

$c : \text{perfect complexes over }R \longrightarrow K_0(R)$

with the following properties

1. $c(K[n]) = (-1)^ nc(K)$ for a perfect complex $K$,

2. if $K \to L \to M \to K$ is a distinguished triangle of perfect complexes, then $c(L) = c(K) + c(M)$,

3. if $K$ is represented by a finite complex $M^\bullet$ consisting of finite projective modules, then $c(K) = \sum (-1)^ i[M_ i]$.

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

$0 \to L^\bullet \to C(f)^\bullet \to K^\bullet  \to 0$

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

$0 = c(C(f)^\bullet ) = c(L^\bullet ) + c(K^\bullet ) = c(L^\bullet ) - c(K^\bullet )$

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

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

Lemma 15.109.2. Let $R$ be a ring. Let $D_{perf}(R)$ be the derived category of perfect objects, see Lemma 15.73.1. The map $c$ of Lemma 15.109.1 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 $M$ 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 $M \to M' \to M'' \to M$, 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 $[M]$ 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)

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

and induction show that

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

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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