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.
Lemma 15.119.1. Let $R$ be a ring. There is a map with the following properties
$c(K[n]) = (-1)^ nc(K)$ for a perfect complex $K$,
if $K \to L \to M \to K[1]$ is a distinguished triangle of perfect complexes, then $c(L) = c(K) + c(M)$,
if $K$ is represented by a finite complex $M^\bullet $ consisting of finite projective modules, then $c(K) = \sum (-1)^ i[M_ i]$.
\[ c : \text{perfect complexes over }R \longrightarrow K_0(R) \]
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
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
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
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
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
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
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 15.119.2. 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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)