The Stacks project

Lemma 15.104.5. 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 $K \to L \to M \to K[1]$ 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

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

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. $\square$


Comments (0)


Post a comment

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AFY. Beware of the difference between the letter 'O' and the digit '0'.