The Stacks project

Lemma 15.117.2. Let $R$ be a ring. Let $D_{perf}(R)$ be the derived category of perfect objects, see Lemma 15.77.1. The map $c$ of Lemma 15.117.1 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)

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

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)


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Your email address will not be published.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 0FCU. Beware of the difference between the letter 'O' and the digit '0'.