Lemma 42.56.9. Let $X$ be a Noetherian regular scheme. Let $Z \subset X$ be a closed subscheme. The maps constructed in Remarks 42.56.7 and 42.56.8 are mutually inverse and we get $K'_0(Z) = K_0(D_{Z, perf}(\mathcal{O}_ X))$.
Proof. Clearly the composition
\[ K_0(\textit{Coh}(Z)) \longrightarrow K_0(D_{Z, perf}(\mathcal{O}_ X)) \longrightarrow K_0(\textit{Coh}(Z)) \]
is the identity map. Thus it suffices to show the first arrow is surjective. Let $E$ be an object of $D_{Z, perf}(\mathcal{O}_ X)$. Recall that $D_{perf}(\mathcal{O}_ X) = D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.11.8. Hence the cohomologies $H^ i(E)$ are coherent, can be viewed as objects of $D_{Z, perf}(\mathcal{O}_ X)$, and only a finite number are nonzero. Using the distinguished triangles of canonical truncations the reader sees that
\[ [E] = \sum (-1)^ i[H^ i(E)[0]] \]
in $K_0(D_{Z, perf}(\mathcal{O}_ X))$. Then it suffices to show that $[\mathcal{F}[0]]$ is in the image of the map for any coherent $\mathcal{O}_ X$-module set theoretically supported on $Z$. Since we can find a finite filtration on $\mathcal{F}$ whose subquotients are $\mathcal{O}_ Z$-modules, the proof is complete. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)