Lemma 36.38.1. Let $X$ be a Noetherian scheme. Then
\[ K_0(\textit{Coh}(\mathcal{O}_ X)) = K_0(D^ b(\textit{Coh}(\mathcal{O}_ X)) = K_0(D^ b_{\textit{Coh}}(\mathcal{O}_ X)) \]
Proof. The first equality is Derived Categories, Lemma 13.28.2. We have $K_0(\textit{Coh}(\mathcal{O}_ X)) = K_0(D^ b_{\textit{Coh}}(\mathcal{O}_ X))$ by Derived Categories, Lemma 13.28.5. This proves the lemma. (We can also use that $D^ b(\textit{Coh}(\mathcal{O}_ X)) = D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Proposition 36.11.2 to see the second equality.) $\square$
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