Lemma 36.38.3. Let $X = \mathop{\mathrm{Spec}}(R)$ be an affine scheme. Then $K_0(X) = K_0(R)$ and if $R$ is Noetherian then $K'_0(X) = K'_0(R)$.

**Proof.**
Recall that $K'_0(R)$ and $K_0(R)$ have been defined in Algebra, Section 10.55.

By More on Algebra, Lemma 15.119.2 we have $K_0(R) = K_0(D_{perf}(R))$. By Lemmas 36.10.7 and 36.3.5 we have $D_{perf}(R) = D_{perf}(\mathcal{O}_ X)$. This proves the equality $K_0(R) = K_0(X)$.

The equality $K'_0(R) = K'_0(X)$ holds because $\textit{Coh}(\mathcal{O}_ X)$ is equivalent to the category of finite $R$-modules by Cohomology of Schemes, Lemma 30.9.1. Moreover it is clear that $K'_0(R)$ is the zeroth K-group of the category of finite $R$-modules from the definitions. $\square$

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