Remark 36.38.6. Let $X$ be a scheme. The K-group $K_0(X)$ is canonically a commutative ring. Namely, using the derived tensor product

$\otimes = \otimes ^\mathbf {L}_{\mathcal{O}_ X} : D_{perf}(\mathcal{O}_ X) \times D_{perf}(\mathcal{O}_ X) \longrightarrow D_{perf}(\mathcal{O}_ X)$

and Derived Categories, Lemma 13.28.6 we obtain a bilinear multiplication. Since $K \otimes L \cong L \otimes K$ we see that this product is commutative. Since $(K \otimes L) \otimes M = K \otimes (L \otimes M)$ we see that this product is associative. Finally, the unit of $K_0(X)$ is the element $1 = [\mathcal{O}_ X]$.

If $\textit{Vect}(X)$ and $K_0(\textit{Vect}(X))$ are as above, then it is clearly the case that $K_0(\textit{Vect}(X))$ also has a ring structure: if $\mathcal{E}$ and $\mathcal{F}$ are finite locally free $\mathcal{O}_ X$-modules, then we set

$[\mathcal{E}] \cdot [\mathcal{F}] = [\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}]$

The reader easily verifies that this indeed defines a bilinear commutative, associative product. Details omitted. The map

$K_0(\textit{Vect}(X)) \longrightarrow K_0(X)$

constructed above is a ring map with these definitions.

Now assume $X$ is Noetherian. The derived tensor product also produces a map

$\otimes = \otimes ^\mathbf {L}_{\mathcal{O}_ X} : D_{perf}(\mathcal{O}_ X) \times D^ b_{\textit{Coh}}(\mathcal{O}_ X) \longrightarrow D^ b_{\textit{Coh}}(\mathcal{O}_ X)$

Again using Derived Categories, Lemma 13.28.6 we obtain a bilinear multiplication $K_0(X) \times K'_0(X) \to K'_0(X)$ since $K'_0(X) = K_0(D^ b_{\textit{Coh}}(\mathcal{O}_ X))$ by Lemma 36.38.1. The reader easily shows that this gives $K'_0(X)$ the structure of a module over the ring $K_0(X)$.

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