Lemma 51.2.1. Let $A$ be a ring and let $I$ be a finitely generated ideal. Set $Z = V(I) \subset X = \mathop{\mathrm{Spec}}(A)$. For $K \in D(A)$ corresponding to $\widetilde{K} \in D_\mathit{QCoh}(\mathcal{O}_ X)$ via Derived Categories of Schemes, Lemma 36.3.5 there is a functorial isomorphism

$R\Gamma _ Z(K) = R\Gamma _ Z(X, \widetilde{K})$

where on the left we have Dualizing Complexes, Equation (47.9.0.1) and on the right we have the functor of Cohomology, Section 20.34.

Proof. By Cohomology, Lemma 20.34.5 there exists a distinguished triangle

$R\Gamma _ Z(X, \widetilde{K}) \to R\Gamma (X, \widetilde{K}) \to R\Gamma (U, \widetilde{K}) \to R\Gamma _ Z(X, \widetilde{K})[1]$

where $U = X \setminus Z$. We know that $R\Gamma (X, \widetilde{K}) = K$ by Derived Categories of Schemes, Lemma 36.3.5. Say $I = (f_1, \ldots , f_ r)$. Then we obtain a finite affine open covering $\mathcal{U} : U = D(f_1) \cup \ldots \cup D(f_ r)$. By Derived Categories of Schemes, Lemma 36.9.4 the alternating Čech complex $\text{Tot}(\check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \widetilde{K^\bullet }))$ computes $R\Gamma (U, \widetilde{K})$ where $K^\bullet$ is any complex of $A$-modules representing $K$. Working through the definitions we find

$R\Gamma (U, \widetilde{K}) = \text{Tot}\left( K^\bullet \otimes _ A (\prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r})\right)$

It is clear that $K^\bullet = R\Gamma (X, \widetilde{K^\bullet }) \to R\Gamma (U, \widetilde{K}^\bullet )$ is induced by the diagonal map from $A$ into $\prod A_{f_ i}$. Hence we conclude that

$R\Gamma _ Z(X, \mathcal{F}^\bullet ) = \text{Tot}\left( K^\bullet \otimes _ A (A \to \prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r})\right)$

By Dualizing Complexes, Lemma 47.9.1 this complex computes $R\Gamma _ Z(K)$ and we see the lemma holds. $\square$

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