Lemma 47.9.3. Let $A \to B$ be a ring homomorphism and let $I \subset A$ be a finitely generated ideal. Set $J = IB$. Let $Z = V(I)$ and $Y = V(J)$. Then

functorially in $K \in D(A)$.

Lemma 47.9.3. Let $A \to B$ be a ring homomorphism and let $I \subset A$ be a finitely generated ideal. Set $J = IB$. Let $Z = V(I)$ and $Y = V(J)$. Then

\[ R\Gamma _ Z(K) \otimes _ A^\mathbf {L} B = R\Gamma _ Y(K \otimes _ A^\mathbf {L} B) \]

functorially in $K \in D(A)$.

**Proof.**
Write $I = (f_1, \ldots , f_ r)$. Then $J$ is generated by the images $g_1, \ldots , g_ r \in B$ of $f_1, \ldots , f_ r$. Then we have

\[ (A \to \prod A_{f_{i_0}} \to \ldots \to A_{f_1\ldots f_ r}) \otimes _ A B = (B \to \prod B_{g_{i_0}} \to \ldots \to B_{g_1\ldots g_ r}) \]

as complexes of $B$-modules. Represent $K$ by a K-flat complex $K^\bullet $ of $A$-modules. Since the total complexes associated to

\[ K^\bullet \otimes _ A (A \to \prod A_{f_{i_0}} \to \ldots \to A_{f_1\ldots f_ r}) \otimes _ A B \]

and

\[ K^\bullet \otimes _ A B \otimes _ B (B \to \prod B_{g_{i_0}} \to \ldots \to B_{g_1\ldots g_ r}) \]

represent the left and right hand side of the displayed formula of the lemma (see Lemma 47.9.1) we conclude. $\square$

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