Lemma 51.21.3. In the situation above, for $q \geq q(A, I)$ and any $A$-module $M$ we have

in $D(A)$.

Lemma 51.21.3. In the situation above, for $q \geq q(A, I)$ and any $A$-module $M$ we have

\[ R\Gamma (X, Lp^*\widetilde{M}(q)) \cong M \otimes _ A^\mathbf {L} I^ q \]

in $D(A)$.

**Proof.**
Choose a free resolution $F_\bullet \to M$. Then $\widetilde{F}_\bullet $ is a flat resolution of $\widetilde{M}$. Hence $Lp^*\widetilde{M}$ is given by the complex $p^*\widetilde{F}_\bullet $. Thus $Lp^*\widetilde{M}(q)$ is given by the complex $p^*\widetilde{F}_\bullet (q)$. Since $p^*\widetilde{F}_ i(q)$ are right acyclic for $\Gamma (X, -)$ by our choice of $q \geq q(A, I)$ and since we have $\Gamma (X, p^*\widetilde{F}_ i(q)) = I^ qF_ i$ by our choice of $q \geq q(A, I)$, we get that $R\Gamma (X, Lp^*\widetilde{M}(q))$ is given by the complex with terms $I^ qF_ i$ by Derived Categories of Schemes, Lemma 36.4.3. The result follows as the complex $I^ qF_\bullet $ computes $M \otimes _ A^\mathbf {L} I^ q$ by definition.
$\square$

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