Lemma 52.6.17. Let $\mathcal{C}$ be a site. Let $\mathcal{O} \to \mathcal{O}'$ be a homomorphism of sheaves of rings. Let $\mathcal{I} \subset \mathcal{O}$ be a finite type sheaf of ideals. If $\mathcal{O} \to \mathcal{O}'$ is flat and $\mathcal{O}/\mathcal{I} \cong \mathcal{O}'/\mathcal{I}\mathcal{O}'$, then the restriction functor $D(\mathcal{O}') \to D(\mathcal{O})$ induces an equivalence $D_{comp}(\mathcal{O}', \mathcal{I}\mathcal{O}') \to D_{comp}(\mathcal{O}, \mathcal{I})$.

Proof. Lemma 52.6.7 implies restriction $r : D(\mathcal{O}') \to D(\mathcal{O})$ sends $D_{comp}(\mathcal{O}', \mathcal{I}\mathcal{O}')$ into $D_{comp}(\mathcal{O}, \mathcal{I})$. We will construct a quasi-inverse $E \mapsto E'$.

Let $K \to \mathcal{O}$ be the morphism of $D(\mathcal{O})$ constructed in Lemma 52.6.11. Set $K' = K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}'$ in $D(\mathcal{O}')$. Then $K' \to \mathcal{O}'$ is a map in $D(\mathcal{O}')$ which satisfies the conclusions of Lemma 52.6.11 with respect to $\mathcal{I}' = \mathcal{I}\mathcal{O}'$. The map $K \to r(K')$ is a quasi-isomorphism by Lemma 52.6.16. Now, for $E \in D_{comp}(\mathcal{O}, \mathcal{I})$ we set

$E' = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(r(K'), E)$

viewed as an object in $D(\mathcal{O}')$ using the $\mathcal{O}'$-module structure on $K'$. Since $E$ is derived complete we have $E = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K, E)$, see proof of Proposition 52.6.12. On the other hand, since $K \to r(K')$ is an isomorphism in we see that there is an isomorphism $E \to r(E')$ in $D(\mathcal{O})$. To finish the proof we have to show that, if $E = r(M')$ for an object $M'$ of $D_{comp}(\mathcal{O}', \mathcal{I}')$, then $E' \cong M'$. To get a map we use

$M' = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}'}(\mathcal{O}', M') \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(r(\mathcal{O}'), r(M')) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(r(K'), r(M')) = E'$

where the second arrow uses the map $K' \to \mathcal{O}'$. To see that this is an isomorphism, one shows that $r$ applied to this arrow is the same as the isomorphism $E \to r(E')$ above. Details omitted. $\square$

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