Lemma 15.91.13. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K, L \in D(A)$. Then

$R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)^\wedge = R\mathop{\mathrm{Hom}}\nolimits _ A(K, L^\wedge ) = R\mathop{\mathrm{Hom}}\nolimits _ A(K^\wedge , L^\wedge )$

Proof. By Lemma 15.91.10 we know that derived completion is given by $R\mathop{\mathrm{Hom}}\nolimits _ A(C, -)$ for some $C \in D(A)$. Then

\begin{align*} R\mathop{\mathrm{Hom}}\nolimits _ A(C, R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)) & = R\mathop{\mathrm{Hom}}\nolimits _ A(C \otimes _ A^\mathbf {L} K, L) \\ & = R\mathop{\mathrm{Hom}}\nolimits _ A(K, R\mathop{\mathrm{Hom}}\nolimits _ A(C, L)) \end{align*}

by Lemma 15.73.1. This proves the first equation. The map $K \to K^\wedge$ induces a map

$R\mathop{\mathrm{Hom}}\nolimits _ A(K^\wedge , L^\wedge ) \to R\mathop{\mathrm{Hom}}\nolimits _ A(K, L^\wedge )$

which is an isomorphism in $D(A)$ by definition of the derived completion as the left adjoint to the inclusion functor. $\square$

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