Lemma 15.94.7. Let $I$ be an ideal in a Noetherian ring $A$. Let ${}^\wedge$ denote derived completion with respect to $I$. Let $K \in D^-(A)$.

1. If $M$ is a finite $A$-module, then $(K \otimes _ A^\mathbf {L} M)^\wedge = K^\wedge \otimes _ A^\mathbf {L} M$.

2. If $L \in D(A)$ is pseudo-coherent, then $(K \otimes _ A^\mathbf {L} L)^\wedge = K^\wedge \otimes _ A^\mathbf {L} L$.

Proof. Let $L$ be as in (2). We may represent $K$ by a bounded above complex $P^\bullet$ of free $A$-modules. We may represent $L$ by a bounded above complex $F^\bullet$ of finite free $A$-modules. Since $\text{Tot}(P^\bullet \otimes _ A F^\bullet )$ represents $K \otimes _ A^\mathbf {L} L$ we see that $(K \otimes _ A^\mathbf {L} L)^\wedge$ is represented by

$\text{Tot}((P^\bullet )^\wedge \otimes _ A F^\bullet )$

where $(P^\bullet )^\wedge$ is the complex whose terms are the usual $=$ derived completions $(P^ n)^\wedge$, see for example Proposition 15.94.2 and Lemma 15.94.6. This proves (2). Part (1) is a special case of (2). $\square$

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