**Proof.**
For an $A$-module $M$ denote $M^\wedge $ the derived completion and $\mathop{\mathrm{lim}}\nolimits M/I^ nM$ the usual completion. Assume $M$ is finite. The system $\text{Tor}^ A_ i(M, A/I^ n)$ is pro-zero for $i > 0$, see Lemma 15.27.3. Since $\text{Tor}_ i^ A(M \otimes _ A N, A/I^ n) = \text{Tor}_ i^ A(M, A/I^ n) \otimes _ A N$ as $N$ is flat, the same is true for the system $\text{Tor}^ A_ i(M \otimes _ A N, A/I^ n)$. By Lemma 15.94.3 we conclude $R\mathop{\mathrm{lim}}\nolimits (M \otimes _ A N) \otimes _ A^\mathbf {L} A/I^ n$ only has cohomology in degree $0$ given by the usual completion $\mathop{\mathrm{lim}}\nolimits M \otimes _ A N/ I^ n(M \otimes _ A N)$. This proves (1). Part (2) follows from (1) and the fact that $M_ f = M \otimes _ A A_ f$.
$\square$

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