Lemma 15.93.2. Let $A$ be a ring and $f \in A$. Set $I = (f)$. In this situation we have the naive derived completion $K \mapsto K' = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A/f^ nA)$ and the derived completion

\[ K \mapsto K^\wedge = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A)) \]

of Lemma 15.91.18. The natural transformation of functors $K^\wedge \to K'$ is an isomorphism if and only if the $f$-power torsion of $A$ is bounded.

**Proof.**
If the $f$-power torsion is bounded, then the pro-objects $\{ (f^ n : A \to A)\} $ and $\{ A/f^ nA\} $ are isomorphic by Lemma 15.93.1. Hence the functors are isomorphic by Lemma 15.86.10. Conversely, we see from Lemma 15.87.11 that the condition is exactly that

\[ R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} A[f^ n]) \]

is zero for all $K \in D(A)$. Here the maps of the system $(A[f^ n])$ are given by multiplication by $f$. Taking $K = A$ and $K = \bigoplus _{i \in \mathbf{N}} A$ we see from Lemma 15.86.13 this implies $(A[f^ n])$ is zero as a pro-object, i.e., $f^{n - 1}A[f^ n] = 0$ for some $n$, i.e., $A[f^{n - 1}] = A[f^ n]$, i.e., the $f$-power torsion is bounded.
$\square$

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