Lemma 15.93.8 (Bhatt). Let $I$ be a finitely generated ideal in a ring $A$. Let $M$ be a derived complete $A$-module. If $M$ is an $I$-power torsion module, then $I^ nM = 0$ for some $n$.

**Proof.**
Say $I = (f_1, \ldots , f_ r)$. It suffices to show that for each $i$ there is an $n_ i$ such that $f_ i^{n_ i}M = 0$. Hence we may assume that $I = (f)$ is a principal ideal. Let $B = \mathbf{Z}[x] \to A$ be the ring map sending $x$ to $f$. By Lemma 15.91.23 we see that $M$ is derived complete as a $B$-module with respect to the ideal $(x)$. After replacing $A$ by $B$, we may assume that $f$ is a nonzerodivisor in $A$.

Assume $I = (f)$ with $f \in A$ a nonzerodivisor. According to Example 15.93.3 there exists a short exact sequence

where $K$ and $L$ are $I$-adically complete $A$-modules whose $f$-torsion is zero^{1}. Consider $K$ and $L$ as topological modules with the $I$-adic topology. Then $u$ is continuous. Let

Since $M$ is $f$-power torsion we see that $L = \bigcup L_ n$. Let $N_ n$ be the closure of $L_ n$ in $L$. By Lemma 15.36.4 we see that $N_ n$ is open in $L$ for some $n$. Fix such an $n$. Since $f^{n + m} : L \to L$ is a continuous open map, and since $f^{n + m} L_ n \subset u(f^ m K)$ we conclude that the closure of $u(f^ mK)$ is open for all $m \geq 1$. Thus by Lemma 15.36.5 we conclude that $u$ is open. Hence $f^ tL \subset \mathop{\mathrm{Im}}(u)$ for some $t$ and we conclude that $f^ t$ annihilates $M$ as desired. $\square$

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