Lemma 10.96.4. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $0 \to M \to N \to Q \to 0$ be an exact sequence of $R$-modules such that $Q$ is annihilated by a power of $I$. Then completion produces an exact sequence $0 \to M^\wedge \to N^\wedge \to Q \to 0$.

Proof. Say $I^ c Q = 0$. Then $Q/I^ nQ = Q$ for $n \geq c$. On the other hand, it is clear that $I^ nM \subset M \cap I^ nN \subset I^{n - c}M$ for $n \geq c$. Thus $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/(M \cap I^ n N)$. Apply Lemma 10.87.1 to the system of exact sequences

$0 \to M/(M \cap I^ n N) \to N/I^ n N \to Q \to 0$

for $n \geq c$ to conclude. $\square$

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