Lemma 10.98.2. Let $I \subset A$ be a finitely generated ideal of a ring. Let $(M_ n)$ be an inverse system of $A$-modules with $M_ n = M_{n + 1}/I^ nM_{n + 1}$. Set $M = \mathop{\mathrm{lim}}\nolimits M_ n$. Then $M/I^ nM = M_ n$ and $M$ is $I$-adically complete.

**Proof.**
By Lemma 10.98.1 we see that $M$ is $I$-adically complete. Since the transition maps are surjective, the maps $M \to M_ n$ are surjective. Consider the inverse system of short exact sequences

defining $N_ n$. Since $M_ n = M_{n + 1}/I^ nM_{n + 1}$ the map $N_{n + 1} + I^ nM \to N_ n$ is surjective. Hence $N_{n + 1}/(N_{n + 1} \cap I^{n + 1}M) \to N_ n/(N_ n \cap I^ nM)$ is surjective. Taking the inverse limit of the short exact sequences

we obtain an exact sequence

Since $M$ is $I$-adically complete we conclude that $\mathop{\mathrm{lim}}\nolimits N_ n/(N_ n \cap I^ nM) = 0$ and hence by the surjectivity of the transition maps we get $N_ n/(N_ n \cap I^ nM) = 0$ for all $n$. Thus $M_ n = M/I^ nM$ as desired. $\square$

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