Lemma 15.100.5 (09BC)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.100: Systems of modules
Lemma 15.100.5 (cite)

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Lemma 15.100.5. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$ , $N$ be finite $A$ -modules. Set $M_ n = M/I^ nM$ and $N_ n = N/I^ nN$ . If $M_ n \cong N_ n$ for all $n$ , then $M^\wedge \cong N^\wedge$ as $A^\wedge$ -modules.

Proof. By Lemma 15.100.4 the system $(\text{Isom}_ A(M_ n, N_ n))$ is Mittag-Leffler. By assumption each of the sets $\text{Isom}_ A(M_ n, N_ n)$ is nonempty. Hence $\mathop{\mathrm{lim}}\nolimits \text{Isom}_ A(M_ n, N_ n)$ is nonempty. Since $\mathop{\mathrm{lim}}\nolimits \text{Isom}_ A(M_ n, N_ n) = \text{Isom}_{A^\wedge }(M^\wedge , N^\wedge )$ we obtain an isomorphism. $\square$

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