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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