Taken from an unpublished note of Lenstra and de Smit.

Lemma 10.96.5. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Denote $K_ n = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ nM)$. Then $M^\wedge$ is $I$-adically complete if and only if $K_ n$ is equal to $I^ nM^\wedge$ for all $n \geq 1$.

Proof. The module $I^ n M^\wedge$ is contained in $K_ n$. Thus for each $n \geq 1$ there is a canonical exact sequence

$0 \to K_ n/I^ nM^\wedge \to M^\wedge /I^ nM^\wedge \to M/I^ nM \to 0.$

As $I^ nM^\wedge$ maps onto $I^ nM/I^{n + 1}M$ we see that $K_{n + 1} + I^ n M^\wedge = K_ n$. Thus the inverse system $\{ K_ n/I^ n M^\wedge \} _{n \geq 1}$ has surjective transition maps. By Lemma 10.87.1 we see that there is a short exact sequence

$0 \to \mathop{\mathrm{lim}}\nolimits _ n K_ n/I^ n M^\wedge \to (M^\wedge )^\wedge \to M^\wedge \to 0$

Hence $M^\wedge$ is complete if and only if $K_ n/I^ n M^\wedge = 0$ for all $n \geq 1$. $\square$

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