Lemma 10.96.5 (0318)—The Stacks project

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 $n \geq 1$. $\square$

Comments (3)

Comment #9971 by Rankeya on February 07, 2025 at 12:04

The induced surjection $(M^\vee)^\vee \twoheadrightarrow M^\vee$ , which exists by the above proof even when $M^\vee$ is not assumed to be $I$ -adically complete, is a right inverse of the canonical map $M^\vee \to (M^\vee)^\vee$ . Thus, $M^\vee$ is always $I$ -adically separated. It may be nice to point this out explicitly somewhere in this section?

Comment #9972 by Rankeya on February 07, 2025 at 12:06

Sorry, $M^\vee$ (resp. $(M^\vee)^\vee$ ) should be $M^$ (resp. $(M^)^$ ) in my comment above.

Comment #9973 by Rankeya on February 07, 2025 at 12:50

Let's try this again: $M^\vee$ (resp. $(M^\vee)^\vee$ ) should be $M^\wedge$ (resp. $(M^\wedge)^\wedge$ ). Sorry for multiple comments. I couldn't figure out how to edit the first comment.

Comments (3)

Post a comment