The Stacks project

Taken from an unpublished note of Lenstra and de Smit.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 $n \geq 1$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0318. Beware of the difference between the letter 'O' and the digit '0'.