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 $n \geq 1$.
Taken from an unpublished note of Lenstra and de Smit.
Proof.
The module $I^ n M^\wedge $ is contained in $K_ n$. Thus for each $n \geq 1$ there is a canonical exact sequence
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
Hence $M^\wedge $ is complete if and only if $K_ n/I^ n M^\wedge = 0$ for all $n \geq 1$.
$\square$
\[ 0 \to K_ n/I^ nM^\wedge \to M^\wedge /I^ nM^\wedge \to M/I^ nM \to 0. \]
\[ 0 \to \mathop{\mathrm{lim}}\nolimits _ n K_ n/I^ n M^\wedge \to (M^\wedge )^\wedge \to M^\wedge \to 0 \]
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)