The Stacks project

Lemma 10.96.8. Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. If $M$ is $J$-adically complete and $I$ is finitely generated, then $M$ is $I$-adically complete.

Proof. Assume $M$ is $J$-adically complete and $I$ is finitely generated. We have $\bigcap I^ nM = 0$ because $\bigcap J^ nM = 0$. By Lemma 10.96.7 it suffices to prove the surjectivity of $M \to \mathop{\mathrm{lim}}\nolimits M/I^ nM$ in case $I$ is generated by a single element. Say $I = (f)$. Let $x_ n \in M$ with $x_{n + 1} - x_ n \in f^ nM$. We have to show there exists an $x \in M$ such that $x_ n - x \in f^ nM$ for all $n$. As $x_{n + 1} - x_ n \in J^ nM$ and as $M$ is $J$-adically complete, there exists an element $x \in M$ such that $x_ n - x \in J^ nM$. Replacing $x_ n$ by $x_ n - x$ we may assume that $x_ n \in J^ nM$. To finish the proof we will show that this implies $x_ n \in I^ nM$. Namely, write $x_ n - x_{n + 1} = f^ nz_ n$. Then

\[ x_ n = f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots ) \]

The sum $z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots $ converges in $M$ as $f^ c \in J^ c$. The sum $f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots )$ converges in $M$ to $x_ n$ because the partial sums equal $x_ n - x_{n + c}$ and $x_{n + c} \in J^{n + c}M$. $\square$


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Your email address will not be published.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 090T. Beware of the difference between the letter 'O' and the digit '0'.