The Stacks project

Lemma 15.93.8 (Bhatt). Let $I$ be a finitely generated ideal in a ring $A$. Let $M$ be a derived complete $A$-module. If $M$ is an $I$-power torsion module, then $I^ nM = 0$ for some $n$.

Proof. Say $I = (f_1, \ldots , f_ r)$. It suffices to show that for each $i$ there is an $n_ i$ such that $f_ i^{n_ i}M = 0$. Hence we may assume that $I = (f)$ is a principal ideal. Let $B = \mathbf{Z}[x] \to A$ be the ring map sending $x$ to $f$. By Lemma 15.91.23 we see that $M$ is derived complete as a $B$-module with respect to the ideal $(x)$. After replacing $A$ by $B$, we may assume that $f$ is a nonzerodivisor in $A$.

Assume $I = (f)$ with $f \in A$ a nonzerodivisor. According to Example 15.93.3 there exists a short exact sequence

\[ 0 \to K \xrightarrow {u} L \to M \to 0 \]

where $K$ and $L$ are $I$-adically complete $A$-modules whose $f$-torsion is zero1. Consider $K$ and $L$ as topological modules with the $I$-adic topology. Then $u$ is continuous. Let

\[ L_ n = \{ x \in L \mid f^ n x \in u(K)\} \]

Since $M$ is $f$-power torsion we see that $L = \bigcup L_ n$. Let $N_ n$ be the closure of $L_ n$ in $L$. By Lemma 15.36.4 we see that $N_ n$ is open in $L$ for some $n$. Fix such an $n$. Since $f^{n + m} : L \to L$ is a continuous open map, and since $f^{n + m} L_ n \subset u(f^ m K)$ we conclude that the closure of $u(f^ mK)$ is open for all $m \geq 1$. Thus by Lemma 15.36.5 we conclude that $u$ is open. Hence $f^ tL \subset \mathop{\mathrm{Im}}(u)$ for some $t$ and we conclude that $f^ t$ annihilates $M$ as desired. $\square$

[1] For the proof it is enough to show that there exists a sequence $K \xrightarrow {u} L \to M \to 0$ where $K$ and $L$ are $I$-adically complete $A$-modules. This can be shown by choosing a presentation $F_1 \to F_0 \to M \to 0$ with $F_ i$ free and then setting $K$ and $L$ equal to the $f$-adic completions of $F_1$ and $F_0$. Namely, as $f$ is a nonzerodivisor these completions will be the derived completions and the sequence will remain exact.

Comments (0)

There are also:

  • 2 comment(s) on Section 15.93: Derived completion for a principal ideal

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Your email address will not be published.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