The Stacks project

Lemma 15.93.7. Let $f \in A$ be an element of a ring. Set $J = \bigcap f^ nA$. Let $M$ be an $A$-module derived complete with respect to $f$. Then $JM' = 0$ where $M' = \mathop{\mathrm{Ker}}(M \to \mathop{\mathrm{lim}}\nolimits M/f^ nM)$. In particular, if $A$ is derived complete then $J$ is an ideal of square zero.

Proof. Take $x \in M'$ and $g \in J$. For every $n \geq 1$ we may write $x = f^ n x_ n$. Since $g$ is in $f^ nA$ we see that the element $y_ n = gx_ n$ in $M'$ is independent of the choice of $x_ n$. In particular, we may take $x_ n = fx_{n + 1}$ and we find that $y_ n = fy_{n + 1}$. Thus we obtain a map $A_ f \to M$ sending $1/f^ n$ to $y_ n$. This map has to be zero as $M$ is derived complete (Lemma 15.91.1) and hence $y_ n = 0$ for all $n$. Since $gx = gfx_1 = fy_1$ this completes the proof. $\square$

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Your email address will not be published.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 0G3G. Beware of the difference between the letter 'O' and the digit '0'.