Lemma 15.93.9. 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. $\square$
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Your email address will not be published.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)
There are also: