Example 15.93.3. Let $A$ be a ring. Let $f \in A$ be a nonzerodivisor. An example to keep in mind is $A = \mathbf{Z}_ p$ and $f = p$. Let $M$ be an $A$-module. Claim: $M$ is derived complete with respect to $f$ if and only if there exists a short exact sequence

$0 \to K \to L \to M \to 0$

where $K, L$ are $f$-adically complete modules whose $f$-torsion is zero. Namely, if there is a such a short exact sequence, then

$M \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A) = (K/f^ nK \to L/f^ nL)$

because $f$ is a nonzerodivisor on $K$ and $L$ and we conclude that $R\mathop{\mathrm{lim}}\nolimits (M \otimes _ A^\mathbf {L} (A \xrightarrow {f^ n} A))$ is quasi-isomorphic to $K \to L$, i.e., $M$. This shows that $M$ is derived complete by Lemma 15.91.17. Conversely, suppose that $M$ is derived complete. Choose a surjection $F \to M$ where $F$ is a free $A$-module. Since $f$ is a nonzerodivisor on $F$ the derived completion of $F$ is $L = \mathop{\mathrm{lim}}\nolimits F/f^ nF$. Note that $L$ is $f$-torsion free: if $(x_ n)$ with $x_ n \in F$ represents an element $\xi$ of $L$ and $f\xi = 0$, then $x_ n = x_{n + 1} + f^ nz_ n$ and $fx_ n = f^ ny_ n$ for some $z_ n, y_ n \in F$. Then $f^ n y_ n = fx_ n = fx_{n + 1} + f^{n + 1}z_ n = f^{n + 1}y_{n + 1} + f^{n + 1}z_ n$ and since $f$ is a nonzerodivisor on $F$ we see that $y_ n \in fF$ which implies that $x_ n \in f^ nF$, i.e., $\xi = 0$. Since $L$ is the derived completion, the universal property gives a map $L \to M$ factoring $F \to M$. Let $K = \mathop{\mathrm{Ker}}(L \to M)$ be the kernel. Again $K$ is $f$-torsion free, hence the derived completion of $K$ is $\mathop{\mathrm{lim}}\nolimits K/f^ nK$. On the other hand, both $M$ and $L$ are derived complete, hence $K$ is too by Lemma 15.91.6. It follows that $K = \mathop{\mathrm{lim}}\nolimits K/f^ nK$ and the claim is proved.

## 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Your email address will not be published.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 09AT. Beware of the difference between the letter 'O' and the digit '0'.