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

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

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: