Lemma 15.94.4. Let $A$ be a Noetherian ring and $I \subset A$ an ideal. Let $K$ be an object of $D(A)$ such that $H^ n(K)$ a finite $A$-module for all $n \in \mathbf{Z}$. Then the cohomology modules $H^ n(K^\wedge )$ of the derived completion are the $I$-adic completions of the cohomology modules $H^ n(K)$.

**Proof.**
The complex $\tau _{\leq m}K$ is pseudo-coherent for all $m$ by Lemma 15.64.17. Thus $\tau _{\leq m}K$ is represented by a bounded above complex $P^\bullet $ of finite free $A$-modules. Then $\tau _{\leq m}K \otimes _ A^\mathbf {L} A/I^ n = P^\bullet /I^ nP^\bullet $. Hence $(\tau _{\leq m}K)^\wedge = R\mathop{\mathrm{lim}}\nolimits P^\bullet /I^ nP^\bullet $ (Proposition 15.94.2) and since the $R\mathop{\mathrm{lim}}\nolimits $ is just given by termwise $\mathop{\mathrm{lim}}\nolimits $ (Lemma 15.87.1) and since $I$-adic completion is an exact functor on finite $A$-modules (Algebra, Lemma 10.97.2) we conclude the result holds for $\tau _{\leq m}K$. Hence the result holds for $K$ as derived completion has finite cohomological dimension, see Lemma 15.91.20.
$\square$

## 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Your email address will not be published.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).

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.

## Comments (0)