The Stacks project

Lemma 15.91.20. Let $A$ be a ring and let $I \subset A$ be an ideal which can be generated by $r$ elements. Then derived completion has finite cohomological dimension:

  1. Let $K \to L$ be a morphism in $D(A)$ such that $H^ i(K) \to H^ i(L)$ is an isomorphism for $i \geq 1$ and surjective for $i = 0$. Then $H^ i(K^\wedge ) \to H^ i(L^\wedge )$ is an isomorphism for $i \geq 1$ and surjective for $i = 0$.

  2. Let $K \to L$ be a morphism of $D(A)$ such that $H^ i(K) \to H^ i(L)$ is an isomorphism for $i \leq -1$ and injective for $i = 0$. Then $H^ i(K^\wedge ) \to H^ i(L^\wedge )$ is an isomorphism for $i \leq -r - 1$ and injective for $i = -r$.

Proof. Say $I$ is generated by $f_1, \ldots , f_ r$. For any $K \in D(A)$ by Lemma 15.91.18 we have $K^\wedge = R\mathop{\mathrm{lim}}\nolimits K \otimes _ A^\mathbf {L} K_ n$ where $K_ n$ is the Koszul complex on $f_1^ n, \ldots , f_ r^ n$ and hence we obtain a short exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{i - 1}(K \otimes _ A^\mathbf {L} K_ n) \to H^ i(K^\wedge ) \to \mathop{\mathrm{lim}}\nolimits H^ i(K \otimes _ A^\mathbf {L} K_ n) \to 0 \]

by Lemma 15.87.4.

Proof of (1). Pick a distinguished triangle $K \to L \to C \to K[1]$. Then $H^ i(C) = 0$ for $i \geq 0$. Since $K_ n$ is sitting in degrees $\leq 0$ we see that $H^ i(C \otimes _ A^\mathbf {L} K_ n) = 0$ for $i \geq 0$ and that $H^{-1}(C \otimes _ A^\mathbf {L} K_ n) = H^{-1}(C) \otimes _ A A/(f_1^ n, \ldots , f_ r^ n)$ is a system with surjective transition maps. The displayed equation above shows that $H^ i(C^\wedge ) = 0$ for $i \geq 0$. Applying the distinguished triangle $K^\wedge \to L^\wedge \to C^\wedge \to K^\wedge [1]$ we get (1).

Proof of (2). Pick a distinguished triangle $K \to L \to C \to K[1]$. Then $H^ i(C) = 0$ for $i < 0$. Since $K_ n$ is sitting in degrees $-r, \ldots , 0$ we see that $H^ i(C \otimes _ A^\mathbf {L} K_ n) = 0$ for $i < -r$. The displayed equation above shows that $H^ i(C^\wedge ) = 0$ for $i < r$. Applying the distinguished triangle $K^\wedge \to L^\wedge \to C^\wedge \to K^\wedge [1]$ we get (2). $\square$


Comments (0)

There are also:

  • 14 comment(s) on Section 15.91: Derived Completion

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Your email address will not be published.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