The Stacks project

Lemma 15.64.7. Let $R$ be a ring. Let $K^\bullet $ be a complex of $R$-modules. Let $m \in \mathbf{Z}$.

  1. If $H^ i(K^\bullet ) = 0$ for all $i \geq m$, then $K^\bullet $ is $m$-pseudo-coherent.

  2. If $H^ i(K^\bullet ) = 0$ for $i > m$ and $H^ m(K^\bullet )$ is a finite $R$-module, then $K^\bullet $ is $m$-pseudo-coherent.

  3. If $H^ i(K^\bullet ) = 0$ for $i > m + 1$, the module $H^{m + 1}(K^\bullet )$ is of finite presentation, and $H^ m(K^\bullet )$ is of finite type, then $K^\bullet $ is $m$-pseudo-coherent.

Proof. It suffices to prove (3). Set $M = H^{m + 1}(K^\bullet )$. Note that $\tau _{\geq m + 1}K^\bullet $ is quasi-isomorphic to $M[- m - 1]$. By Lemma 15.64.4 we see that $M[- m - 1]$ is $m$-pseudo-coherent. Since we have the distinguished triangle

\[ (\tau _{\leq m}K^\bullet , K^\bullet , \tau _{\geq m + 1}K^\bullet ) \]

(Derived Categories, Remark 13.12.4) by Lemma 15.64.2 it suffices to prove that $\tau _{\leq m}K^\bullet $ is pseudo-coherent. By assumption $H^ m(\tau _{\leq m}K^\bullet )$ is a finite type $R$-module. Hence we can find a finite free $R$-module $E$ and a map $E \to \mathop{\mathrm{Ker}}(d_ K^ m)$ such that the composition $E \to \mathop{\mathrm{Ker}}(d_ K^ m) \to H^ m(\tau _{\leq m}K^\bullet )$ is surjective. Then $E[-m] \to \tau _{\leq m}K^\bullet $ witnesses the fact that $\tau _{\leq m}K^\bullet $ is $m$-pseudo-coherent. $\square$


Comments (0)

There are also:

  • 11 comment(s) on Section 15.64: Pseudo-coherent modules, I

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Your email address will not be published.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 064W. Beware of the difference between the letter 'O' and the digit '0'.