The Stacks project

Lemma 15.63.19. Let $R$ be a coherent ring (Algebra, Definition 10.89.1). Let $K \in D^-(R)$. The following are equivalent

  1. $K$ is $m$-pseudo-coherent,

  2. $H^ m(K)$ is a finite $R$-module and $H^ i(K)$ is coherent for $i > m$, and

  3. $H^ m(K)$ is a finite $R$-module and $H^ i(K)$ is finitely presented for $i > m$.

Thus $K$ is pseudo-coherent if and only if $H^ i(K)$ is a coherent module for all $i$.

Proof. Recall that an $R$-module $M$ is coherent if and only if it is of finite presentation (Algebra, Lemma 10.89.4). This explains the equivalence of (2) and (3). If so and if we choose an exact sequence $0 \to N \to R^{\oplus m} \to M \to 0$, then $N$ is coherent by Algebra, Lemma 10.89.3. Thus in this case, repeating this procedure with $N$ we find a resolution

\[ \ldots \to R^{\oplus n} \to R^{\oplus m} \to M \to 0 \]

by finite free $R$-modules. In other words, $M$ is pseudo-coherent. The equivalence of (1) and (2) follows from this and Lemmas 15.63.11 and 15.63.4. The final assertion follows from the equivalence of (1) and (2) combined with Lemma 15.63.5. $\square$


Comments (0)

There are also:

  • 3 comment(s) on Section 15.63: Pseudo-coherent modules

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Your email address will not be published.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