The Stacks project

Lemma 15.64.17. Let $R$ be a Noetherian ring. Then

  1. A complex of $R$-modules $K^\bullet $ is $m$-pseudo-coherent if and only if $K^\bullet \in D^{-}(R)$ and $H^ i(K^\bullet )$ is a finite $R$-module for $i \geq m$.

  2. A complex of $R$-modules $K^\bullet $ is pseudo-coherent if and only if $K^\bullet \in D^{-}(R)$ and $H^ i(K^\bullet )$ is a finite $R$-module for all $i$.

  3. An $R$-module is pseudo-coherent if and only if it is finite.

Proof. In Algebra, Lemma 10.71.1 we have seen that any finite $R$-module is pseudo-coherent. On the other hand, a pseudo-coherent module is finite, see Lemma 15.64.4. Hence (3) holds. Suppose that $K^\bullet $ is an $m$-pseudo-coherent complex. Then there exists a bounded complex of finite free $R$-modules $E^\bullet $ such that $H^ i(K^\bullet )$ is isomorphic to $H^ i(E^\bullet )$ for $i > m$ and such that $H^ m(K^\bullet )$ is a quotient of $H^ m(E^\bullet )$. Thus it is clear that each $H^ i(K^\bullet )$, $i \geq m$ is a finite module. The converse implication in (1) follows from Lemma 15.64.10 and part (3). Part (2) follows from (1) and Lemma 15.64.5. $\square$

Comments (0)

There are also:

  • 8 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Your email address will not be published.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 066E. Beware of the difference between the letter 'O' and the digit '0'.