The Stacks project

Lemma 37.57.8. Let $f : X \to S$ be a finite morphism of schemes. Then $f$ is pseudo-coherent if and only if $f_*\mathcal{O}_ X$ is pseudo-coherent as an $\mathcal{O}_ S$-module.

Proof. Translated into algebra this lemma says the following: If $R \to A$ is a finite ring map, then $R \to A$ is pseudo-coherent as a ring map (which means by definition that $A$ as an $A$-module is pseudo-coherent relative to $R$) if and only if $A$ is pseudo-coherent as an $R$-module. This follows from the more general More on Algebra, Lemma 15.81.5. $\square$


Comments (0)


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Your email address will not be published.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