The Stacks project

Lemma 38.17.1. Let $f : X \to S$ be a morphism of schemes which is of finite type. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ X$-module.

  1. If the support of $\mathcal{F}$ is proper over $S$, then $\mathcal{F}$ is universally pure relative to $S$.

  2. If $f$ is proper, then $\mathcal{F}$ is universally pure relative to $S$.

  3. If $f$ is proper, then $X$ is universally pure relative to $S$.

Proof. First we reduce (1) to (2). Namely, let $Z \subset X$ be the scheme theoretic support of $\mathcal{F}$. Let $i : Z \to X$ be the corresponding closed immersion and write $\mathcal{F} = i_*\mathcal{G}$ for some finite type quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{G}$, see Morphisms, Section 29.5. In case (1) $Z \to S$ is proper by assumption. Thus by Lemma 38.16.7 case (1) reduces to case (2).

Assume $f$ is proper. Let $(g : T \to S, t' \leadsto t, \xi )$ be an impurity of $\mathcal{F}$ above $s \in S$. Since $f$ is proper, it is universally closed. Hence $f_ T : X_ T \to T$ is closed. Since $f_ T(\xi ) = t'$ this implies that $t \in f(\overline{\{ \xi \} })$ which is a contradiction. $\square$


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Your email address will not be published.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 05K3. Beware of the difference between the letter 'O' and the digit '0'.