The Stacks project

Lemma 36.32.3. Let $f : X \to S$ be a flat, proper morphism of finite presentation. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation, flat over $S$. Fix $i, r \in \mathbf{Z}$. Then there exists an open subscheme $U \subset S$ with the following property: A morphism $T \to S$ factors through $U$ if and only if $Rf_{T, *}\mathcal{F}_ T$ is isomorphic to a finite locally free module of rank $r$ placed in degree $i$.

Proof. By cohomology and base change (more precisely by Lemma 36.30.4) the object $K = Rf_*\mathcal{F}$ is a perfect object of the derived category of $S$ whose formation commutes with arbitrary base change. Thus this lemma follows immediately from Lemma 36.31.3. $\square$

Comments (1)

Comment #7876 by qyk on

Is the condition that is flat redundant? We can use (2) in \href{}{Lemma 0B91}, which does not need flatness of .

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Your email address will not be published.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 0B9S. Beware of the difference between the letter 'O' and the digit '0'.