The Stacks project

Lemma 77.11.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module.

  1. If $f$ is of finite presentation, $\mathcal{F}$ is an $\mathcal{O}_ X$-module of finite presentation, and $\mathcal{F}$ is pure relative to $Y$, then there exists a universal flattening $Y' \to Y$ of $\mathcal{F}$. Moreover $Y' \to Y$ is a monomorphism of finite presentation.

  2. If $f$ is of finite presentation and $X$ is pure relative to $Y$, then there exists a universal flattening $Y' \to Y$ of $X$. Moreover $Y' \to Y$ is a monomorphism of finite presentation.

  3. If $f$ is proper and of finite presentation and $\mathcal{F}$ is an $\mathcal{O}_ X$-module of finite presentation, then there exists a universal flattening $Y' \to Y$ of $\mathcal{F}$. Moreover $Y' \to Y$ is a monomorphism of finite presentation.

  4. If $f$ is proper and of finite presentation then there exists a universal flattening $Y' \to Y$ of $X$.

Proof. These statements follow immediately from Theorem 77.11.7 applied to $F_0 = F_{flat}$ and the fact that if $f$ is proper then $\mathcal{F}$ is automatically pure over the base, see Lemma 77.3.6. $\square$

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Your email address will not be published.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 0CX2. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 77.11.8 as pdf