The Stacks project

The definition of an algebraic space is fppf local.

Lemma 80.11.1. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Let $\{ S_ i \to S\} _{i \in I}$ be a covering of $(\mathit{Sch}/S)_{fppf}$. Assume that

  1. $F$ is a sheaf,

  2. each $F_ i = h_{S_ i} \times F$ is an algebraic space, and

  3. $\coprod _{i \in I} F_ i$ is an algebraic space (see Spaces, Lemma 65.8.4).

Then $F$ is an algebraic space.

Proof. Consider the morphism $\coprod F_ i \to F$. This is the base change of $\coprod S_ i \to S$ via $F \to S$. Hence it is representable, locally of finite presentation, flat and surjective by our definition of an fppf covering and Lemma 80.4.2. Thus Theorem 80.10.1 applies to show that $F$ is an algebraic space. $\square$

Comments (1)

Comment #1225 by David Corwin on

Suggested slogan: The definition of an algebraic space is fppf local

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Your email address will not be published.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 04SK. Beware of the difference between the letter 'O' and the digit '0'.