The Stacks project

Lemma 34.9.11. Let $T$ be a scheme. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a family of morphisms of schemes with target $T$. Assume that

  1. each $f_ i$ is flat, and

  2. every affine scheme $Z$ and morphism $h : Z \to T$ there exists a standard fpqc covering $\{ Z_ j \to Z\} _{j = 1, \ldots , n}$ which refines the family $\{ T_ i \times _ T Z \to Z\} _{i \in I}$.

Then $\{ f_ i : T_ i \to T\} _{i \in I}$ is an fpqc covering of $T$.

Proof. Let $T = \bigcup U_\alpha $ be an affine open covering. For each $\alpha $ the pullback family $\{ T_ i \times _ T U_\alpha \to U_\alpha \} $ can be refined by a standard fpqc covering, hence is an fpqc covering by Lemma 34.9.4. As $\{ U_\alpha \to T\} $ is an fpqc covering we conclude that $\{ T_ i \to T\} $ is an fpqc covering by Lemma 34.9.5. $\square$


Comments (2)

Comment #3203 by Dario WeiƟmann on

Typo: 'a fpqc covering'

There are also:

  • 6 comment(s) on Section 34.9: The fpqc topology

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Your email address will not be published.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 03LB. Beware of the difference between the letter 'O' and the digit '0'.