The Stacks project

Proof. Let $X \to Y$ be a smooth morphism of algebraic spaces. By definition this means there exists a diagram as in Lemma 67.22.1 with $h$ smooth and surjective vertical arrow $a$. By Morphisms, Lemma 29.34.8 $h$ is flat. Hence $X \to Y$ is flat by definition. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 67.37: Smooth morphisms

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Your email address will not be published.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



View Lemma 67.37.7 as pdf