The Stacks project

Unramified morphisms are the same as formally unramified morphism that are locally of finite type.

Lemma 37.6.8. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

  1. The morphism $f$ is unramified (resp. G-unramified), and

  2. the morphism $f$ is locally of finite type (resp. locally of finite presentation) and formally unramified.

Proof. Use Lemma 37.6.7 and Morphisms, Lemma 29.35.2. $\square$

Comments (1)

Comment #1284 by on

Suggested slogan: Unramified morphisms are the same as formally unramified morphism that are locally of finite type.

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Your email address will not be published.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 02HE. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 37.6.8 as pdf