The Stacks project

Lemma 76.48.7. Let $S$ be a scheme. A Koszul-regular immersion of algebraic spaces over $S$ is a local complete intersection morphism.

Proof. Let $i : X \to Y$ be a Koszul-regular immersion of algebraic spaces over $S$. By definition there exists a surjective ├ętale morphism $V \to Y$ where $V$ is a scheme such that $X \times _ Y V$ is a scheme and the base change $X \times _ Y V \to V$ is a Koszul-regular immersion of schemes. By More on Morphisms, Lemma 37.62.9 we see that $X \times _ Y V \to V$ is a local complete intersection morphism. From Definition 76.48.1 we conclude that $i$ is a local complete intersection morphism of algebraic spaces. $\square$


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Your email address will not be published.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