Lemma 101.36.5. An immersion is unramified.

**Proof.**
Let $j : \mathcal{Z} \to \mathcal{X}$ be an immersion of algebraic stacks. Since $j$ is representable, it is DM by Lemma 101.4.3. On the other hand, if $X \to \mathcal{X}$ is a smooth and surjective morphism where $X$ is a scheme, then $Z = \mathcal{Z} \times _\mathcal {X} X$ is a locally closed subscheme of $X$. Hence $Z \to X$ is unramified (Morphisms, Lemmas 29.35.7 and 29.35.8) and we conclude that $j$ is unramified from the definition.
$\square$

## 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Your email address will not be published.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.

## Comments (0)