The Stacks project

Lemma 33.16.8. Let $f : X \to Y$ be a morphism of schemes locally of finite type over a base scheme $S$. Let $x \in X$ be a point. Set $y = f(x)$ and assume that $\kappa (y) = \kappa (x)$. Then the following are equivalent

  1. $\text{d}f : T_{X/S, x} \longrightarrow T_{Y/S, y}$ is injective, and

  2. $f$ is unramified at $x$.

Proof. The morphism $f$ is locally of finite type by Morphisms, Lemma 29.15.8. The map $\text{d}f$ is injective, if and only if $\Omega _{Y/S, y} \otimes \kappa (y) \to \Omega _{X/S, x} \otimes \kappa (x)$ is surjective (Lemma 33.16.6). The exact sequence $f^*\Omega _{Y/S} \to \Omega _{X/S} \to \Omega _{X/Y} \to 0$ (Morphisms, Lemma 29.32.9) then shows that this happens if and only if $\Omega _{X/Y, x} \otimes \kappa (x) = 0$. Hence the result follows from Morphisms, Lemma 29.35.14. $\square$

Comments (2)

Comment #2159 by Ariyan on

Typo: The last sentence "Hence the result by Morphisms" should be "Hence the result follows from.."

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Your email address will not be published.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 0B2G. Beware of the difference between the letter 'O' and the digit '0'.