The Stacks project

Lemma 48.18.2. In Situation 48.16.1 let $f : X \to Y$ be an ├ętale morphism of $\textit{FTS}_ S$. Then $f^! \cong f^*$ as functors on $D^+_\mathit{QCoh}(\mathcal{O}_ Y)$.

Proof. We are going to use that an ├ętale morphism is flat, syntomic, and a local complete intersection morphism (Morphisms, Lemma 29.34.10 and 29.34.12 and More on Morphisms, Lemma 37.54.8). By Lemma 48.17.8 it suffices to show $f^!\mathcal{O}_ Y = \mathcal{O}_ X$. By Lemma 48.17.10 we know that $f^!\mathcal{O}_ Y$ is an invertible module. Consider the commutative diagram

\[ \xymatrix{ X \times _ Y X \ar[r]_{p_2} \ar[d]_{p_1} & X \ar[d]^ f \\ X \ar[r]^ f & Y } \]

and the diagonal $\Delta : X \to X \times _ Y X$. Since $\Delta $ is an open immersion (by Morphisms, Lemmas 29.33.13 and 29.34.5), by Lemma 48.17.1 we have $\Delta ^! = \Delta ^*$. By Lemma 48.16.3 we have $\Delta ^! \circ p_1^! \circ f^! = f^!$. By Lemma 48.18.1 applied to the diagram we have $p_1^!\mathcal{O}_ X = p_2^*f^!\mathcal{O}_ Y$. Hence we conclude

\[ f^!\mathcal{O}_ X = \Delta ^!p_1^!f^!\mathcal{O}_ Y = \Delta ^*(p_1^*f^!\mathcal{O}_ Y \otimes p_1^!\mathcal{O}_ X) = \Delta ^*(p_2^*f^!\mathcal{O}_ Y \otimes p_1^*f^!\mathcal{O}_ Y) = (f^!\mathcal{O}_ Y)^{\otimes 2} \]

where in the second step we have used Lemma 48.17.8 once more. Thus $f^!\mathcal{O}_ Y = \mathcal{O}_ X$ as desired. $\square$


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Your email address will not be published.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