Lemma 48.18.2. In Situation 48.16.1 let $f : X \to Y$ be an étale 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 étale morphism is flat, syntomic, and a local complete intersection morphism (Morphisms, Lemma 29.36.10 and 29.36.12 and More on Morphisms, Lemma 37.62.8). By Lemma 48.17.9 it suffices to show $f^!\mathcal{O}_ Y = \mathcal{O}_ X$. By Lemma 48.17.11 we know that $f^!\mathcal{O}_ Y$ is an invertible module. Consider the commutative diagram

and the diagonal $\Delta : X \to X \times _ Y X$. Since $\Delta $ is an open immersion (by Morphisms, Lemmas 29.35.13 and 29.36.5), 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

where in the second step we have used Lemma 48.17.9 once more. Thus $f^!\mathcal{O}_ Y = \mathcal{O}_ X$ as desired. $\square$

## 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Your email address will not be published.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 (2)

Comment #6629 by Jefferson Baudin on

Comment #6860 by Johan on

There are also: