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

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

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

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

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

Comment #6629 by Jefferson Baudin on

In the last equation, the first term should be $f^!\mathcal{O}_Y$ instead of $f^!\mathcal{O}_X$

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