Lemma 48.15.7. Let $S$ be a Noetherian scheme. Let $f : X \to S$ be a smooth proper morphism of relative dimension $d$. Let $a$ be the right adjoint of $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ S)$ as in Lemma 48.3.1. Then there is an isomorphism

\[ \wedge ^ d \Omega _{X/S}[d] \longrightarrow a(\mathcal{O}_ S) \]

in $D(\mathcal{O}_ X)$.

**Proof.**
Set $\omega _{X/S}^\bullet = a(\mathcal{O}_ S)$ as in Remark 48.12.5. Let $c$ be the right adjoint of Lemma 48.3.1 for $\Delta : X \to X \times _ S X$. Because $\Delta $ is the diagonal of a smooth morphism it is a Koszul-regular immersion, see Divisors, Lemma 31.22.11. In particular, $\Delta $ is a perfect proper morphism (More on Morphisms, Lemma 37.61.7) and we obtain

\begin{align*} \mathcal{O}_ X & = c(L\text{pr}_1^*\omega _{X/S}^\bullet ) \\ & = L\Delta ^*(L\text{pr}_1^*\omega _{X/S}^\bullet ) \otimes _{\mathcal{O}_ X}^\mathbf {L} c(\mathcal{O}_{X \times _ S X}) \\ & = \omega _{X/S}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} c(\mathcal{O}_{X \times _ S X}) \\ & = \omega _{X/S}^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} \wedge ^ d(\mathcal{N}_\Delta )[-d] \end{align*}

The first equality is (48.12.8.1) because $\omega _{X/S}^\bullet = a(\mathcal{O}_ S)$. The second equality by Lemma 48.13.3. The third equality because $\text{pr}_1 \circ \Delta = \text{id}_ X$. The fourth equality by Lemma 48.15.6. Observe that $\wedge ^ d(\mathcal{N}_\Delta )$ is an invertible $\mathcal{O}_ X$-module. Hence $\wedge ^ d(\mathcal{N}_\Delta )[-d]$ is an invertible object of $D(\mathcal{O}_ X)$ and we conclude that $a(\mathcal{O}_ S) = \omega _{X/S}^\bullet = \wedge ^ d(\mathcal{C}_\Delta )[d]$. Since the conormal sheaf $\mathcal{C}_\Delta $ of $\Delta $ is $\Omega _{X/S}$ by Morphisms, Lemma 29.32.7 the proof is complete.
$\square$

## Comments (4)

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