Lemma 48.15.7 (0BRT)—The Stacks project

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)

Comment #5903 by Qingyuan Jiang on January 26, 2021 at 11:07

It seems that the condition " $S$ being noetherian" could be relaxed by " $S$ being quasi-compact quasi-separated" -- It appears that the only place the noetherian condition is used is the second equality of the above equation; It is referred to Lem. 48.13.3 which is stated under noetherian condition. However we know this result holds without noetherian condition, see for example Prop. 2.1 in Lipman--Neeman's paper "Quasi-perfect scheme-maps and boundedness of the twisted inverse image functor"; So in particular it holds for $\Delta$ as long as $S$ is quasi-compact and quasi-separated.

P.S. By the same reference, the noetherian condition of both Lem. 48.13.1 and 48.13.3 could be replaced by "quasi-compact quasi-separated". P.P.S. Also in the second line below the equation $p$ should be $\mathrm{pr}_1$ .

Comment #5907 by Johan on January 26, 2021 at 19:32

You are very likely correct (I did not check everything you said). We mention in at least two places in this chapter that the generality we're working with could be improved, see the introduction to Section 48.13 and the footnote in Lemma 48.4.4. But it would require substantial additional work to do so. I recall that the Stacks project doesn't allow outside references in proofs.

For this particular lemma we could also use absolute Noetherian reduction and use that base change for the relative dualizing complex works well for flat proper morphisms of finite presentation to prove the lemma directly (with little extra work) in the generalization you mention. But somehow I doubt this would be worth it (see below).

Of course, this chapter tries to do just enough so (a) it is useful later and (b) the proofs don't get extremely long. Because the correct version of the comparison between relative dualizing complexes and modules of relative differentials requires a much, much longer discussion which one can find in the literature.

I will fix the typo later. Thanks for you comment!

Comment #5908 by Qingyuan Jiang on January 27, 2021 at 16:43

Hi Johan, I see, it makes sense. Thanks for the reply and the detailed explanation!

Comment #6104 by Johan on April 29, 2021 at 22:04

OK, the typo is now fixed here.

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