Lemma 107.19.1. Let $X \to S$ be a family of curves with Cohen-Macaulay fibres equidimensional of dimension $1$ (Lemma 107.8.2). Then $\omega _{X/S}^\bullet = \omega _{X/S}[1]$ where $\omega _{X/S}$ is a pseudo-coherent $\mathcal{O}_ X$-module flat over $S$ whose formation commutes with arbitrary base change.

Proof. We urge the reader to deduce this directly from the discussion above of what happens after base change to a field. Our proof will use a somewhat cumbersome reduction to the Noetherian schemes case.

Once we show $\omega _{X/S}^\bullet = \omega _{X/S}[1]$ with $\omega _{X/S}$ flat over $S$, the statement on base change will follow as we already know that formation of $\omega _{X/S}^\bullet$ commutes with arbitrary base change. Moreover, the pseudo-coherence will be automatic as $\omega _{X/S}^\bullet$ is pseudo-coherent by definition. Vanishing of the other cohomology sheaves and flatness may be checked étale locally. Thus we may assume $f : X \to S$ is a morphism of schemes with $S$ affine (see discussion above). Write $S = \mathop{\mathrm{lim}}\nolimits S_ i$ as a cofiltered limit of affine schemes $S_ i$ of finite type over $\mathbf{Z}$. Since $\mathcal{C}\! \mathit{urves}^{CM, 1}$ is locally of finite presentation over $\mathbf{Z}$ (as an open substack of $\mathcal{C}\! \mathit{urves}$, see Lemmas 107.8.2 and 107.5.3), we can find an $i$ and a family of curves $X_ i \to S_ i$ whose pullback is $X \to S$ (Limits of Stacks, Lemma 100.3.5). After increasing $i$ if necessary we may assume $X_ i$ is a scheme, see Limits of Spaces, Lemma 68.5.11. Since formation of $\omega _{X/S}^\bullet$ commutes with arbitrary base change, we may replace $S$ by $S_ i$. Doing so we may and do assume $S_ i$ is Noetherian. Then $f$ is clearly a Cohen-Macaulay morphism (More on Morphisms, Definition 37.20.1) by our assumption on the fibres. Also then $\omega _{X/S}^\bullet = f^!\mathcal{O}_ S$ by the very construction of $f^!$ in Duality for Schemes, Section 48.16. Thus the lemma by Duality for Schemes, Lemma 48.23.3. $\square$

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