The Stacks project

Lemma 48.21.1. Let $S$ be a Noetherian scheme and let $\omega _ S^\bullet $ be a dualizing complex. Let $X$ be a scheme of finite type over $S$ and let $\omega _ X^\bullet $ be the dualizing complex normalized relative to $\omega _ S^\bullet $. If $x \in X$ is a closed point lying over a closed point $s$ of $S$, then $\omega _{X, x}^\bullet $ is a normalized dualizing complex over $\mathcal{O}_{X, x}$ provided that $\omega _{S, s}^\bullet $ is a normalized dualizing complex over $\mathcal{O}_{S, s}$.

Proof. We may replace $X$ by an affine neighbourhood of $x$, hence we may and do assume that $f : X \to S$ is separated. Then $\omega _ X^\bullet = f^!\omega _ S^\bullet $. We have to show that $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\kappa (x), \omega _{X, x}^\bullet )$ is sitting in degree $0$. Let $i_ x : x \to X$ denote the inclusion morphism which is a closed immersion as $x$ is a closed point. Hence $R\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\kappa (x), \omega _{X, x}^\bullet )$ represents $i_ x^!\omega _ X^\bullet $ by Lemma 48.17.4. Consider the commutative diagram

\[ \xymatrix{ x \ar[r]_{i_ x} \ar[d]_\pi & X \ar[d]^ f \\ s \ar[r]^{i_ s} & S } \]

By Morphisms, Lemma 29.20.3 the extension $\kappa (s) \subset \kappa (x)$ is finite and hence $\pi $ is a finite morphism. We conclude that

\[ i_ x^!\omega _ X^\bullet = i_ x^! f^! \omega _ S^\bullet = \pi ^! i_ s^! \omega _ S^\bullet \]

Thus if $\omega _{S, s}^\bullet $ is a normalized dualizing complex over $\mathcal{O}_{S, s}$, then $i_ s^!\omega _ S^\bullet = \kappa (s)[0]$ by the same reasoning as above. We have

\[ R\pi _*(\pi ^!(\kappa (s)[0])) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ s}(R\pi _*(\kappa (x)[0]), \kappa (s)[0]) = \widetilde{\mathop{\mathrm{Hom}}\nolimits _{\kappa (s)}(\kappa (x), \kappa (s))} \]

The first equality by Example 48.3.9 applied with $L = \kappa (x)[0]$. The second equality holds because $\pi _*$ is exact. Thus $\pi ^!(\kappa (s)[0])$ is supported in degree $0$ and we win. $\square$


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