
Lemma 46.23.5. Let $X/A$ with dualizing module $\omega _ X$ be as in Example 46.23.1. Let $d = \dim (X_ s)$ be the dimension of the closed fibre. If $\dim (X) = d + \dim (A)$, then the dualizing module $\omega _ X$ represents the functor

$\mathcal{F} \longmapsto \mathop{\mathrm{Hom}}\nolimits _ A(H^ d(X, \mathcal{F}), \omega _ A)$

on the category of coherent $\mathcal{O}_ X$-modules.

Proof. We have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, \omega _ X) & = \mathop{\mathrm{Ext}}\nolimits ^{-\dim (X)}_ X(\mathcal{F}, \omega _ X^\bullet ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}[\dim (X)], \omega _ X^\bullet ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}[\dim (X)], f^!(\omega _ A^\bullet )) \\ & = \mathop{\mathrm{Hom}}\nolimits _ S(Rf_*\mathcal{F}[\dim (X)], \omega _ A^\bullet ) \\ & = \mathop{\mathrm{Hom}}\nolimits _ A(H^ d(X, \mathcal{F}), \omega _ A) \end{align*}

The first equality because $H^ i(\omega _ X^\bullet ) = 0$ for $i < -\dim (X)$, see Lemma 46.23.4 and Derived Categories, Lemma 13.27.3. The second equality is follows from the definition of Ext groups. The third equality is our choice of $\omega _ X^\bullet$. The fourth equality holds because $f^!$ is the right adjoint of Lemma 46.3.1 for $f$, see Section 46.20. The final equality holds because $R^ if_*\mathcal{F}$ is zero for $i > d$ (Cohomology of Schemes, Lemma 29.20.9) and $H^ j(\omega _ A^\bullet )$ is zero for $j < -\dim (A)$. $\square$

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