# The Stacks Project

## Tag 0AWP

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 $j < -\dim(A)$. $\square$

The code snippet corresponding to this tag is a part of the file duality.tex and is located in lines 5757–5768 (see updates for more information).

\begin{lemma}
\label{lemma-dualizing-module-proper-over-A}
Let $X/A$ with dualizing module $\omega_X$ be as in
Example \ref{example-proper-over-local}.
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
$$\mathcal{F} \longmapsto \Hom_A(H^d(X, \mathcal{F}), \omega_A)$$
on the category of coherent $\mathcal{O}_X$-modules.
\end{lemma}

\begin{proof}
We have
\begin{align*}
\Hom_X(\mathcal{F}, \omega_X)
& =
\Ext^{-\dim(X)}_X(\mathcal{F}, \omega_X^\bullet) \\
& =
\Hom_X(\mathcal{F}[\dim(X)], \omega_X^\bullet) \\
& =
\Hom_X(\mathcal{F}[\dim(X)], f^!(\omega_A^\bullet)) \\
& =
\Hom_S(Rf_*\mathcal{F}[\dim(X)], \omega_A^\bullet) \\
& =
\Hom_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 \ref{lemma-vanishing-good-dualizing} and
Derived Categories, Lemma \ref{derived-lemma-negative-exts}.
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 \ref{lemma-twisted-inverse-image} for
$f$, see Section \ref{section-duality}.
The final equality holds because $R^if_*\mathcal{F}$ is zero
for $i > d$ (Cohomology of Schemes, Lemma
\ref{coherent-lemma-higher-direct-images-zero-above-dimension-fibre})
and $H^j(\omega_A^\bullet)$ is zero for $j < -\dim(A)$.
\end{proof}

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