Lemma 54.7.6. In Situation 54.7.1 assume $X$ is normal and $A$ Nagata. Then

is zero. This uses $D(A) = D_\mathit{QCoh}(\mathcal{O}_ S)$ to think of $Rf_*\mathcal{O}_ X$ as an object of $D(A)$.

Lemma 54.7.6. In Situation 54.7.1 assume $X$ is normal and $A$ Nagata. Then

\[ \mathop{\mathrm{Hom}}\nolimits _{D(A)}(\kappa [-1], Rf_*\mathcal{O}_ X) \]

is zero. This uses $D(A) = D_\mathit{QCoh}(\mathcal{O}_ S)$ to think of $Rf_*\mathcal{O}_ X$ as an object of $D(A)$.

**Proof.**
By adjointness of $Rf_*$ and $Lf^*$ such a map is the same thing as a map $\alpha : Lf^*\kappa [-1] \to \mathcal{O}_ X$. Note that

\[ H^ i(Lf^*\kappa [-1]) = \left\{ \begin{matrix} 0
& \text{if}
& i > 1
\\ \mathcal{O}_{X_ s}
& \text{if}
& i = 1
\\ \text{some }\mathcal{O}_{X_ s}\text{-module}
& \text{if}
& i \leq 0
\end{matrix} \right. \]

Since $\mathop{\mathrm{Hom}}\nolimits (H^0(Lf^*\kappa [-1]), \mathcal{O}_ X) = 0$ as $\mathcal{O}_ X$ is torsion free, the spectral sequence for $\mathop{\mathrm{Ext}}\nolimits $ (Cohomology on Sites, Example 21.31.1) implies that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(Lf^*\kappa [-1], \mathcal{O}_ X)$ is equal to $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{O}_{X_ s}, \mathcal{O}_ X)$. We conclude that $\alpha : Lf^*\kappa [-1] \to \mathcal{O}_ X$ is given by an extension

\[ 0 \to \mathcal{O}_ X \to \mathcal{E} \to \mathcal{O}_{X_ s} \to 0 \]

By Lemma 54.7.5 the pullback of this extension via the surjection $\mathcal{O}_ X \to \mathcal{O}_{X_ s}$ is zero (since this pullback is clearly split over $f^{-1}(U)$). Thus $1 \in \mathcal{O}_{X_ s}$ lifts to a global section $s$ of $\mathcal{E}$. Multiplying $s$ by the ideal sheaf $\mathcal{I}$ of $X_ s$ we obtain an $\mathcal{O}_ X$-module map $c_ s : \mathcal{I} \to \mathcal{O}_ X$. Applying $f_*$ we obtain an $A$-linear map $f_*c_ s : \mathfrak m \to A$. Since $A$ is a Noetherian normal local domain this map is given by multiplication by an element $a \in A$. Changing $s$ into $s - a$ we find that $s$ is annihilated by $\mathcal{I}$ and the extension is trivial as desired. $\square$

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