**Proof.**
In this proof we will use the identification $D(A) = D_\mathit{QCoh}(\mathcal{O}_ S)$ to identify quasi-coherent $\mathcal{O}_ S$-modules with $A$-modules. Moreover, we may assume that $\omega _ A^\bullet $ is normalized, see Dualizing Complexes, Section 47.16. Since $X$ is a Noetherian normal $2$-dimensional scheme it is Cohen-Macaulay (Properties, Lemma 28.12.7). Thus $\omega _ X^\bullet = \omega _ X[2]$ (Duality for Schemes, Lemma 48.23.1 and the normalization in Duality for Schemes, Example 48.22.1). If the proposition is false, then we can find a nonzero map $R^1f_*\omega _ X \to \kappa $. In other words we obtain a nonzero map $\alpha : Rf_*\omega _ X^\bullet \to \kappa [1]$. Applying $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A^\bullet )$ we get a nonzero map

\[ \beta : \kappa [-1] \longrightarrow Rf_*\mathcal{O}_ X \]

which is impossible by Lemma 54.7.6. To see that $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A^\bullet )$ does what we said, first note that

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa [1], \omega _ A^\bullet ) = R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa , \omega _ A^\bullet )[-1] = \kappa [-1] \]

as $\omega _ A^\bullet $ is normalized and we have

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(Rf_*\omega _ X^\bullet , \omega _ A^\bullet ) = Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\omega _ X^\bullet , \omega _ X^\bullet ) = Rf_*\mathcal{O}_ X \]

The first equality by Duality for Schemes, Example 48.3.9 and the fact that $\omega _ X^\bullet = f^!\omega _ A^\bullet $ by construction, and the second equality because $\omega _ X^\bullet $ is a dualizing complex for $X$ (which goes back to Duality for Schemes, Lemma 48.17.7).
$\square$

## Comments (0)