The Stacks project

Proof. There is a trace map $Rf_*\omega _ X \to \omega _ A$, see Duality for Schemes, Section 48.7. By Grauert-Riemenschneider (Resolution of Surfaces, Proposition 54.7.8) we have $R^1f_*\omega _ X = 0$. Thus the trace map is a map $f_*\omega _ X \to \omega _ A$. Then we can consider

where the first map comes from the map $\mathcal{O}_ X \to \mathcal{O}_ X(E) = \omega _ X$ which is assumed to exist in the statement of the lemma. The composition is an isomorphism by Divisors, Lemma 31.2.11 as it is an isomorphism over the punctured spectrum of $A$ (by the assumption in the lemma and the fact that $f$ is an isomorphism over the punctured spectrum) and $A$ and $\omega _ A$ are $A$-modules of depth $2$ (by Algebra, Lemma 10.157.4 and Dualizing Complexes, Lemma 47.17.5). Hence $f_*\omega _ X \to \omega _ A$ is surjective whence an isomorphism. Thus $Rf_*\omega _ X = \omega _ A$ which by duality implies $Rf_*\mathcal{O}_ X = \mathcal{O}_{\mathop{\mathrm{Spec}}(A)}$. Whence $H^1(X, \mathcal{O}_ X) = 0$ which implies that $A$ defines a rational singularity (see discussion in Resolution of Surfaces, Section 54.8 in particular Lemmas 54.8.7 and 54.8.1). If $f$ is minimal, then $E = 0$ because the map $f^*\omega _ A \to \omega _ X$ is surjective by a repeated application of Resolution of Surfaces, Lemma 54.9.7 and $\omega _ A \cong A$ as we've seen above. $\square$