Lemma 58.28.1. Let A be a Noetherian normal local domain of dimension 2. Assume A is Nagata, has a dualizing module \omega _ A, and has a resolution of singularities f : X \to \mathop{\mathrm{Spec}}(A). Let \omega _ X be as in Resolution of Surfaces, Remark 54.7.7. If \omega _ X \cong \mathcal{O}_ X(E) for some effective Cartier divisor E \subset X supported on the exceptional fibre, then A defines a rational singularity. If f is a minimal resolution, then E = 0.
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
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