Lemma 54.8.4. Let (A, \mathfrak m, \kappa ) be a local normal Nagata domain of dimension 2 which defines a rational singularity. Let A \subset B be a local extension of domains with the same fraction field which is essentially of finite type such that \dim (B) = 2 and B normal. Then B defines a rational singularity.
Proof. Choose a finite type A-algebra C such that B = C_\mathfrak q for some prime \mathfrak q \subset C. After replacing C by the image of C in B we may assume that C is a domain with fraction field equal to the fraction field of A. Then we can choose a closed immersion \mathop{\mathrm{Spec}}(C) \to \mathbf{A}^ n_ A and take the closure in \mathbf{P}^ n_ A to conclude that B is isomorphic to \mathcal{O}_{X, x} for some closed point x \in X of a projective modification X \to \mathop{\mathrm{Spec}}(A). (Morphisms, Lemma 29.52.1, shows that \kappa (x) is finite over \kappa and then Morphisms, Lemma 29.20.2 shows that x is a closed point.) Let \nu : X^\nu \to X be the normalization. Since A is Nagata the morphism \nu is finite (Morphisms, Lemma 29.54.11). Thus X^\nu is projective over A by More on Morphisms, Lemma 37.50.2. Since B = \mathcal{O}_{X, x} is normal, we see that \mathcal{O}_{X, x} = (\nu _*\mathcal{O}_{X^\nu })_ x. Hence there is a unique point x^\nu \in X^\nu lying over x and \mathcal{O}_{X^\nu , x^\nu } = \mathcal{O}_{X, x}. Thus we may assume X is normal and projective over A. Let Y \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) = \mathop{\mathrm{Spec}}(B) be a modification with Y normal. We have to show that H^1(Y, \mathcal{O}_ Y) = 0. By Limits, Lemma 32.21.1 we can find a morphism of schemes g : X' \to X which is an isomorphism over X \setminus \{ x\} such that X' \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}) is isomorphic to Y. Then g is a modification as it is proper by Limits, Lemma 32.21.2. The local ring of X' at a point of x' is either isomorphic to the local ring of X at g(x') if g(x') \not= x and if g(x') = x, then the local ring of X' at x' is isomorphic to the local ring of Y at the corresponding point. Hence we see that X' is normal as both X and Y are normal. Thus H^1(X', \mathcal{O}_{X'}) = 0 by our assumption on A. By Lemma 54.8.1 we have R^1g_*\mathcal{O}_{X'} = 0. Clearly this means that H^1(Y, \mathcal{O}_ Y) = 0 as desired. \square
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