The Stacks project

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. $\square$