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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)