The Stacks project

54.9 Rational singularities

In this section we reduce from rational singular points to Gorenstein rational singular points. See [Lipman-rational] and [Mattuck].

Situation 54.9.1. Here $(A, \mathfrak m, \kappa )$ be a local normal Nagata domain of dimension $2$ which defines a rational singularity. Let $s$ be the closed point of $S = \mathop{\mathrm{Spec}}(A)$ and $U = S \setminus \{ s\} $. Let $f : X \to S$ be a modification with $X$ normal. We denote $C_1, \ldots , C_ r$ the irreducible components of the special fibre $X_ s$ of $f$.

Lemma 54.9.2. In Situation 54.9.1. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then

  1. $H^ p(X, \mathcal{F}) = 0$ for $p \not\in \{ 0, 1\} $, and

  2. $H^1(X, \mathcal{F}) = 0$ if $\mathcal{F}$ is globally generated.

Proof. Part (1) follows from Cohomology of Schemes, Lemma 30.20.9. If $\mathcal{F}$ is globally generated, then there is a surjection $\bigoplus _{i \in I} \mathcal{O}_ X \to \mathcal{F}$. By part (1) and the long exact sequence of cohomology this induces a surjection on $H^1$. Since $H^1(X, \mathcal{O}_ X) = 0$ as $S$ has a rational singularity, and since $H^1(X, -)$ commutes with direct sums (Cohomology, Lemma 20.19.1) we conclude. $\square$

Lemma 54.9.3. In Situation 54.9.1 assume $E = X_ s$ is an effective Cartier divisor. Let $\mathcal{I}$ be the ideal sheaf of $E$. Then $H^0(X, \mathcal{I}^ n) = \mathfrak m^ n$ and $H^1(X, \mathcal{I}^ n) = 0$.

Proof. We have $H^0(X, \mathcal{O}_ X) = A$, see discussion following Situation 54.7.1. Then $\mathfrak m \subset H^0(X, \mathcal{I}) \subset H^0(X, \mathcal{O}_ X)$. The second inclusion is not an equality as $X_ s \not= \emptyset $. Thus $H^0(X, \mathcal{I}) = \mathfrak m$. As $\mathcal{I}^ n = \mathfrak m^ n\mathcal{O}_ X$ our Lemma 54.9.2 shows that $H^1(X, \mathcal{I}^ n) = 0$.

Choose generators $x_1, \ldots , x_{\mu + 1}$ of $\mathfrak m$. These define global sections of $\mathcal{I}$ which generate it. Hence a short exact sequence

\[ 0 \to \mathcal{F} \to \mathcal{O}_ X^{\oplus \mu + 1} \to \mathcal{I} \to 0 \]

Then $\mathcal{F}$ is a finite locally free $\mathcal{O}_ X$-module of rank $\mu $ and $\mathcal{F} \otimes \mathcal{I}$ is globally generated by Constructions, Lemma 27.13.9. Hence $\mathcal{F} \otimes \mathcal{I}^ n$ is globally generated for all $n \geq 1$. Thus for $n \geq 2$ we can consider the exact sequence

\[ 0 \to \mathcal{F} \otimes \mathcal{I}^{n - 1} \to (\mathcal{I}^{n - 1})^{\oplus \mu + 1} \to \mathcal{I}^ n \to 0 \]

Applying the long exact sequence of cohomology using that $H^1(X, \mathcal{F} \otimes \mathcal{I}^{n - 1}) = 0$ by Lemma 54.9.2 we obtain that every element of $H^0(X, \mathcal{I}^ n)$ is of the form $\sum x_ i a_ i$ for some $a_ i \in H^0(X, \mathcal{I}^{n - 1})$. This shows that $H^0(X, \mathcal{I}^ n) = \mathfrak m^ n$ by induction. $\square$

Lemma 54.9.4. In Situation 54.9.1 the blowup of $\mathop{\mathrm{Spec}}(A)$ in $\mathfrak m$ is normal.

Proof. Let $X' \to \mathop{\mathrm{Spec}}(A)$ be the blowup, in other words

\[ X' = \text{Proj}(A \oplus \mathfrak m \oplus \mathfrak m^2 \oplus \ldots ). \]

is the Proj of the Rees algebra. This in particular shows that $X'$ is integral and that $X' \to \mathop{\mathrm{Spec}}(A)$ is a projective modification. Let $X$ be the normalization of $X'$. Since $A$ is Nagata, we see that $\nu : X \to X'$ is finite (Morphisms, Lemma 29.54.11). Let $E' \subset X'$ be the exceptional divisor and let $E \subset X$ be the inverse image. Let $\mathcal{I}' \subset \mathcal{O}_{X'}$ and $\mathcal{I} \subset \mathcal{O}_ X$ be their ideal sheaves. Recall that $\mathcal{I}' = \mathcal{O}_{X'}(1)$ (Divisors, Lemma 31.32.13). Observe that $\mathcal{I} = \nu ^*\mathcal{I}'$ and that $E$ is an effective Cartier divisor (Divisors, Lemma 31.13.13). We are trying to show that $\nu $ is an isomorphism. As $\nu $ is finite, it suffices to show that $\mathcal{O}_{X'} \to \nu _*\mathcal{O}_ X$ is an isomorphism. If not, then we can find an $n \geq 0$ such that

\[ H^0(X', (\mathcal{I}')^ n) \not= H^0(X', (\nu _*\mathcal{O}_ X) \otimes (\mathcal{I}')^ n) \]

for example because we can recover quasi-coherent $\mathcal{O}_{X'}$-modules from their associated graded modules, see Properties, Lemma 28.28.3. By the projection formula we have

\[ H^0(X', (\nu _*\mathcal{O}_ X) \otimes (\mathcal{I}')^ n) = H^0(X, \nu ^*(\mathcal{I}')^ n) = H^0(X, \mathcal{I}^ n) = \mathfrak m^ n \]

the last equality by Lemma 54.9.3. On the other hand, there is clearly an injection $\mathfrak m^ n \to H^0(X', (\mathcal{I}')^ n)$. Since $H^0(X', (\mathcal{I}')^ n)$ is torsion free we conclude equality holds for all $n$, hence $X = X'$. $\square$

Lemma 54.9.5. In Situation 54.9.1. Let $X$ be the blowup of $\mathop{\mathrm{Spec}}(A)$ in $\mathfrak m$. Let $E \subset X$ be the exceptional divisor. With $\mathcal{O}_ X(1) = \mathcal{I}$ as usual and $\mathcal{O}_ E(1) = \mathcal{O}_ X(1)|_ E$ we have

  1. $E$ is a proper Cohen-Macaulay curve over $\kappa $.

  2. $\mathcal{O}_ E(1)$ is very ample

  3. $\deg (\mathcal{O}_ E(1)) \geq 1$ and equality holds only if $A$ is a regular local ring,

  4. $H^1(E, \mathcal{O}_ E(n)) = 0$ for $n \geq 0$, and

  5. $H^0(E, \mathcal{O}_ E(n)) = \mathfrak m^ n/\mathfrak m^{n + 1}$ for $n \geq 0$.

Proof. Since $\mathcal{O}_ X(1)$ is very ample by construction, we see that its restriction to the special fibre $E$ is very ample as well. By Lemma 54.9.4 the scheme $X$ is normal. Then $E$ is Cohen-Macaulay by Divisors, Lemma 31.15.6. Lemma 54.9.3 applies and we obtain (4) and (5) from the exact sequences

\[ 0 \to \mathcal{I}^{n + 1} \to \mathcal{I}^ n \to i_*\mathcal{O}_ E(n) \to 0 \]

and the long exact cohomology sequence. In particular, we see that

\[ \deg (\mathcal{O}_ E(1)) = \chi (E, \mathcal{O}_ E(1)) - \chi (E, \mathcal{O}_ E) = \dim (\mathfrak m/\mathfrak m^2) - 1 \]

by Varieties, Definition 33.44.1. Thus (3) follows as well. $\square$

Lemma 54.9.6. In Situation 54.9.1 assume $A$ has a dualizing complex $\omega _ A^\bullet $. With $\omega _ X$ the dualizing module of $X$, the trace map $H^0(X, \omega _ X) \to \omega _ A$ is an isomorphism and consequently there is a canonical map $f^*\omega _ A \to \omega _ X$.

Proof. By Grauert-Riemenschneider (Proposition 54.7.8) we see that $Rf_*\omega _ X = f_*\omega _ X$. By duality we have a short exact sequence

\[ 0 \to f_*\omega _ X \to \omega _ A \to \mathop{\mathrm{Ext}}\nolimits ^2_ A(R^1f_*\mathcal{O}_ X, \omega _ A) \to 0 \]

(for example see proof of Lemma 54.8.8) and since $A$ defines a rational singularity we obtain $f_*\omega _ X = \omega _ A$. $\square$

Lemma 54.9.7. In Situation 54.9.1 assume $A$ has a dualizing complex $\omega _ A^\bullet $ and is not regular. Let $X$ be the blowup of $\mathop{\mathrm{Spec}}(A)$ in $\mathfrak m$ with exceptional divisor $E \subset X$. Let $\omega _ X$ be the dualizing module of $X$. Then

  1. $\omega _ E = \omega _ X|_ E \otimes \mathcal{O}_ E(-1)$,

  2. $H^1(X, \omega _ X(n)) = 0$ for $n \geq 0$,

  3. the map $f^*\omega _ A \to \omega _ X$ of Lemma 54.9.6 is surjective.

Proof. We will use the results of Lemma 54.9.5 without further mention. Observe that $\omega _ E = \omega _ X|_ E \otimes \mathcal{O}_ E(-1)$ by Duality for Schemes, Lemmas 48.14.2 and 48.9.7. Thus $\omega _ X|_ E = \omega _ E(1)$. Consider the short exact sequences

\[ 0 \to \omega _ X(n + 1) \to \omega _ X(n) \to i_*\omega _ E(n + 1) \to 0 \]

By Algebraic Curves, Lemma 53.6.4 we see that $H^1(E, \omega _ E(n + 1)) = 0$ for $n \geq 0$. Thus we see that the maps

\[ \ldots \to H^1(X, \omega _ X(2)) \to H^1(X, \omega _ X(1)) \to H^1(X, \omega _ X) \]

are surjective. Since $H^1(X, \omega _ X(n))$ is zero for $n \gg 0$ (Cohomology of Schemes, Lemma 30.16.2) we conclude that (2) holds.

By Algebraic Curves, Lemma 53.6.7 we see that $\omega _ X|_ E = \omega _ E \otimes \mathcal{O}_ E(1)$ is globally generated. Since we seen above that $H^1(X, \omega _ X(1)) = 0$ the map $H^0(X, \omega _ X) \to H^0(E, \omega _ X|_ E)$ is surjective. We conclude that $\omega _ X$ is globally generated hence (3) holds because $\Gamma (X, \omega _ X) = \omega _ A$ is used in Lemma 54.9.6 to define the map. $\square$

Lemma 54.9.8. Let $(A, \mathfrak m, \kappa )$ be a local normal Nagata domain of dimension $2$ which defines a rational singularity. Assume $A$ has a dualizing complex. Then there exists a finite sequence of blowups in singular closed points

\[ X = X_ n \to X_{n - 1} \to \ldots \to X_1 \to X_0 = \mathop{\mathrm{Spec}}(A) \]

such that $X_ i$ is normal for each $i$ and such that the dualizing sheaf $\omega _ X$ of $X$ is an invertible $\mathcal{O}_ X$-module.

Proof. The dualizing module $\omega _ A$ is a finite $A$-module whose stalk at the generic point is invertible. Namely, $\omega _ A \otimes _ A K$ is a dualizing module for the fraction field $K$ of $A$, hence has rank $1$. Thus there exists a blowup $b : Y \to \mathop{\mathrm{Spec}}(A)$ such that the strict transform of $\omega _ A$ with respect to $b$ is an invertible $\mathcal{O}_ Y$-module, see Divisors, Lemma 31.35.3. By Lemma 54.5.3 we can choose a sequence of normalized blowups

\[ X_ n \to X_{n - 1} \to \ldots \to X_1 \to \mathop{\mathrm{Spec}}(A) \]

such that $X_ n$ dominates $Y$. By Lemma 54.9.4 and arguing by induction each $X_ i \to X_{i - 1}$ is simply a blowing up.

We claim that $\omega _{X_ n}$ is invertible. Since $\omega _{X_ n}$ is a coherent $\mathcal{O}_{X_ n}$-module, it suffices to see its stalks are invertible modules. If $x \in X_ n$ is a regular point, then this is clear from the fact that regular schemes are Gorenstein (Dualizing Complexes, Lemma 47.21.3). If $x$ is a singular point of $X_ n$, then each of the images $x_ i \in X_ i$ of $x$ is a singular point (because the blowup of a regular point is regular by Lemma 54.3.2). Consider the canonical map $f_ n^*\omega _ A \to \omega _{X_ n}$ of Lemma 54.9.6. For each $i$ the morphism $X_{i + 1} \to X_ i$ is either a blowup of $x_ i$ or an isomorphism at $x_ i$. Since $x_ i$ is always a singular point, it follows from Lemma 54.9.7 and induction that the maps $f_ i^*\omega _ A \to \omega _{X_ i}$ is always surjective on stalks at $x_ i$. Hence

\[ (f_ n^*\omega _ A)_ x \longrightarrow \omega _{X_ n, x} \]

is surjective. On the other hand, by our choice of $b$ the quotient of $f_ n^*\omega _ A$ by its torsion submodule is an invertible module $\mathcal{L}$. Moreover, the dualizing module is torsion free (Duality for Schemes, Lemma 48.22.3). It follows that $\mathcal{L}_ x \cong \omega _{X_ n, x}$ and the proof is complete. $\square$

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