The Stacks project

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. $\square$

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

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

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. $\square$

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

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

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

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 $X = X'$. $\square$

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

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

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

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

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

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

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

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. $\square$

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

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

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. $\square$