The Stacks project

Proof. We will use the observations made following Situation 54.7.1 without further mention. As $X$ is normal and $g$ is dominant and birational, we have $g_*\mathcal{O}_{X'} = \mathcal{O}_ X$, see for example More on Morphisms, Lemma 37.53.6. Since the fibres of $g$ have dimension $\leq 1$, we have $R^ pg_*\mathcal{O}_{X'} = 0$ for $p > 1$, see for example Cohomology of Schemes, Lemma 30.20.9. The support of $R^1g_*\mathcal{O}_{X'}$ is contained in the set of points of $X$ where the fibres of $g'$ have dimension $\geq 1$. Thus it is contained in the set of images of those irreducible components $C' \subset X'_ s$ which map to points of $X_ s$ which is a finite set of closed points (recall that $X'_ s \to X_ s$ is a morphism of proper $1$-dimensional schemes over $\kappa $). Then $R^1g_*\mathcal{O}_{X'}$ is globally generated by Cohomology of Schemes, Lemma 30.9.10. Using the morphism $f : X \to S$ and the references above we find that $H^ p(X, \mathcal{F}) = 0$ for $p > 1$ for any coherent $\mathcal{O}_ X$-module $\mathcal{F}$. Hence the short exact sequence of the lemma is a consequence of the Leray spectral sequence for $g$ and $\mathcal{O}_{X'}$, see Cohomology, Lemma 20.13.4. $\square$

Proof. By the short exact sequence $ 0 \to \mathcal{O}_ X \xrightarrow {a} \mathcal{O}_ X \to \mathcal{O}_ Z \to 0 $ we see that

Here $N[a] = \{ n \in N \mid an = 0\} $ for an $A$-module $N$. Thus if $a$ divides $b$, then $M_{X, a} \subset M_{X, b}$. Suppose that for some $c \in A$ the modules $M_{X, c}$ have bounded length. Then for every $X$ we have an exact sequence

where the second arrow is given by multiplication by $c$. Hence we see that $M_{X, c^2}$ has bounded length as well. Thus it suffices to find a $c \in A$ for which the lemma is true such that $a$ divides $c^ n$ for some $n > 0$. By More on Algebra, Lemma 15.125.6 we may assume $A/(a)$ is a reduced ring.

Assume that $A/(a)$ is reduced. Let $A/(a) \subset B$ be the normalization of $A/(a)$ in its quotient ring. Because $A$ is Nagata, we see that $\mathop{\mathrm{Coker}}(A \to B)$ is finite. We claim the length of this finite module is a bound. To see this, consider $f : X \to \mathop{\mathrm{Spec}}(A)$ as in the lemma and let $Z' \subset Z$ be the scheme theoretic closure of $Z \cap f^{-1}(U)$. Then $Z' \to \mathop{\mathrm{Spec}}(A/(a))$ is finite for example by Varieties, Lemma 33.17.2. Hence $Z' = \mathop{\mathrm{Spec}}(B')$ with $A/(a) \subset B' \subset B$. On the other hand, we claim the map

is injective. Namely, if $s \in H^0(Z, \mathcal{O}_ Z)$ is in the kernel, then the restriction of $s$ to $f^{-1}(U) \cap Z$ is zero. Hence the image of $s$ in $H^1(X, \mathcal{O}_ X)$ vanishes in $H^1(f^{-1}(U), \mathcal{O}_ X)$. By Lemma 54.7.5 we see that $s$ comes from an element $\tilde s$ of $A$. But by assumption $\tilde s$ maps to zero in $B'$ which implies that $s = 0$. Putting everything together we see that $M_{X, a}$ is a subquotient of $B'/A$, namely not every element of $B'$ extends to a global section of $\mathcal{O}_ Z$, but in any case the length of $M_{X, a}$ is bounded by the length of $B/A$. $\square$

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. $\square$

Proof. We choose a modification $X \to \mathop{\mathrm{Spec}}(A)$ with $X$ normal which maximizes the length of $H^1(X, \mathcal{O}_ X)$. By Lemma 54.8.1 for any further modification $g : X' \to X$ with $X'$ normal we have $R^1g_*\mathcal{O}_{X'} = 0$ and $H^1(X, \mathcal{O}_ X) = H^1(X', \mathcal{O}_{X'})$.

Let $x \in X$ be a closed point. We will show that $\mathcal{O}_{X, x}$ defines a rational singularity. Let $Y \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ 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. By maximality we have $R^1g_*\mathcal{O}_{X'} = 0$ (see first paragraph). Clearly this means that $H^1(Y, \mathcal{O}_ Y) = 0$ as desired.

The conclusion is that we've found one normal modification $X$ of $\mathop{\mathrm{Spec}}(A)$ such that the local rings of $X$ at closed points all define rational singularities. Then we choose a sequence of normalized blowups $X_ n \to \ldots \to X_1 \to \mathop{\mathrm{Spec}}(A)$ such that $X_ n$ dominates $X$, see Lemma 54.5.3. For a closed point $x' \in X_ n$ mapping to $x \in X$ we can apply Lemma 54.8.4 to the ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{X_ n, x'}$ to see that $\mathcal{O}_{X_ n, x'}$ defines a rational singularity. $\square$

Proof. Let $n$ be the degree of the fraction field extension $L/K$. Let $\text{Trace}_{L/K} : L \to K$ be the trace. Since the extension is finite separable the trace pairing $(h, g) \mapsto \text{Trace}_{L/K}(fg)$ is a nondegenerate bilinear form on $L$ over $K$. See Fields, Lemma 9.20.7. Pick $b_1, \ldots , b_ n \in B$ which form a basis of $L$ over $K$. By the above $d = \det (\text{Trace}_{L/K}(b_ ib_ j)) \in A$ is nonzero.

Let $Y \to \mathop{\mathrm{Spec}}(B)$ be a modification with $Y$ normal. We can find a $U$-admissible blowup $X'$ of $\mathop{\mathrm{Spec}}(A)$ such that the strict transform $Y'$ of $Y$ is finite over $X'$, see More on Flatness, Lemma 38.31.2. Picture

After replacing $X'$ and $Y'$ by their normalizations we may assume that $X'$ and $Y'$ are normal modifications of $\mathop{\mathrm{Spec}}(A)$ and $\mathop{\mathrm{Spec}}(B)$. In this way we reduce to the case where there exists a commutative diagram

with $X$ and $Y$ normal modifications of $\mathop{\mathrm{Spec}}(A)$ and $\mathop{\mathrm{Spec}}(B)$ and $\pi $ finite.

The trace map on $L$ over $K$ extends to a map of $\mathcal{O}_ X$-modules $\text{Trace} : \pi _*\mathcal{O}_ Y \to \mathcal{O}_ X$. Consider the map

This map is injective (because it is injective in the generic point) and there is a map

whose composition with $\Phi $ has matrix $\text{Trace}(b_ ib_ j)$. Hence the cokernel of $\Phi $ is annihilated by $d$. Thus we see that we have an exact sequence

Since the right hand side is bounded by assumption, it suffices to show that the $d$-torsion in $H^1(Y, \mathcal{O}_ Y)$ is bounded. This is the content of Lemma 54.8.2 and (54.8.2.1). $\square$

Proof. (The assumption that $A$ be Nagata is not necessary for this proof, but we've only defined the notion of rational singularity in the case of Nagata $2$-dimensional normal local domains.) Let $X \to \mathop{\mathrm{Spec}}(A)$ be a modification with $X$ normal. By Lemma 54.4.2 we can dominate $X$ by a scheme $X_ n$ which is the last in a sequence

of blowing ups in closed points. By Lemma 54.3.2 the schemes $X_ i$ are regular, in particular normal (Algebra, Lemma 10.157.5). By Lemma 54.8.1 we have $H^1(X, \mathcal{O}_ X) \subset H^1(X_ n, \mathcal{O}_{X_ n})$. Thus it suffices to prove $H^1(X_ n, \mathcal{O}_{X_ n}) = 0$. Using Lemma 54.8.1 again, we see that it suffices to prove $R^1(X_ i \to X_{i - 1})_*\mathcal{O}_{X_ i} = 0$ for $i = 1, \ldots , n$. This follows from Lemma 54.3.4. $\square$

Proof. For $X \to \mathop{\mathrm{Spec}}(A)$ as in the statement we have to bound $H^1(X, \mathcal{O}_ X)$. Let $\omega _ X$ be the dualizing module of $X$ as in the statement of Grauert-Riemenschneider (Proposition 54.7.8). The trace map is the map $Rf_*\omega _ X \to \omega _ A$ described in Duality for Schemes, Section 48.7. By Grauert-Riemenschneider we have $Rf_*\omega _ X = f_*\omega _ X$ thus the trace map indeed produces a map $\Gamma (X, \omega _ X) \to \omega _ A$. By duality we have $Rf_*\omega _ X = R\mathop{\mathrm{Hom}}\nolimits _ A(Rf_*\mathcal{O}_ X, \omega _ A)$ (this uses that $\omega _ X[2]$ is the dualizing complex on $X$ normalized relative to $\omega _ A[2]$, see Duality for Schemes, Lemma 48.20.9 or more directly Section 48.19 or even more directly Example 48.3.9). The distinguished triangle

is transformed by $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A)$ into the short exact sequence

(and $\mathop{\mathrm{Ext}}\nolimits _ A^ i(R^1f_*\mathcal{O}_ X, \omega _ A) = 0$ for $i \not= 2$; this will follow from the discussion below as well). Since $R^1f_*\mathcal{O}_ X$ is supported in $\{ \mathfrak m\} $, the local duality theorem tells us that

is the Matlis dual of $R^1f_*\mathcal{O}_ X$ (and the other ext groups are zero), see Dualizing Complexes, Lemma 47.18.4. By the equivalence of categories inherent in Matlis duality (Dualizing Complexes, Proposition 47.7.8), if $R^1f_*\mathcal{O}_ X$ is not annihilated by $d$, then neither is the $\mathop{\mathrm{Ext}}\nolimits ^2$ above. Hence we see that $H^1(X, \mathcal{O}_ X)$ is annihilated by $d$. Thus the required boundedness follows from Lemma 54.8.2 and (54.8.2.1). $\square$

Proof. Observe that $A$ is Gorenstein (Dualizing Complexes, Lemma 47.21.3) and hence the invertible module $\omega _ A$ does indeed serve as a dualizing module. Moreover, any $X$ as in the lemma has an invertible dualizing module $\omega _ X$ as $X$ is regular (hence Gorenstein) and proper over $A$, see Remark 54.7.7 and Lemma 54.3.2. Suppose we have constructed the map $\varphi _ X : (\Omega ^ r_{X/A_0})^{**} \to \omega _ X$ and suppose that $b : X' \to X$ is a blowup in a closed point. Set $\Omega ^ r_ X = (\Omega ^ r_{X/A_0})^{**}$ and $\Omega ^ r_{X'} = (\Omega ^ r_{X'/A_0})^{**}$. Since $\omega _{X'} = b^!(\omega _ X)$ a map $\Omega ^ r_{X'} \to \omega _{X'}$ is the same thing as a map $Rb_*(\Omega ^ r_{X'}) \to \omega _ X$. See discussion in Remark 54.7.7 and Duality for Schemes, Section 48.19. Thus in turn it suffices to produce a map

The sheaves $\Omega ^ r_{X'}$ and $\Omega ^ r_ X$ are invertible, see Divisors, Lemma 31.12.15. Consider the exact sequence

A local calculation shows that $\Omega _{X'/X}$ is isomorphic to an invertible module on the exceptional divisor $E$, see Lemma 54.3.6. It follows that either

see Divisors, Lemma 31.15.13. (The second possibility never happens in characteristic zero, but can happen in characteristic $p$.) In both cases we see that $R^1b_*(\Omega ^ r_{X'}) = 0$ and $b_*(\Omega ^ r_{X'}) = \Omega ^ r_ X$ by Lemma 54.3.4. $\square$

Proof. We have $A = k[[x, y]]$. Write $L = K[x]/(x^ p - f)$ for some $f \in A$ and denote $g \in B$ the congruence class of $x$, i.e., the element such that $g^ p = f$. By Algebra, Lemma 10.158.2 we see that $\text{d}f$ is nonzero in $\Omega _{K/\mathbf{F}_ p}$. By More on Algebra, Lemma 15.46.5 there exists a subfield $k^ p \subset k' \subset k$ with $p^ e = [k : k'] < \infty $ such that $\text{d}f$ is nonzero in $\Omega _{K/K_0}$ where $K_0$ is the fraction field of $A_0 = k'[[x^ p, y^ p]] \subset A$. Then

is finite free of rank $e + 2$. Set $\omega _ A = \Omega ^{e + 2}_{A/A_0}$. Consider the canonical map

of Lemma 54.2.4. By duality this determines a map

Claim: the cokernel of $c$ is annihilated by a nonzero element of $B$.

Since $\text{d}f$ is nonzero in $\Omega _{A/A_0}$ we can find $\eta _1, \ldots , \eta _{e + 1} \in \Omega _{A/A_0}$ such that $\theta = \eta _1 \wedge \ldots \wedge \eta _{e + 1} \wedge \text{d}f$ is nonzero in $\omega _ A = \Omega ^{e + 2}_{A/A_0}$. To prove the claim we will construct elements $\omega _ i$ of $\Omega ^{e + 2}_{B/A_0}$, $i = 0, \ldots , p - 1$ which are mapped to $\varphi _ i \in \omega _ B = \mathop{\mathrm{Hom}}\nolimits _ A(B, \omega _ A)$ with $\varphi _ i(g^ j) = \delta _{ij}\theta $ for $j = 0, \ldots , p - 1$. Since $\{ 1, g, \ldots , g^{p - 1}\} $ is a basis for $L/K$ this proves the claim. We set $\eta = \eta _1 \wedge \ldots \wedge \eta _{e + 1}$ so that $\theta = \eta \wedge \text{d}f$. Set $\omega _ i = \eta \wedge g^{p - 1 - i}\text{d}g$. Then by construction we have

by the explicit description of the trace map in Lemma 54.2.2.

Let $Y \to \mathop{\mathrm{Spec}}(B)$ be a normal modification. Exactly as in the proof of Lemma 54.8.6 we can reduce to the case where $Y$ is finite over a modification $X$ of $\mathop{\mathrm{Spec}}(A)$. By Lemma 54.4.2 we may even assume $X \to \mathop{\mathrm{Spec}}(A)$ is the result of a sequence of blowing ups in closed points. Picture:

We may apply Lemma 54.2.4 to $\pi $ and we obtain the first arrow in

and the second arrow is from Lemma 54.8.9 (because $f$ is a sequence of blowups in closed points). By duality for the finite morphism $\pi $ this corresponds to a map

extending the map $c$ above. Hence we see that the image of $\Gamma (Y, \omega _ Y) \to \omega _ B$ contains the image of $c$. By our claim we see that the cokernel is annihilated by a fixed nonzero element of $B$. We conclude by Lemma 54.8.8. $\square$