The Stacks project

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51.7 Vanishing

In this section we will often work in the following setting. Recall that a modification is a proper birational morphism between integral schemes (Morphisms, Definition 28.48.11).

Situation 51.7.1. Here $(A, \mathfrak m, \kappa )$ be a local Noetherian normal domain of dimension $2$. 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. We denote $C_1, \ldots , C_ r$ the irreducible components of the special fibre $X_ s$ of $f$.

By Varieties, Lemma 32.17.3 the morphism $f$ defines an isomorphism $f^{-1}(U) \to U$. The special fibre $X_ s$ is proper over $\mathop{\mathrm{Spec}}(\kappa )$ and has dimension at most $1$ by Varieties, Lemma 32.19.3. By Stein factorization (More on Morphisms, Lemma 36.46.6) we have $f_*\mathcal{O}_ X = \mathcal{O}_ S$ and the special fibre $X_ s$ is geometrically connected over $\kappa $. If $X_ s$ has dimension $0$, then $f$ is finite (More on Morphisms, Lemma 36.38.5) and hence an isomorphism (Morphisms, Lemma 28.51.8). We will discard this uninteresting case and we conclude that $\dim (C_ i) = 1$ for $i = 1, \ldots , r$.

Lemma 51.7.2. In Situation 51.7.1 there exists a $U$-admissible blowup $X' \to S$ which dominates $X$.

Proof. This is a special case of More on Flatness, Lemma 37.31.4. $\square$

Lemma 51.7.3. In Situation 51.7.1 there exists a nonzero $f \in \mathfrak m$ such that for every $i = 1, \ldots , r$ there exist

  1. a closed point $x_ i \in C_ i$ with $x_ i \not\in C_ j$ for $j \not= i$,

  2. a factorization $f = g_ i f_ i$ of $f$ in $\mathcal{O}_{X, x_ i}$ such that $g_ i \in \mathfrak m_{x_ i}$ maps to a nonzero element of $\mathcal{O}_{C_ i, x_ i}$.

Proof. We will use the observations made following Situation 51.7.1 without further mention. Pick a closed point $x_ i \in C_ i$ which is not in $C_ j$ for $j \not= i$. Pick $g_ i \in \mathfrak m_{x_ i}$ which maps to a nonzero element of $\mathcal{O}_{C_ i, x_ i}$. Since the fraction field of $A$ is the fraction field of $\mathcal{O}_{X_ i, x_ i}$ we can write $g_ i = a_ i/b_ i$ for some $a_ i, b_ i \in A$. Take $f = \prod a_ i$. $\square$

Lemma 51.7.4. In Situation 51.7.1 assume $X$ is normal. Let $Z \subset X$ be a nonempty effective Cartier divisor such that $Z \subset X_ s$ set theoretically. Then the conormal sheaf of $Z$ is not trivial. More precisely, there exists an $i$ such that $C_ i \subset Z$ and $\deg (\mathcal{C}_{Z/X}|_{C_ i}) > 0$.

Proof. We will use the observations made following Situation 51.7.1 without further mention. Let $f$ be a function as in Lemma 51.7.3. Let $\xi _ i \in C_ i$ be the generic point. Let $\mathcal{O}_ i$ be the local ring of $X$ at $\xi _ i$. Then $\mathcal{O}_ i$ is a discrete valuation ring. Let $e_ i$ be the valuation of $f$ in $\mathcal{O}_ i$, so $e_ i > 0$. Let $h_ i \in \mathcal{O}_ i$ be a local equation for $Z$ and let $d_ i$ be its valuation. Then $d_ i \geq 0$. Choose and fix $i$ with $d_ i/e_ i$ maximal (then $d_ i > 0$ as $Z$ is not empty). Replace $f$ by $f^{d_ i}$ and $Z$ by $e_ iZ$. This is permissible, by the relation $\mathcal{O}_ X(e_ i Z) = \mathcal{O}_ X(Z)^{\otimes e_ i}$, the relation between the conormal sheaf and $\mathcal{O}_ X(Z)$ (see Divisors, Lemmas 30.14.4 and 30.14.2, and since the degree gets multiplied by $e_ i$, see Varieties, Lemma 32.43.7. Let $\mathcal{I}$ be the ideal sheaf of $Z$ so that $\mathcal{C}_{Z/X} = \mathcal{I}|_ Z$. Consider the image $\overline{f}$ of $f$ in $\Gamma (Z, \mathcal{O}_ Z)$. By our choices above we see that $\overline{f}$ vanishes in the generic points of irreducible components of $Z$ (these are all generic points of $C_ j$ as $Z$ is contained in the special fibre). On the other hand, $Z$ is $(S_1)$ by Divisors, Lemma 30.15.6. Thus the scheme $Z$ has no embedded associated points and we conclude that $\overline{f} = 0$ (Divisors, Lemmas 30.4.3 and 30.5.6). Hence $f$ is a global section of $\mathcal{I}$ which generates $\mathcal{I}_{\xi _ i}$ by construction. Thus the image $s_ i$ of $f$ in $\Gamma (C_ i, \mathcal{I}|_{C_ i})$ is nonzero. However, our choice of $f$ guarantees that $s_ i$ has a zero at $x_ i$. Hence the degree of $\mathcal{I}|_{C_ i}$ is $>0$ by Varieties, Lemma 32.43.12. $\square$

Lemma 51.7.5. In Situation 51.7.1 assume $X$ is normal and $A$ Nagata. The map

\[ H^1(X, \mathcal{O}_ X) \longrightarrow H^1(f^{-1}(U), \mathcal{O}_ X) \]

is injective.

Proof. Let $0 \to \mathcal{O}_ X \to \mathcal{E} \to \mathcal{O}_ X \to 0$ be the extension corresponding to a nontrivial element $\xi $ of $H^1(X, \mathcal{O}_ X)$ (Cohomology, Lemma 20.6.1). Let $\pi : P = \mathbf{P}(\mathcal{E}) \to X$ be the projective bundle associated to $\mathcal{E}$. The surjection $\mathcal{E} \to \mathcal{O}_ X$ defines a section $\sigma : X \to P$ whose conormal sheaf is isomorphic to $\mathcal{O}_ X$ (Divisors, Lemma 30.31.6). If the restriction of $\xi $ to $f^{-1}(U)$ is trivial, then we get a map $\mathcal{E}|_{f^{-1}(U)} \to \mathcal{O}_{f^{-1}(U)}$ splitting the injection $\mathcal{O}_ X \to \mathcal{E}$. This defines a second section $\sigma ' : f^{-1}(U) \to P$ disjoint from $\sigma $. Since $\xi $ is nontrivial we conclude that $\sigma '$ cannot extend to all of $X$ and be disjoint from $\sigma $. Let $X' \subset P$ be the scheme theoretic image of $\sigma '$ (Morphisms, Definition 28.6.2). Picture

\[ \xymatrix{ & X' \ar[rd]_ g \ar[r] & P \ar[d]_\pi \\ f^{-1}(U) \ar[ru]_{\sigma '} \ar[rr] & & X \ar@/_/[u]_\sigma } \]

The morphism $P \setminus \sigma (X) \to X$ is affine. If $X' \cap \sigma (X) = \emptyset $, then $X' \to X$ is both affine and proper, hence finite (Morphisms, Lemma 28.42.11), hence an isomorphism (as $X$ is normal, see Morphisms, Lemma 28.51.8). This is impossible as mentioned above.

Let $X^\nu $ be the normalization of $X'$. Since $A$ is Nagata, we see that $X^\nu \to X'$ is finite (Morphisms, Lemmas 28.51.10 and 28.17.2). Let $Z \subset X^\nu $ be the pullback of the effective Cartier divisor $\sigma (X) \subset P$. By the above we see that $Z$ is not empty and is contained in the closed fibre of $X^\nu \to S$. Since $P \to X$ is smooth, we see that $\sigma (X)$ is an effective Cartier divisor (Divisors, Lemma 30.22.7). Hence $Z \subset X^\nu $ is an effective Cartier divisor too. Since the conormal sheaf of $\sigma (X)$ in $P$ is $\mathcal{O}_ X$, the conormal sheaf of $Z$ in $X^\nu $ (which is a priori invertible) is $\mathcal{O}_ Z$ by Morphisms, Lemma 28.30.4. This is impossible by Lemma 51.7.4 and the proof is complete. $\square$

Lemma 51.7.6. In Situation 51.7.1 assume $X$ is normal and $A$ Nagata. Then

\[ \mathop{\mathrm{Hom}}\nolimits _{D(A)}(\kappa [-1], Rf_*\mathcal{O}_ X) \]

is zero. This uses $D(A) = D_\mathit{QCoh}(\mathcal{O}_ S)$ to think of $Rf_*\mathcal{O}_ X$ as an object of $D(A)$.

Proof. By adjointness of $Rf_*$ and $Lf^*$ such a map is the same thing as a map $\alpha : Lf^*\kappa [-1] \to \mathcal{O}_ X$. Note that

\[ H^ i(Lf^*\kappa [-1]) = \left\{ \begin{matrix} 0 & \text{if} & i > 1 \\ \mathcal{O}_{X_ s} & \text{if} & i = 1 \\ \text{some }\mathcal{O}_{X_ s}\text{-module} & \text{if} & i \leq 0 \end{matrix} \right. \]

Since $\mathop{\mathrm{Hom}}\nolimits (H^0(Lf^*\kappa [-1]), \mathcal{O}_ X) = 0$ as $\mathcal{O}_ X$ is torsion free, the spectral sequence for $\mathop{\mathrm{Ext}}\nolimits $ (Cohomology on Sites, Example 21.26.1) implies that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(Lf^*\kappa [-1], \mathcal{O}_ X)$ is equal to $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathcal{O}_{X_ s}, \mathcal{O}_ X)$. We conclude that $\alpha : Lf^*\kappa [-1] \to \mathcal{O}_ X$ is given by an extension

\[ 0 \to \mathcal{O}_ X \to \mathcal{E} \to \mathcal{O}_{X_ s} \to 0 \]

By Lemma 51.7.5 the pullback of this extension via the surjection $\mathcal{O}_ X \to \mathcal{O}_{X_ s}$ is zero (since this pullback is clearly split over $f^{-1}(U)$). Thus $1 \in \mathcal{O}_{X_ s}$ lifts to a global section $s$ of $\mathcal{E}$. Multiplying $s$ by the ideal sheaf $\mathcal{I}$ of $X_ s$ we obtain an $\mathcal{O}_ X$-module map $c_ s : \mathcal{I} \to \mathcal{O}_ X$. Applying $f_*$ we obtain an $A$-linear map $f_*c_ s : \mathfrak m \to A$. Since $A$ is a Noetherian normal local domain this map is given by multiplication by an element $a \in A$. Changing $s$ into $s - a$ we find that $s$ is annihilated by $\mathcal{I}$ and the extension is trivial as desired. $\square$

Remark 51.7.7. Let $X$ be an integral Noetherian normal scheme of dimension $2$. In this case the following are equivalent

  1. $X$ has a dualizing complex $\omega _ X^\bullet $,

  2. there is a coherent $\mathcal{O}_ X$-module $\omega _ X$ such that $\omega _ X[n]$ is a dualizing complex, where $n$ can be any integer.

This follows from the fact that $X$ is Cohen-Macaulay (Properties, Lemma 27.12.7) and Duality for Schemes, Lemma 46.24.1. In this situation we will say that $\omega _ X$ is a dualizing module in accordance with Duality for Schemes, Section 46.23. In particular, when $A$ is a Noetherian normal local domain of dimension $2$, then we say $A$ has a dualizing module $\omega _ A$ if the above is true. In this case, if $X \to \mathop{\mathrm{Spec}}(A)$ is a normal modification, then $X$ has a dualizing module too, see Duality for Schemes, Example 46.23.1. In this situation we always denote $\omega _ X$ the dualizing module normalized with respect to $\omega _ A$, i.e., such that $\omega _ X[2]$ is the dualizing complex normalized relative to $\omega _ A[2]$. See Duality for Schemes, Section 46.21.

The Grauert-Riemenschneider vanishing of the next proposition is a formal consequence of Lemma 51.7.6 and the general theory of duality.

Proof. In this proof we will use the identification $D(A) = D_\mathit{QCoh}(\mathcal{O}_ S)$ to identify quasi-coherent $\mathcal{O}_ S$-modules with $A$-modules. Moreover, we may assume that $\omega _ A^\bullet $ is normalized, see Dualizing Complexes, Section 45.16. Since $X$ is a Noetherian normal $2$-dimensional scheme it is Cohen-Macaulay (Properties, Lemma 27.12.7). Thus $\omega _ X^\bullet = \omega _ X[2]$ (Duality for Schemes, Lemma 46.24.1 and the normalization in Duality for Schemes, Example 46.23.1). If the proposition is false, then we can find a nonzero map $R^1f_*\omega _ X \to \kappa $. In other words we obtain a nonzero map $\alpha : Rf_*\omega _ X^\bullet \to \kappa [1]$. Applying $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A^\bullet )$ we get a nonzero map

\[ \beta : \kappa [-1] \longrightarrow Rf_*\mathcal{O}_ X \]

which is impossible by Lemma 51.7.6. To see that $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A^\bullet )$ does what we said, first note that

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa [1], \omega _ A^\bullet ) = R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa , \omega _ A^\bullet )[-1] = \kappa [-1] \]

as $\omega _ A^\bullet $ is normalized and we have

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(Rf_*\omega _ X^\bullet , \omega _ A^\bullet ) = Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\omega _ X^\bullet , \omega _ X^\bullet ) = Rf_*\mathcal{O}_ X \]

The first equality by Duality for Schemes, Lemma 46.3.6 and the fact that $\omega _ X^\bullet = f^!\omega _ A^\bullet $ by construction, and the second equality because $\omega _ X^\bullet $ is a dualizing complex for $X$ (which goes back to Duality for Schemes, Lemma 46.18.6). $\square$

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