Situation 54.7.1. 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 $f$.

## 54.7 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 29.51.11).

By Varieties, Lemma 33.17.3 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 33.19.3. By Stein factorization (More on Morphisms, Lemma 37.53.6) 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 37.44.2) and hence an isomorphism (Morphisms, Lemma 29.54.8). We will discard this uninteresting case and we conclude that $\dim (C_ i) = 1$ for $i = 1, \ldots , r$.

Lemma 54.7.2. In Situation 54.7.1 there exists a $U$-admissible blowup $X' \to S$ which dominates $X$.

**Proof.**
This is a special case of More on Flatness, Lemma 38.31.4.
$\square$

Lemma 54.7.3. In Situation 54.7.1 there exists a nonzero $f \in \mathfrak m$ such that for every $i = 1, \ldots , r$ there exist

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

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 54.7.1 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 54.7.4. In Situation 54.7.1 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 54.7.1 without further mention. Let $f$ be a function as in Lemma 54.7.3. 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 31.14.4 and 31.14.2, and since the degree gets multiplied by $e_ i$, see Varieties, Lemma 33.44.7. 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 31.15.6. Thus the scheme $Z$ has no embedded associated points and we conclude that $\overline{f} = 0$ (Divisors, Lemmas 31.4.3 and 31.5.6). 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 33.44.12.
$\square$

Lemma 54.7.5. In Situation 54.7.1 assume $X$ is normal and $A$ Nagata. The map

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 20.5.1). 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 31.31.6). 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 29.6.2). Picture

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 29.44.11), hence an isomorphism (as $X$ is normal, see Morphisms, Lemma 29.54.8). 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 29.54.10 and 29.18.2). 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 31.22.8). 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 29.31.4. This is impossible by Lemma 54.7.4 and the proof is complete. $\square$

Lemma 54.7.6. In Situation 54.7.1 assume $X$ is normal and $A$ Nagata. Then

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

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 21.32.1) 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

By Lemma 54.7.5 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 54.7.7. Let $X$ be an integral Noetherian normal scheme of dimension $2$. In this case the following are equivalent

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

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 28.12.7) and Duality for Schemes, Lemma 48.23.1. In this situation we will say that $\omega _ X$ is a *dualizing module* in accordance with Duality for Schemes, Section 48.22. 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 48.22.1. 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 48.20.

The Grauert-Riemenschneider vanishing of the next proposition is a formal consequence of Lemma 54.7.6 and the general theory of duality.

Proposition 54.7.8 (Grauert-Riemenschneider). In Situation 54.7.1 assume

$X$ is a normal scheme,

$A$ is Nagata and has a dualizing complex $\omega _ A^\bullet $.

Let $\omega _ X$ be the dualizing module of $X$ (Remark 54.7.7). Then $R^1f_*\omega _ X = 0$.

**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 47.16. Since $X$ is a Noetherian normal $2$-dimensional scheme it is Cohen-Macaulay (Properties, Lemma 28.12.7). Thus $\omega _ X^\bullet = \omega _ X[2]$ (Duality for Schemes, Lemma 48.23.1 and the normalization in Duality for Schemes, Example 48.22.1). 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

which is impossible by Lemma 54.7.6. To see that $R\mathop{\mathrm{Hom}}\nolimits _ A(-, \omega _ A^\bullet )$ does what we said, first note that

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

The first equality by Duality for Schemes, Example 48.3.9 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 48.17.7). $\square$

## 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)