The Stacks project

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

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 $f = \prod a_ i$. $\square$

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. $\square$

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 29.54.11 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. $\square$

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. $\square$

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). $\square$