The Stacks project

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. $\square$