The Stacks project

Proof. Since $\mathrm{Pic}_{X/B} \to B$ is quasi-separated, it suffices to check the uniqueness part of the valuative criterion, see Morphisms of Spaces, Lemma 67.43.2. This immediately reduces to the following problem: given

1. a valuation ring $R$ with fraction field $K$,

2. an algebraic space $X$ proper and flat over $R$ with integral geometric fibre,

3. an element $a \in \mathrm{Pic}_{X/R}(R)$ with $a|_{\mathop{\mathrm{Spec}}(K)} = 0$,

then we have to prove $a = 0$. Applying Morphisms of Stacks, Lemma 101.25.6 to the surjective flat morphism $\mathcal{P}\! \mathit{ic}_{X/R} \to \mathrm{Pic}_{X/R}$ (surjective and flat by Lemma 108.9.1 and Morphisms of Stacks, Lemma 101.28.8) after replacing $R$ by an extension we may assume $a$ is given by an invertible $\mathcal{O}_ X$-module $\mathcal{L}$. Since $a|_{\mathop{\mathrm{Spec}}(K)} = 0$ we find $\mathcal{L}_ K \cong \mathcal{O}_{X_ K}$ by Quot, Lemma 99.11.3.

Denote $f : X \to \mathop{\mathrm{Spec}}(R)$ the structure morphism. Let $\eta , 0 \in \mathop{\mathrm{Spec}}(R)$ be the generic and closed point. Consider the perfect complexes $K = Rf_*\mathcal{L}$ and $M = Rf_*(\mathcal{L}^{\otimes -1})$ on $\mathop{\mathrm{Spec}}(R)$, see Derived Categories of Spaces, Lemma 75.25.4. Consider the functions $\beta _{K, i}, \beta _{M, i} : \mathop{\mathrm{Spec}}(R) \to \mathbf{Z}$ of Derived Categories of Spaces, Lemma 75.26.1 associated to $K$ and $M$. Since the formation of $K$ amd $M$ commutes with base change (see lemma cited above) we find $\beta _{K, 0}(\eta ) = \beta _{M, 0}(\beta ) = 1$ by Spaces over Fields, Lemma 72.14.3 and our assumption on the fibres of $f$. By upper semi-continuity we find $\beta _{K, 0}(0) \geq 1$ and $\beta _{M, 0} \geq 1$. By Spaces over Fields, Lemma 72.14.4 we conclude that the restriction of $\mathcal{L}$ to the special fibre $X_0$ is trivial. In turn this gives $\beta _{K, 0}(0) = \beta _{M, 0} = 1$ as above. Then by More on Algebra, Lemma 15.75.5 we can represent $K$ by a complex of the form

Now $R \to R^{\oplus \beta _{K, 1}(0)}$ is zero because $\beta _{K, 0}(\eta ) = 1$. In other words $K = R \oplus \tau _{\geq 1}(K)$ in $D(R)$ where $\tau _{\geq 1}(K)$ has tor amplitude in $[1, b]$ for some $b \in \mathbf{Z}$. Hence there is a global section $s \in H^0(X, \mathcal{L})$ whose restriction $s_0$ to $X_0$ is nonvanishing (again because formation of $K$ commutes with base change). Then $s : \mathcal{O}_ X \to \mathcal{L}$ is a map of invertible sheaves whose restriction to $X_0$ is an isomorphism and hence is an isomorphism as desired. $\square$