Lemma 108.9.1. The morphism \mathcal{P}\! \mathit{ic}_{X/B} \to \mathrm{Pic}_{X/B} turns the Picard stack into a gerbe over the Picard functor.
Proof. The definition of \mathcal{P}\! \mathit{ic}_{X/B} \to \mathrm{Pic}_{X/B} being a gerbe is given in Morphisms of Stacks, Definition 101.28.1, which in turn refers to Stacks, Definition 8.11.4. To prove it, we will check conditions (2)(a) and (2)(b) of Stacks, Lemma 8.11.3. This follows immediately from Quot, Lemma 99.11.2; here is a detailed explanation.
Condition (2)(a). Suppose that \xi \in \mathrm{Pic}_{X/B}(U) for some scheme U over B. Since \mathrm{Pic}_{X/B} is the fppf sheafification of the rule T \mapsto \mathop{\mathrm{Pic}}\nolimits (X_ T) on schemes over B (Quot, Situation 99.11.1), we see that there exists an fppf covering \{ U_ i \to U\} such that \xi |_{U_ i} corresponds to some invertible module \mathcal{L}_ i on X_{U_ i}. Then (U_ i \to B, \mathcal{L}_ i) is an object of \mathcal{P}\! \mathit{ic}_{X/B} over U_ i mapping to \xi |_{U_ i}.
Condition (2)(b). Suppose that U is a scheme over B and \mathcal{L}, \mathcal{N} are invertible modules on X_ U which map to the same element of \mathrm{Pic}_{X/B}(U). Then there exists an fppf covering \{ U_ i \to U\} such that \mathcal{L}|_{X_{U_ i}} is isomorphic to \mathcal{N}|_{X_{U_ i}}. Thus we find isomorphisms between (U \to B, \mathcal{L})|_{U_ i} \to (U \to B, \mathcal{N})|_{U_ i} as desired. \square
Comments (0)