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The Stacks project

108.9 Properties of the Picard functor

Let f : X \to B be a morphism of algebraic spaces which is flat, proper, and of finite presentation such that moreover for every T/B the canonical map

\mathcal{O}_ T \longrightarrow f_{T, *}\mathcal{O}_{X_ T}

is an isomorphism. Then the Picard functor \mathrm{Pic}_{X/B} is an algebraic space, see Quot, Proposition 99.11.8. There is a closed relationship with the Picard stack.

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

Lemma 108.9.2. The diagonal of \mathrm{Pic}_{X/B} over B is a quasi-compact immersion.

Proof. The diagonal is an immersion by Quot, Lemma 99.11.9. To finish we show that the diagonal is quasi-compact. The diagonal of \mathcal{P}\! \mathit{ic}_{X/B} is quasi-compact by Lemma 108.8.1 and \mathcal{P}\! \mathit{ic}_{X/B} is a gerbe over \mathrm{Pic}_{X/B} by Lemma 108.9.1. We conclude by Morphisms of Stacks, Lemma 101.28.14. \square

Lemma 108.9.3. The morphism \mathrm{Pic}_{X/B} \to B is quasi-separated and locally of finite presentation.

Proof. To check \mathrm{Pic}_{X/B} \to B is quasi-separated we have to show that its diagonal is quasi-compact. This is immediate from Lemma 108.9.2. Since the morphism \mathcal{P}\! \mathit{ic}_{X/B} \to \mathrm{Pic}_{X/B} is surjective, flat, and locally of finite presentation (by Lemma 108.9.1 and Morphisms of Stacks, Lemma 101.28.8) it suffices to prove that \mathcal{P}\! \mathit{ic}_{X/B} \to B is locally of finite presentation, see Morphisms of Stacks, Lemma 101.27.12. This follows from Lemma 108.8.2. \square

Lemma 108.9.4. Assume the geometric fibres of X \to B are integral in addition to the other assumptions in this section. Then \mathrm{Pic}_{X/B} \to B is separated.

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 15.75.6 we can represent K by a complex of the form

\ldots \to 0 \to R \to R^{\oplus \beta _{K, 1}(0)} \to R^{\oplus \beta _{K, 2}(0)} \to \ldots

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

Lemma 108.9.5. Assume f : X \to B has relative dimension \leq 1 in addition to the other assumptions in this section. Then \mathrm{Pic}_{X/B} \to B is smooth.

Proof. By Lemma 108.8.5 we know that \mathcal{P}\! \mathit{ic}_{X/B} \to B is smooth. The morphism \mathcal{P}\! \mathit{ic}_{X/B} \to \mathrm{Pic}_{X/B} is surjective and smooth by combining Lemma 108.9.1 with Morphisms of Stacks, Lemma 101.33.8. Thus if U is a scheme and U \to \mathcal{P}\! \mathit{ic}_{X/B} is surjective and smooth, then U \to \mathrm{Pic}_{X/B} is surjective and smooth and U \to B is surjective and smooth (because these properties are preserved by composition). Thus \mathrm{Pic}_{X/B} \to B is smooth for example by Descent on Spaces, Lemma 74.8.3. \square


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