The Stacks project

Proof. Since the fibre category $\mathcal{P}\! \mathit{ic}_{X/B, T}$ of the Picard stack $\mathcal{P}\! \mathit{ic}_{X/B}$ over $T$ is the category of invertible sheaves on $X_ T$ (see Section 99.10 and Examples of Stacks, Section 95.16) this is immediate from the definitions. $\square$

Proof. We may replace $B$ by $T$ and $X$ by $X_ T$ and assume that $B = T$ to simplify the notation. Let $\mathcal{N}$ be an invertible $\mathcal{O}_ B$-module. If $f^*\mathcal{N} \cong \mathcal{O}_ X$, then we see that $f_*f^*\mathcal{N} \cong f_*\mathcal{O}_ X \cong \mathcal{O}_ B$ by assumption. Since $\mathcal{N}$ is locally trivial, we see that the canonical map $\mathcal{N} \to f_*f^*\mathcal{N}$ is locally an isomorphism (because $\mathcal{O}_ B \to f_*f^*\mathcal{O}_ B$ is an isomorphism by assumption). Hence we conclude that $\mathcal{N} \to f_*f^*\mathcal{N} \to \mathcal{O}_ B$ is an isomorphism and we see that $\mathcal{N}$ is trivial. This proves the first arrow is injective.

Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module which is in the kernel of $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathrm{Pic}_{X/B}(B)$. Then there exists an fppf covering $\{ B_ i \to B\} $ such that $\mathcal{L}$ pulls back to the trivial invertible sheaf on $X_{B_ i}$. Choose a trivializing section $s_ i$. Then $\text{pr}_0^*s_ i$ and $\text{pr}_1^*s_ j$ are both trivialising sections of $\mathcal{L}$ over $X_{B_ i \times _ B B_ j}$ and hence differ by a multiplicative unit

(equality by our assumption on pushforward of structure sheaves). Of course these elements satisfy the cocycle condition on $B_ i \times _ B B_ j \times _ B B_ k$, hence they define a descent datum on invertible sheaves for the fppf covering $\{ B_ i \to B\} $. By Descent, Proposition 35.5.2 there is an invertible $\mathcal{O}_ B$-module $\mathcal{N}$ with trivializations over $B_ i$ whose associated descent datum is $\{ f_{ij}\} $. (The proposition applies because $B$ is a scheme by the replacement performed at the start of the proof.) Then $f^*\mathcal{N} \cong \mathcal{L}$ as the functor from descent data to modules is fully faithful. $\square$

Proof. Denote $K(T) = \mathop{\mathrm{Ker}}(\sigma _ T^* : \mathop{\mathrm{Pic}}\nolimits (X_ T) \to \mathop{\mathrm{Pic}}\nolimits (T))$. Since $\sigma $ is a section of $f$ we see that $\mathop{\mathrm{Pic}}\nolimits (X_ T)$ is the direct sum of $\mathop{\mathrm{Pic}}\nolimits (T)$ and $K(T)$. Thus by Lemma 99.11.3 we see that $K(T) \subset \mathrm{Pic}_{X/B}(T)$ for all $T$. Moreover, it is clear from the construction that $\mathrm{Pic}_{X/B}$ is the sheafification of the presheaf $K$. To finish the proof it suffices to show that $K$ satisfies the sheaf condition for fppf coverings which we do in the next paragraph.

Let $\{ T_ i \to T\} $ be an fppf covering. Let $\mathcal{L}_ i$ be elements of $K(T_ i)$ which map to the same elements of $K(T_ i \times _ T T_ j)$ for all $i$ and $j$. Choose an isomorphism $\alpha _ i : \mathcal{O}_{T_ i} \to \sigma _{T_ i}^*\mathcal{L}_ i$ for all $i$. Choose an isomorphism

If the map

is not equal to multiplication by $1$ but some $u_{ij}$, then we can scale $\varphi _{ij}$ by $u_{ij}^{-1}$ to correct this. Having done this, consider the self map

which is given by multiplication by some section $f_{ijk}$ of the structure sheaf of $X_{T_ i \times _ T T_ j \times _ T T_ k}$. By our choice of $\varphi _{ij}$ we see that the pullback of this map by $\sigma $ is equal to multiplication by $1$. By our assumption on functions on $X$, we see that $f_{ijk} = 1$. Thus we obtain a descent datum for the fppf covering $\{ X_{T_ i} \to X\} $. By Descent on Spaces, Proposition 74.4.1 there is an invertible $\mathcal{O}_{X_ T}$-module $\mathcal{L}$ and an isomorphism $\alpha : \mathcal{O}_ T \to \sigma _ T^*\mathcal{L}$ whose pullback to $X_{T_ i}$ recovers $(\mathcal{L}_ i, \alpha _ i)$ (small detail omitted). Thus $\mathcal{L}$ defines an object of $K(T)$ as desired. $\square$

Proof. We already know that $\mathcal{P}\! \mathit{ic}_{X/B}$ is a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$ by Examples of Stacks, Lemma 95.16.1. Let us show descent for objects for $\mathcal{P}\! \mathit{ic}_{X/B, \sigma }$. Let $\{ T_ i \to T\} $ be an fppf covering and let $\xi _ i = (T_ i, h_ i, \mathcal{L}_ i, \alpha _ i)$ be an object of $\mathcal{P}\! \mathit{ic}_{X/B, \sigma }$ lying over $T_ i$, and let $\varphi _{ij} : \text{pr}_0^*\xi _ i \to \text{pr}_1^*\xi _ j$ be a descent datum. Applying the result for $\mathcal{P}\! \mathit{ic}_{X/B}$ we see that we may assume we have an object $(T, h, \mathcal{L})$ of $\mathcal{P}\! \mathit{ic}_{X/B}$ over $T$ which pulls back to $\xi _ i$ for all $i$. Then we get

Since the maps $\varphi _{ij}$ are compatible with the $\alpha _ i$ we see that $\alpha _ i$ and $\alpha _ j$ pullback to the same map on $T_ i \times _ T T_ j$. By descent of quasi-coherent sheaves (Descent, Proposition 35.5.2, we see that the $\alpha _ i$ are the restriction of a single map $\alpha : \mathcal{O}_ T \to \sigma _ T^*\mathcal{L}$ as desired. We omit the proof of descent for morphisms. $\square$

Proof. Let $T$ be a scheme and let $(\mathit{Sch}/T)_{fppf} \to \mathcal{P}\! \mathit{ic}_{X/B}$ be given by the object $\xi = (T, h, \mathcal{L})$ of $\mathcal{P}\! \mathit{ic}_{X/B}$ over $T$. We have to show that

is representable by a scheme $V$ and that the corresponding morphism $V \to T$ is surjective and smooth. See Algebraic Stacks, Sections 94.6, 94.9, and 94.10. The forgetful functor $\mathcal{P}\! \mathit{ic}_{X/B, \sigma } \to \mathcal{P}\! \mathit{ic}_{X/B}$ is faithful on fibre categories and for $T'/T$ the set of isomorphism classes is the set of isomorphisms

See Algebraic Stacks, Lemma 94.9.2. We know this functor is representable by an affine scheme $U$ of finite presentation over $T$ by Proposition 99.4.3 (applied to $\text{id} : T \to T$ and $\mathcal{O}_ T$ and $\sigma ^*\mathcal{L}$). Working Zariski locally on $T$ we may assume that $\sigma _ T^*\mathcal{L}$ is isomorphic to $\mathcal{O}_ T$ and then we see that our functor is representable by $\mathbf{G}_ m \times T$ over $T$. Hence $U \to T$ Zariski locally on $T$ looks like the projection $\mathbf{G}_ m \times T \to T$ which is indeed smooth and surjective. $\square$

Proof. If $\xi = (T, h, \mathcal{L}, \alpha )$ is an object of $\mathcal{P}\! \mathit{ic}_{X/B, \sigma }$ over $T$, then an automorphism $\varphi $ of $\xi $ is given by multiplication with an invertible global section $u$ of the structure sheaf of $X_ T$ such that moreover $\sigma _ T^*u = 1$. Then $u = 1$ by our assumption that $\mathcal{O}_ T \to f_{T, *}\mathcal{O}_{X_ T}$ is an isomorphism. Hence $\mathcal{P}\! \mathit{ic}_{X/B, \sigma }$ is fibred in setoids over $(\mathit{Sch}/S)_{fppf}$. Given $T$ and $h : T \to B$ the set of isomorphism classes of pairs $(\mathcal{L}, \alpha )$ is the same as the set of isomorphism classes of $\mathcal{L}$ with $\sigma _ T^*\mathcal{L} \cong \mathcal{O}_ T$ (isomorphism not specified). This is clear because any two choices of $\alpha $ differ by a global unit on $T$ and this is the same thing as a global unit on $X_ T$. $\square$

Proof. There exists a surjective, flat, finitely presented morphism $B' \to B$ of algebraic spaces such that the base change $X' = X \times _ B B'$ over $B'$ has a section: namely, we can take $B' = X$. Observe that $\mathrm{Pic}_{X'/B'} = B' \times _ B \mathrm{Pic}_{X/B}$. Hence $\mathrm{Pic}_{X'/B'} \to \mathrm{Pic}_{X/B}$ is representable by algebraic spaces, surjective, flat, and finitely presented. Hence, if we can show that $\mathrm{Pic}_{X'/B'}$ is an algebraic space, then it follows that $\mathrm{Pic}_{X/B}$ is an algebraic space by Bootstrap, Theorem 80.10.1. In this way we reduce to the case described in the next paragraph.

In addition to the assumptions of the proposition, assume that we have a section $\sigma : B \to X$. By Proposition 99.10.2 we see that $\mathcal{P}\! \mathit{ic}_{X/B}$ is an algebraic stack. By Lemma 99.11.6 and Algebraic Stacks, Lemma 94.15.4 we see that $\mathcal{P}\! \mathit{ic}_{X/B, \sigma }$ is an algebraic stack. By Lemma 99.11.7 and Algebraic Stacks, Lemma 94.8.2 we see that $T \mapsto \mathop{\mathrm{Ker}}(\sigma _ T^* : \mathop{\mathrm{Pic}}\nolimits (X_ T) \to \mathop{\mathrm{Pic}}\nolimits (T))$ is an algebraic space. By Lemma 99.11.4 this functor is the same as $\mathrm{Pic}_{X/B}$. $\square$

Proof. Let $T$ be a scheme over $B$ and let $s, t \in \mathrm{Pic}_{X/B}(T)$. We want to show that there exists a locally closed subscheme $Z \subset T$ such that $s|_ Z = t|_ Z$ and such that a morphism $T' \to T$ factors through $Z$ if and only if $s|_{T'} = t|_{T'}$.

We first reduce the general problem to the case where $s$ and $t$ come from invertible modules on $X_ T$. We suggest the reader skip this step. Choose an fppf covering $\{ T_ i \to T\} _{i \in I}$ such that $s|_{T_ i}$ and $t|_{T_ i}$ come from $\mathop{\mathrm{Pic}}\nolimits (X_{T_ i})$ for all $i$. Suppose that we can show the result for all the pairs $s|_{T_ i}, t|_{T_ i}$. Then we obtain locally closed subschemes $Z_ i \subset T_ i$ with the desired universal property. It follows that $Z_ i$ and $Z_ j$ have the same scheme theoretic inverse image in $T_ i \times _ T T_ j$. This determines a descend datum on $Z_ i/T_ i$. Since $Z_ i \to T_ i$ is locally quasi-finite, it follows from More on Morphisms, Lemma 37.57.1 that we obtain a locally quasi-finite morphism $Z \to T$ recovering $Z_ i \to T_ i$ by base change. Then $Z \to T$ is an immersion by Descent, Lemma 35.24.1. Finally, because $\mathrm{Pic}_{X/B}$ is an fppf sheaf, we conclude that $s|_ Z = t|_ Z$ and that $Z$ satisfies the universal property mentioned above.

Assume $s$ and $t$ come from invertible modules $\mathcal{V}$, $\mathcal{W}$ on $X_ T$. Set $\mathcal{L} = \mathcal{V} \otimes \mathcal{W}^{\otimes -1}$ We are looking for a locally closed subscheme $Z$ of $T$ such that $T' \to T$ factors through $Z$ if and only if $\mathcal{L}_{X_{T'}}$ is the pullback of an invertible sheaf on $T'$, see Lemma 99.11.3. Hence the existence of $Z$ follows from More on Morphisms of Spaces, Lemma 76.53.1. $\square$