Given any scheme $X$ we denote $\mathop{\mathrm{Pic}}\nolimits (X)$ the set of isomorphism classes of invertible $\mathcal{O}_ X$-modules. See Modules, Definition 17.25.9. Given a morphism $f : X \to Y$ of schemes, pullback defines a group homomorphism $\mathop{\mathrm{Pic}}\nolimits (Y) \to \mathop{\mathrm{Pic}}\nolimits (X)$. The assignment $X \leadsto \mathop{\mathrm{Pic}}\nolimits (X)$ is a contravariant functor from the category of schemes to the category of abelian groups. This functor is not representable, but it turns out that a relative variant of this construction sometimes is representable.
Let us define the Picard functor for a morphism of schemes $f : X \to S$. The idea behind our construction is that we'll take it to be the sheaf $R^1f_*\mathbf{G}_ m$ where we use the fppf topology to compute the higher direct image. Unwinding the definitions this leads to the following more direct definition.
Definition 44.4.1. Let $\mathit{Sch}_{fppf}$ be a big site as in Topologies, Definition 34.7.8. Let $f : X \to S$ be a morphism of this site. The Picard functor $\mathrm{Pic}_{X/S}$ is the fppf sheafification of the functor If this functor is representable, then we denote $\underline{\mathrm{Pic}}_{X/S}$ a scheme representing it.
\[ (\mathit{Sch}/S)_{fppf} \longrightarrow \textit{Sets},\quad T \longmapsto \mathop{\mathrm{Pic}}\nolimits (X_ T) \]
An often used remark is that if $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, then $\mathrm{Pic}_{X_ T/T}$ is the restriction of $\mathrm{Pic}_{X/S}$ to $(\mathit{Sch}/T)_{fppf}$. It turns out to be nontrivial to see what the value of $\mathrm{Pic}_{X/S}$ is on schemes $T$ over $S$. Here is a lemma that helps with this task.
Lemma 44.4.2. Let $f : X \to S$ be as in Definition 44.4.1. If $\mathcal{O}_ T \to f_{T, *}\mathcal{O}_{X_ T}$ is an isomorphism for all $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, then is an exact sequence for all $T$.
\[ 0 \to \mathop{\mathrm{Pic}}\nolimits (T) \to \mathop{\mathrm{Pic}}\nolimits (X_ T) \to \mathrm{Pic}_{X/S}(T) \]
Proof. We may replace $S$ by $T$ and $X$ by $X_ T$ and assume that $S = T$ to simplify the notation. Let $\mathcal{N}$ be an invertible $\mathcal{O}_ S$-module. If $f^*\mathcal{N} \cong \mathcal{O}_ X$, then we see that $f_*f^*\mathcal{N} \cong f_*\mathcal{O}_ X \cong \mathcal{O}_ S$ 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 $\mathcal{O}_ S \to f_*f^*\mathcal{O}_ S$ is an isomorphism by assumption). Hence we conclude that $\mathcal{N} \to f_*f^*\mathcal{N} \to \mathcal{O}_ S$ 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 $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathrm{Pic}_{X/S}(S)$. Then there exists an fppf covering $\{ S_ i \to S\} $ such that $\mathcal{L}$ pulls back to the trivial invertible sheaf on $X_{S_ i}$. 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 $X_{S_ i \times _ S S_ j}$ 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 $S_ i \times _ S S_ j \times _ S S_ k$, hence they define a descent datum on invertible sheaves for the fppf covering $\{ S_ i \to S\} $. By Descent, Proposition 35.5.2 there is an invertible $\mathcal{O}_ S$-module $\mathcal{N}$ with trivializations over $S_ i$ whose associated descent datum is $\{ f_{ij}\} $. Then $f^*\mathcal{N} \cong \mathcal{L}$ as the functor from descent data to modules is fully faithful (see proposition cited above). $\square$
Lemma 44.4.3. Let $f : X \to S$ be as in Definition 44.4.1. Assume $f$ has a section $\sigma $ and that $\mathcal{O}_ T \to f_{T, *}\mathcal{O}_{X_ T}$ is an isomorphism for all $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. Then is a split exact sequence with splitting given by $\sigma _ T^* : \mathop{\mathrm{Pic}}\nolimits (X_ T) \to \mathop{\mathrm{Pic}}\nolimits (T)$.
\[ 0 \to \mathop{\mathrm{Pic}}\nolimits (T) \to \mathop{\mathrm{Pic}}\nolimits (X_ T) \to \mathrm{Pic}_{X/S}(T) \to 0 \]
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 44.4.2 we see that $K(T) \subset \mathrm{Pic}_{X/S}(T)$ for all $T$. Moreover, it is clear from the construction that $\mathrm{Pic}_{X/S}$ 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 regular function $f_{ijk}$ on the scheme $X_{T_ i \times _ T T_ j \times _ T T_ k}$. 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, Proposition 35.5.2 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. $\square$
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