In the proof for the second part of theorem 2.5 found in Kleiman's paper https://arxiv.org/pdf/math/0504020.pdf, he claimed that any $\lambda\in \text{Pic}_{(X/S)(\text{fppf})}(T)$ can be represented by an invertible sheaf $\lambda^{'}\in\text{Pic}(X_{T^{'}})$ where $T^{'}\rightarrow T$ is an fppf covering with the condition that $X_{p_{1}}^{\ast}\lambda^{'}\simeq X_{p_{2}}^{\ast}\lambda^{'}$ on $T^{'}\times_{T}T^{\prime}$ without taking further fppf covering $T^{''}\rightarrow T^{'}\times_{T}T^{'}$ as $\text{Pic}_{(X/S)}$ is separated. My question is, even if $\text{Pic}_{(X/S)}$ is separated, there should still be a line bundle $\mathcal{N}$ on $T^{'}\times _{T}T^{'}$ such that $X_{p_{1}}^{\ast}\lambda^{'}\simeq X_{p_{2}}^{\ast}\lambda^{'}\otimes f_{T{'}\times _{T}T^{'}}^{\ast}\mathcal{N}$ as we are looking at the relative Picard functor. In this case, we take a zariski trivializing cover $T^{''}\rightarrow T^{'}\times _{T}T^{'}$ to get rid of $\mathcal{N}$.

I was trying to think of $\text{Pic}_{(X/S)(\text{fppf})}$ as the sheafificcation of $\text{Pic}_{X}$ but in this case $\text{Pic}_{X}$ is not separated. For your convenience Kleiman fixed a morphism $f:X\rightarrow S$ and if $T$ is an $S$-scheme $f_{T}:X_{T}\rightarrow T$ is the base change of $f$.


1 Answer 1


When $f:X\rightarrow S$ has a section $g$ and $\mathscr{O}_{S}\simeq f_{\ast}\mathscr{O}_{X}$ holds universally, lemma 2.10 shows that the functor $F$ $T\mapsto\{(\mathcal{L},u)\}/\simeq$ where $\mathcal{L}$ is a line bundle on $X_{T}$ and $u:\mathscr{O}_{T}\simeq g_{T}^{\ast}\mathcal{L}$ is a choice of isomorphism is isomorphic to the relative picard functor $\text{Proj}_{(X/S)}$. It is enough to show that the functor $F$ is a sheaf in the fppf topology. Then descent of line bundles apply. Or, we may assume the invertible sheaf $\lambda^{'}\in\text{Pic}(X_{T^{'}})$ has a $g$-rigidification. Then $X_{p_{1}}^{\ast}\lambda^{'}$ and $X_{p_{2}}^{\ast}\lambda^{'}$ have $g$-rigidifications, too. Pulling back the condition \begin{equation*}X_{p_{1}}^{\ast}\lambda^{'}\simeq X_{p_{2}}^{\ast}\lambda^{'}\otimes f_{T^{\prime}\times_{T}T^{\prime}}^{\ast}\mathcal{N}\end{equation*} along $g_{T^{'}\times_{T}T^{'}}$ yields

\begin{equation*}\mathscr{O}_{T^{'}\times_{T}T^{'}}\simeq g^{\ast}_{T^{'}\times _{T^{'}}T}X_{p_{1}}^{\ast}\lambda^{'}\simeq g_{T^{\prime}\times_{T}T^{\prime}}^{\ast}X_{p_{2}}^{\ast}\lambda^{\prime}\otimes g_{T^{\prime}\times_{T}T^{\prime}}^{\ast}f^{\ast}_{T^{\prime}\times_{T}T^{\prime}}\mathcal{N}\simeq\mathcal{N}\end{equation*}

Therefore in this case $X_{p_{1}}^{\ast}\lambda^{\prime}$ and $X_{p_{2}}^{\ast}\lambda^{\prime}$ are indeed isomorphic.


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