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Let $X \to S$ be a morphism of schemes. The relative Picard functor from schemes over $S$ to abelian groups is usually defined by the formula $T \mapsto \text{Pic}(X \times_S T)/\text{Pic}(T)$ (e.g. in Kleiman's article and the book Neron Models by Bosch et al.). In favorable situations, this functor is representable, and in some less favorable situations its fppf-sheafification is.

There seems to be something wrong with this formula. Namely, it makes it look like $\text{Pic}(T)$ is a subgroup of $\text{Pic}(X \times_S T)$, which it need not be. For example, let $S = \text{Spec } \mathbb{R}$, $X = \text{Spec } \mathbb{C}$, and for $T$ take $\mathbb{P}^1_{\mathbb{R}}$ minus a degree two point. Then $\text{Pic}(T) \cong \mathbb{Z}/2\mathbb{Z}$, but $X \times_S T$ is $\mathbb{P}^1_{\mathbb{C}}$ minus two points and therefore has trivial Picard group.

So I have two questions. First, is it appropriate to simply take the cokernel of the map $\text{Pic}(T) \to \text{Pic}(X \times_S T)$ as the definition of the relative Picard functor? Second, under what hypotheses on $f : X \to S$ is $\text{Pic}(T) \to \text{Pic}(X \times_S T)$ injective for all $T$? Proposition 4 in Chapter 8 of Neron Models says that this is true e.g. if $f$ is quasi-compact and quasi-separated, and $f_*\mathcal{O}_X = \mathcal{O}_S$ universally. But what if $X$ is just, say, a geometrically connected variety over a field?

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I have seen the definition $\mathrm{Pic}_{X/S}(T) = \mathrm{coker}(\mathrm{Pic}(T) \to \mathrm{Pic}(X \times_S T))$ in several places, for example in Tamme's lectures on abelian schemes. This group arises naturally in the Leray spectral sequence. – Martin Brandenburg Mar 14 '13 at 21:27
Oops, I guess no one can read that equation. There really needs to either be a sample feature or an edit comment button. Having neither is absurd. – Matt Mar 14 '13 at 23:51
@Matt (though there is no @-tag support either): just copy the comment, delete it, and paste into a new comment to edit. Screws with the order if the conversation is moving quickly, though. – Ryan Reich Mar 15 '13 at 1:11
If the map is $f:X\to S$, then you could define $Pic_{X/S}(T)=H^0(T, R^1f_{T*}\mathbb{G}_m)$. The notation is really that we're defining the functor of points of something. When doing that it is pretty standard to say the $T$-points are blah up to equivalence where equivalence means blah. So that notation is just meant to say the points are elements of $Pic(X\times T)$ where two line bundles are equivalent if they differ by the pullback of something from $T$. The quotient is by an equivalence relation and not necessarily a subgroup. Thanks Ryan, that was the obvious thing to try. – Matt Mar 15 '13 at 1:43
Justin, it doesn't seem particularly natural to consider the relative Picard functor away from the proper setting. Are you aware of any non-proper examples? (Note that generalized Jacobians are Picard schemes of proper curves with singularities.) The only "natural" hypotheses I'm aware of which ensure that $f_{\ast}O_X = O_S$ universally are that $f$ is proper and finitely presented with geometrically reduced and geometrically connected fibers. – user30379 Mar 15 '13 at 4:07

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