Let $X$ be a smooth projective curve over a field $k$. In chapter 8 of the book Neron Models by Bosch et al., there is a general result (namely Proposition 4) which implies that if $X$ admits a rational point, then the Picard functor which sends a $k$scheme $T$ to $\text{Pic}(X \times T)/\text{Pic}(T)$ is already a (say, fppf) sheaf. What if $X$ does not admit a rational point? Is there a reasonably simple example which demonstrates that fppfsheafification of this functor is necessary? Let's assume that the Brauer group of $k$ vanishes (e.g. $k$ finite) so that this is not an obstruction.
1 Answer
There are such examples over some fields with trivial Brauer group, but not over finite fields. Let me explain why.
Over a finite field $k$, any geometrically integral variety $X$ has a zerocycle of degree $1$. Indeed, by the LangWeil estimates, it has a rational point over the field extensions of $k$ of degrees $2^n$ and $3^n$ for $n\gg 0$, and these degrees are prime to each other. If $X$ is moreover proper, it follows that the Picard functor $T\mapsto \text{Pic}(X_T)/\text{Pic}(T)$ is already a fppf sheaf (and is moreover representable). When $X$ has a rational point, the argument given in [Néron models, 8.1, p. 203] uses the fact that the rational point induces a retraction of the natural pullback map $\text{Br}(T)\to \text{Br}(X_T)$. But if there is only a zerocycle of degree 1, it is still possible to construct such a retraction using restriction and corestriction maps for the Brauer groups, so that the argument still works.
On the other hand, suppose that $k=\mathbb{C}((t))$: this field has trivial Brauer group because it is $C_1$ by Lang's theorem. Let $C/k$ be a smooth geometrically connected genus $1$ curve whose minimal regular model over $\mathbb{C}[[t]]$ has a multiple special fiber, so that $C$ has no rational point. Then I claim that the functor $T\mapsto \text{Pic}(X_T)/\text{Pic}(T)$ is not a fppf sheaf. Indeed, if it were, it would be representable by the Picard scheme of $C$. In particular, there would be a Poincaré line bundle $\mathcal{L}$ on $C\times E$, where $E$ is the jacobian of $C$. Let $\bar{k}/k$ be an algebraic closure. Then $\mathcal{L}_{\bar{k}}$ is also a Poincaré line bundle realizing $C_{\bar{k}}$ as the jacobian of $E_{\bar{k}}$: in particular, there exists a unique $x\in C(\bar{k})$ such that $\mathcal{L}_{E_x}$ is trivial. But if $y\in C(\bar{k})$ is in the Galoisorbit of $x$, $\mathcal{L}_{E_y}$ also has a nowhere vanishing section, hence is also trivial, proving that $y=x$. This shows that $x$ induces a rational point on $C$, a contradiction.
As a last comment, if $k=\mathbb{C}((t))$, and $C/k$ is a smooth geometrically connected genus 2 curve, then $C$ necessarily has a degree $1$ zerocycle, so that the Picard functor $T\mapsto \text{Pic}(X_T)/\text{Pic}(T)$ is already a fppf sheaf (and is moreover representable). For this and related results, see [Esnault, Levine, Wittenberg, Index of varieties over Henselian fields and Euler characteristic of coherent sheaves, Example 2.6].

$\begingroup$ Thanks so much, this is very helpful. I see how a rational point induces a retraction of the relevant map of Brauer groups, but would you mind explaining how a degree 1 zerocycle does the same? If it makes things easier, I don't mind restricting to the case of a curve. $\endgroup$ Mar 9, 2013 at 17:31

2$\begingroup$ @Justin: If you have two effective zerocycles, $e:\text{Spec}(E)\to X$ and $f:\text{Spec}(F)\to X$, with $E/k$, resp. $F/k$, a finite field extension of degree $d+1$, resp. $d$, then the norm functors $e_*(\mc{L}):= \text{Nm}_{E/k}e^*\mc{L}$ and $f_*(\mc{L}):= \text{Nm}_{F/k}f^*\mc{L}$ combine to give a retraction via $\mc{L} \mapsto e_*(\mc{L})\otimes f_*(\mc{L})^\vee$. $\endgroup$ Mar 9, 2013 at 17:49

$\begingroup$ @Jason: Thanks! I hope you see this, it's been a week but I just got a chance to think about this again. I don't actually know this "norm functor," and I can't find anything by Googling or looking at my usual sources. Do you have a reference for this? Also, the TeX in your comment is broken: what is \mc supposed to be? $\endgroup$ Mar 16, 2013 at 17:11

1$\begingroup$ Never mind, I figured it out. If we think of the Brauer group as an $H^2$ then the map is corestriction. $\endgroup$ Mar 18, 2013 at 16:53