I'm trying to understand better the relative Picard functor, as defined, for example, in [Kleiman's article](http://arxiv.org/pdf/math/0504020v1.pdf).


Let $X \to S$ be a smooth projective morphism of schemes whose geometric fibres are integral. Let
$$\mathrm{Pic}\_{X/S}(T) = \mathrm{Pic}(X\_T)/\mathrm{Pic}(T)$$
denote the relative Picard functor of $X$, with sheafications
$$\mathrm{Pic}\_{X/S,Zar} \quad \mbox{and} \quad \mathrm{Pic}\_{X/S,et}$$
for the Zariski and etale topology, respectively. Kleiman shows that $\mathrm{Pic}\_{X/S}(S) \neq \mathrm{Pic}\_{X/S,et}(S)$ in general, e.g. when $X$ a conic over a field without a rational point. My question concerns what happens for $\mathrm{Pic}\_{X/S,Zar}$.



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> What is an explicit example for which $$\mathrm{Pic}\_{X/S}(S) \neq \mathrm{Pic}\_{X/S,Zar}(S)$$ ?
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There are various conditions on $X$ if one wants such an inequality hold. For example, $S$ cannot be the spectrum of a local ring, and $X/S$ cannot admit a section for the Zariski topology.



Relative Picard functor for the Zariski topology
Let $Y$ be [Cayley's nodal cubic surface](http://en.wikipedia.org/wiki/Cayley%27s_nodal_cubic_surface) over the complex numbers, given in $\mathbb{P}^3$ by


$X\_0X\_1X\_2 + X\_0X\_1X\_3 + X\_0X\_2X\_3 + X\_1X\_2X\_3 = 0.$


This surface has four simple double points. It also contains six straight lines, each of which passes through two of the double points and is not locally principal there. Let $f \colon X \to Y$ be the minimal desingularisation. Then $X$ is a smooth rational surface, isomorphic to the blow-up of $\mathbb{P}^2$ in six points in a rather special configuration. The six exceptional curves of $X \to \mathbb{P}^2$ are the strict transforms of the six lines on $Y$. I claim that $\DeclareMathOperator{\Pic}{Pic}\Pic\_{X/Y}$ is not a sheaf in the Zariski topology.


To prove this, let $P$ be one of the singular points of $Y$, let $U$ be a neighbourhood of $P$ not containing any of the other three singular points, and let $V$ be $Y \setminus P$. Let $U'$ and $V'$ be the inverse images of $U$ and $V$ respectively in $X$. Let $L$ be one of the lines passing through $P$, and let $L'$ be its strict transform in $X$. The restriction of $L'$ to $U' \cap V'$ lies in $f^{-1}(\Pic(U \cap V))$, since it's the inverse image of the line $L$ which is locally principal on $U \cap V$. Thus the classes $[L'] \in \Pic\_{X/Y}(U)$ and $0 \in \Pic\_{X/Y}(V)$ agree on $U \cap V$. However, these two classes do not glue to give a section of $\Pic\_{X/Y}$ on $U \cup V = Y$, which we can see by looking at intersection numbers with the four $(-2)$-curves $E\_1 = f^{-1}(P),E\_2,E\_3,E\_4$. If the classes did glue, then that section would be represented by a class in $\Pic X$ having intersection number $1$ with $E\_1$ and intersection number $0$ with $E\_2,E\_3,E\_4$. But $\Pic X$ is generated by the classes of the strict transforms of the six lines on $Y$ together with the pull-back of the hyperplane class on $Y$. Because each line on $Y$ passes through precisely two singular points, the sum $\sum\_{i=1}^4 (D \cdot E\_i)$ is even for every $D \in \Pic X$.


So what's actually going on here? Have a look at the Leray spectral sequence for the sheaf $\newcommand{\O}{\mathcal{O}}\O\_X^\times$ in the Zariski topology. Using $f\_\* \O\_X = \O\_Y$, we get an exact sequence


$0 \to \Pic Y \to \Pic X \to \mathrm{R}^1 f\_\* \O\_X^\times\to \mathrm{H}^2(Y,\O\_Y^\times) \to \mathrm{H}^2(X, \O\_X^\times).$


So the obstruction to gluing local sections of $\Pic\_{X/Y}$ is given by $\mathrm{H}^2(Y,\O\_Y^\times)$, the "Zariski Brauer group" of $Y$. This group vanishes if $Y$ is regular: in that case, the Weil-divisor exact sequence


$0 \to \O\_Y^\times \to R\_Y^\times \to \bigoplus\_{Z \in Y^{(1)}} \mathbb{Z}\_Z \to 0$


gives a flabby resolution of $\O\_Y^\times$, showing that it has no cohomology in degrees $>1$. (In particular, this shows $\mathrm{H}^2(X,\O\_X^\times)=0$ in the sequence above.)
So, when $f\_\* \O\_X = \O\_Y$ holds and $Y$ is regular, your $\Pic\_{X/Y}$ will be a Zariski sheaf. If $Y$ is allowed to be singular, there are counterexamples.


For rational singularities, this is all controlled by intersection numbers with the exceptional divisors. See [arXiv:1202.4299](http://arxiv.org/abs/1201.4299).


