I'm trying to understand better the relative Picard functor, as defined, for example, in Kleiman's article.

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 $\mathrm{Pic}_{X/S,Zar}$.

What is an explicit example for which $$\mathrm{Pic}_{X/S}(S) \neq \mathrm{Pic}_{X/S,Zar}(S)$$ ?

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.


1 Answer 1


Let $Y$ be Cayley's 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.

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    $\begingroup$ Thanks very much Martin, this answer is great! It answers everything I wanted, and more. I have an application in mind where the base is regular, so this is perfect for me. $\endgroup$ Mar 29, 2015 at 14:09
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    $\begingroup$ If you voted my question up, then please also do vote Martin's answer up. This is one of the best answers I have ever received on mathoverflow. $\endgroup$ Mar 29, 2015 at 18:10
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    $\begingroup$ Dan, that's very kind of you, but it's still only a partial answer. I'm waiting for somebody to come along with a much simpler example in which $X \to Y$ is not proper and your presheaf isn't even separated. $\endgroup$ Mar 29, 2015 at 18:54

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