# Picard group of hypersurfaces in $\mathbb{P}^r\times\mathbb{P}^s$

Let $$k$$ be an algebraically closed field, say $$k=\mathbb{C}$$. Let $$r,s$$ be sufficiently large integers.

Is it true that, for any irreducible hypersurface $$X$$ of bi-degree $$(d,1)$$ in $$\mathbb{P}^r\times\mathbb{P}^r$$, the Picard group $$\mathrm{Pic}(X)$$ or the divisor class group $$\mathrm{Cl}(X)$$ equals to $$\mathbb{Z}\oplus\mathbb{Z}$$? I am not sure if the Lefschetz holds for such singular hypersurfaces?

• A general hyperusrface of bidegree $(d,1)$ smooth. Are you really interested in singular (non-general) hypersurfaces? Jan 10, 2020 at 12:33
• @Sasha Yes, I am working on some problem which really needs the case for possibly singular hypersurfaces, thanks! By the Leray spectral sequence, it would be ok if codimension of locus whose Picard number greater than 1 has codimension at least two.
– user39380
Jan 10, 2020 at 12:44
• Codimension in the space of parameters? And I guess you mean Picard number greater than 2, right? Jan 10, 2020 at 12:53
• @Sasha Ah, sorry for my being imprecise.. If we consider the projection $\pi\colon\mathbb{P}^r\times\mathbb{P}^s\to\mathbb{P}^s$, then the restriction to $V(f)\to\mathbb{P}^s$ is a fibration in hypersurfaces. If we apply the Leray spectral sequence for the map, we want to delete a codimension 2 subvariety in $Z\subset V(f)$, so that $V(f)-Z\to\mathbb{P}^s-\pi(Z)$ is a fibration in constant Picard number 1. Thanks!
– user39380
Jan 10, 2020 at 13:15
• I still don't understand --- you want the result for any hypersurface, or for some hypersurfaces? Perhaps, it makes sense to edit the question to make this clear. Jan 10, 2020 at 15:23

• For $Pic(X)$ yes, but unless $X$ is factorial, I don't see how you can deduce it for $Cl(X)$. Beware that there are even nodal threefolds in $P^4$ where $Cl(X)\neq \mathbb Z$. Jan 12, 2020 at 15:05
• Following Lazarsfeld (the example you cited), for it is sufficient to have $r+s\ge 4$. Oct 7, 2020 at 13:43