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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?

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    $\begingroup$ A general hyperusrface of bidegree $(d,1)$ smooth. Are you really interested in singular (non-general) hypersurfaces? $\endgroup$ – Sasha Jan 10 at 12:33
  • $\begingroup$ @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. $\endgroup$ – Qixiao Jan 10 at 12:44
  • $\begingroup$ Codimension in the space of parameters? And I guess you mean Picard number greater than 2, right? $\endgroup$ – Sasha Jan 10 at 12:53
  • $\begingroup$ @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! $\endgroup$ – Qixiao Jan 10 at 13:15
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    $\begingroup$ 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. $\endgroup$ – Sasha Jan 10 at 15:23
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Ok, this follows from Lazarsfeld positivity I, Example 3.1.25.

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    $\begingroup$ 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 $Cl(X)\neq \mathbb Z$. $\endgroup$ – wnx Jan 12 at 15:05
  • $\begingroup$ @WalterNeff Yes you are right, that shall not be true for class groups, thanks! $\endgroup$ – Qixiao Jan 13 at 6:21

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