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It is cyclic, generated by $\mathscr{O}(1)$. Indeed this is true for $X$ by the Lefschetz theorem (SGA2, Exp. XII, Cor. 3.7), and the restriction map $\operatorname{Pic}(X)\rightarrow \operatorname{Pic}(X\smallsetminus V) $ is an isomorphism, because the local rings of $X$ are parafactorial by SGA2, Exp. XI, Thm. 3.13).

Edit: This is wrong, as pointed out by @F_L in the comments (thanks!). The mistake is that parafactoriality must be checked at all points of $V$, and not only the closed points. The local ring $\mathscr{O}_{X,v}$ at the generic point $v$ of $V$ must be not parafactorial. I leave the answer since I think the error is instructive.

It is cyclic, generated by $\mathscr{O}(1)$. Indeed this is true for $X$ by the Lefschetz theorem (SGA2, Exp. XII, Cor. 3.7), and the restriction map $\operatorname{Pic}(X)\rightarrow \operatorname{Pic}(X\smallsetminus V) $ is an isomorphism, because the local rings of $X$ are parafactorial by SGA2, Exp. XI, Thm. 3.13).

It is cyclic, generated by $\mathscr{O}(1)$. Indeed this is true for $X$ by the Lefschetz theorem (SGA2, Exp. XII, Cor. 3.7), and the restriction map $\operatorname{Pic}(X)\rightarrow \operatorname{Pic}(X\smallsetminus V) $ is an isomorphism, because the local rings of $X$ are parafactorial by SGA2, Exp. XI, Thm. 3.13).

Edit: This is wrong, as pointed out by @F_L in the comments (thanks!). The mistake is that parafactoriality must be checked at all points of $V$, and not only the closed points. The local ring $\mathscr{O}_{X,v}$ at the generic point $v$ of $V$ must be not parafactorial. I leave the answer since I think the error is instructive.

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abx
  • 38k
  • 3
  • 86
  • 146

It is cyclic, generated by $\mathscr{O}(1)$. Indeed this is true for $X$ by the Lefschetz theorem (SGA2, Exp. XII, Cor. 3.7), and the restriction map $\operatorname{Pic}(X)\rightarrow \operatorname{Pic}(X\smallsetminus V) $ is an isomorphism, because the local rings of $X$ are parafactorial by SGA2, Exp. XI, Thm. 3.13).