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Consider the following cubic hypersurface in $\mathbb{P}^5$:

$$ X = \{z_0z_3z_5-z_1^2z_5-z_0z_4^2+2z_1z_2z_4-z_2^2z_3 = 0\}\subset\mathbb{P}^5 $$

The singular locus of $X$ is the Veronese surface $V\subset X$. I would like to ask if it is known what is the Picard group of $X\setminus V$?

Thank you very much.

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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.

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  • $\begingroup$ Thank you very much for the answer. Is it clear what is the smallest power of $\mathcal{O}(1)$ that is trivial? Do you know if there is an alternative argument not using the fact $X$ is a complete intersection? For instance, using the fact that $X\setminus V$ is an orbit of an algebraic group. $\endgroup$ – Fra Dec 14 '20 at 17:15
  • $\begingroup$ No power of $\mathscr{O}(1)$ is trivial (Lefschetz gives an isomorphism $\operatorname{Pic}(\mathbb{P}^5)\rightarrow \operatorname{Pic}(X)$). And yes, you can probably give a proof using that $X\smallsetminus V$ is homogeneous. $\endgroup$ – abx Dec 14 '20 at 17:40
  • $\begingroup$ Ok. So $Pic(X)\cong\mathbb{Z}$ and it's generated by the hyperplane section. Thank you. $\endgroup$ – Fra Dec 14 '20 at 17:53
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    $\begingroup$ @ abx. Consider the divisor $D$ in $X$ defined by $\{Z_2Z_3-Z_1Z_4 = Z_1Z_2-Z_0Z_4 = Z_1^2-Z_0Z_3 = 0\}$. This is a cone with vertex $[0:\dots:0:1]$ over a $2$-dimensional cubic scroll $S$ contained in $\{Z_5=0\}$. The restriction of $D$ to $X\setminus V$ is Cartier. If $D = X\cap H$ for some hypersurface $H\subset\mathbb{P}^5$ then $H$ must be a hyperplane. On the other hand, if $H$ is a hyperplane such that $H\cap X = D$ then $H$ must be the hyperplane generated by $S$ which does not contain the vetrtex of $D$. So, it seems that $D$ can not be cut out on $X\setminus V$ by any hypersurface. $\endgroup$ – F_L Jan 17 at 13:58
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    $\begingroup$ One can cut out $2D$ on $X\setminus V$ intersecting $X$ with a quadric hypersurface. So it seems that there is a $2$-torsion divisor in $\text{Pic}(X\setminus V)$. $\endgroup$ – F_L Jan 17 at 14:00
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Another way to find $\mathrm{Pic}(X)$ is the following. Note that the cubic $X$ is the symmetric determinantal cubic and it has a resolution of singularities $$ \tilde{X} = \mathbb{P}_{\mathbb{P}^2}(S^2\Omega_{\mathbb{P}^2}(2)). $$ its explicit form implies that $\mathrm{Pic}(\tilde{X}) \cong \mathbb{Z} \oplus \mathbb{Z}$. Furthermore, the exceptional divisor of the contraction $\tilde{X} \to X$ is the subvariety $$ E = \mathbb{P}_{\mathbb{P}^2}(\Omega_{\mathbb{P}^2}(1)) $$ and its embedding into $\tilde{X}$ is the relative double Veronese embedding. Finally, it is easy to check that the class of $E$ in $\mathrm{Pic}(\tilde{X})$ is equal to $$ 2H + 2h, $$ where $h$ is the hyperplane class of $\mathbb{P}^2$ and $H$ is the relative hyperplane class of $\mathbb{P}_{\mathbb{P}^2}(S^2\Omega_{\mathbb{P}^2}(2))$. Therefore $$ \mathrm{Pic}(X \setminus V) = \mathrm{Pic}(\tilde{X} \setminus E) \cong \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}. $$

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  • $\begingroup$ Is it immediate to see that $\tilde{X}$ is the projectivization of $S^2\Omega_{\mathbb{P}^2}(2)$? I can see that that $\tilde{X}$ has a fibration over $\mathbb{P}^2$ with fiber $\mathbb{P}^2$ but not that it is the projectivization of this particular vector bundle. $\endgroup$ – Fra Dec 14 '20 at 21:44

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