Picard group of a cubic hypersurface 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.
 A: 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.
A: 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
$$
\mathrm{Pic}(X \setminus V) = \mathrm{Pic}(\tilde{X} \setminus E) \cong \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}.
$$
