# 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.

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.

• 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. – Fra Dec 14 '20 at 17:15
• 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. – abx Dec 14 '20 at 17:40
• Ok. So $Pic(X)\cong\mathbb{Z}$ and it's generated by the hyperplane section. Thank you. – Fra Dec 14 '20 at 17:53
• @ 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. – F_L Jan 17 at 13:58
• 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 $\text{Pic}(X\setminus V)$. – F_L Jan 17 at 14:00

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 $$\mathrm{Pic}(X \setminus V) = \mathrm{Pic}(\tilde{X} \setminus E) \cong \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}.$$

• 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. – Fra Dec 14 '20 at 21:44