Compare the Picard group of a projective variety to the Picard group of its cone In Hartshorne's exercise 2.6.3, there are the relationships of the divisor class group of a projective variety and the divisor class group of its cone. I want to ask that if we can get similar conclusions?
For example, how can we compute the Picard group of $Proj \,\,k[X_0,X_1,X_2,X_3,X_4]/(X_0X_1-X_2X_3)$ or $Proj\,\, k[X_0,X_1,X_2,X_3]/(X_0X_1-X_2^2)$ ? The main problem is that it has a singular point.
Let me state my question more explicitly.
For example,I want to compute the Picard group of  $Proj \,\,k[X_0,X_1,X_2,X_3,X_4]/(X_0X_1-X_2X_3)$. First the exercise 2.6.3 tells me that its divisor class group is the same as the original  $Cl(Proj \,\,k[X_0,X_1,X_2,X_3]/(X_0X_1-X_2X_3)) \simeq \mathbb{Z}\oplus\mathbb{Z}$. Then this projective cone is normal. So the Catier divisors are the same as the locally principal Weil divisors. To find all Cartier diviosrs, it's easy to see that we only to check all Weil divisors at the point $(0,0,0,0,1)$. I  want to know that:

*

*How to check the divisor locally principal at this point easily?

*How to generalize this process to more varieties?

 A: Let $X \subset \mathbb{P}^n$ be a connected projective variety. Let $\mathcal{O}_X(-1)$ be the restriction of $\mathcal{O}(-1)$ to $X$. Then
$$
Y := \mathbb{P}_X(\mathcal{O}_X \oplus \mathcal{O}(-1))
$$
is a resolution of singularities of the cone $CX$. The projection $\pi \colon Y \to CX$ contracts the divisor
$$
E := \mathbb{P}_X(\mathcal{O}_X) \subset Y
$$
to the vertex of the cone and is an isomorphism on its complement.
Therefore, the morphism $\pi^* \colon \mathrm{Pic}(CX) \to \mathrm{Pic}(Y)$ induces an isomorphism
$$
\mathrm{Pic}(CX) \cong 
\{ L \in \mathrm{Pic}(Y) \mid L\vert_E \cong \mathcal{O}_E \}.
$$
On the other hand, $\mathrm{Pic}(X)$ is freely generated by $p^*\mathrm{Pic}(X)$ and the relative hyperplane class $H$.
Since the projection $p \colon Y \to X$ when restricted to $E$ is an isomorphism, the composition
$$
\mathrm{Pic}(X) \stackrel{p^*}\to 
\mathrm{Pic}(Y) \to 
\mathrm{Pic}(E)
$$
(the second arrow is the restriction) is an isomorphism. Moreover,
$H\vert_E = 0$. Therefore,
$$
\mathrm{Pic}(CX) = \mathbb{Z}H.
$$
