Class group of $k[x,y,z,w]/(xy-zw)$

Let me first recap the setup and known parts. $A = k[x,y,z,w]/(xy-zw)$ and $p = (y,z)A$. In order to show that $Cl(A)\cong \mathbb{Z}$, it suffices to show that for any $n\ge 1$, $p^{(n)}$ is not principal.

Suppose that for some $n \ge 1$, $p^{(n)}$ is principal. Write $S = k[x,y,z,w]$, and let $f \in S$ whose image in $A$ generates $p^{(n)}$. Since $p^n \subset p^{(n)}$ in $A$, $(y,z)^n \subset (f, xy-zw)$ in $S$. We are going to show that this cannot happen.

The containment $(y,z)^n \subset (f, xy-zw)$ in $S$ implies $$ (y^n, z^n) \subset (f, xy-zw) \subset (f, x,w). $$ In $S/ (x,w) \cong k[y,z]$, the ideal $(y^n,z^n) S/(x)$ is contained in the ideal $(f,x,w) S/(x,w) = f S/(x,w)$ whose height (codimension) is at most 1 (Krull's height theorem). This is a contradiction because the height of $(y^n,z^n)S/(x,w) = (y^n,z^n)k[y,z]$ is 2.

Let $A=k[X,Y,Z,W]/(XY-ZW)$, and $\mathfrak p=(y,z)$. As you noticed the divisor class group of $A$ is cyclic generated by $[\mathfrak p]$.

We want to prove that $n[\mathfrak p]=0$ for $n\ge0$ implies $n=0$.

(This leads us to the conclusion that $\operatorname{Cl}(A)\simeq\mathbb Z$.)

Suppose $n\ge1$. Since $n[\mathfrak p]=0$ there is $f\in A$, $f\ne0$ such that $\mathfrak p^{(n)}=fA$. Now extend these ideals to $A[y^{-1}]$ and observe that $f$ is invertible in $A[y^{-1}]$. Since $A[y^{-1}]$ is the localization of the polynomial ring $k[y,z,w]$ at $y$, that is, $A[y^{-1}]=k[y,z,w][y^{-1}]$, it follows that $f=ay^m$ with $a\in k^\times$ and $m\in\mathbb Z$. Thus we have $\mathfrak p^{(n)}=y^mA$. It's easily seen that $m\ge 1$. Since $\mathfrak p^n\subseteq\mathfrak p^{(n)}$ we get $z^n\in y^mA$, that is, $Z^n\in(Y^m,XY-ZW)$. In particular, by sending $Y$ to $0$, we get $Z^n\in(ZW)$, a contradiction.