Put
$$
R=k[x,y,z,w]/(xy-zw).
$$
Since
$$
R/(x,z)\cong k[y,w]
$$
is an <integral domain>, $(x,z)$ is a <prime ideal>. Moreover $\dim R=3$ and $\dim R/(x,z)=2$, so its <height of an ideal> is one. Therefore $D=V(x,z)$ is a <prime Weil divisor>.

Localizing at $x$ eliminates $y$ and gives
$$
R_x\cong k[x,z,w,x^{-1}],
$$
a <unique factorization domain>. The <Nagata theorem for divisor class groups> says that $\operatorname{Cl}(R)$ is generated by the height-one primes containing $x$. Since
$$
R/(x)\cong k[y,z,w]/(zw),
$$
these are $D=V(x,z)$ and $E=V(x,w)$. Both occur with multiplicity one in the <principal divisor>
$$
\operatorname{div}(x)=D+E.
$$
The units of $R_x\cong k[z,w][x,x^{-1}]$ are exactly $c x^n$ with $c\in k^*$ and $n\in\mathbb Z$. Consequently the only relation supplied by localization is $[D]+[E]=0$, and
$$
\operatorname{Cl}(X)
\cong\frac{\mathbb Z[D]\oplus\mathbb Z[E]}{\mathbb Z([D]+[E])}
\cong\mathbb Z[D].
$$
This is the <divisor class group of the three-dimensional affine quadric cone>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 113 / 2 / a / Solution

{title2=$\operatorname{Cl}(k[x,y,z,w]/(xy-zw))\cong\mathbb Z$}

For the three-dimensional affine quadric cone
$$
X=\operatorname{Spec}k[x,y,z,w]/(xy-zw),
$$
the <divisor class group> is infinite cyclic. A generator is either ruling plane $V(x,z)$ or $V(x,w)$, and the <principal divisor> of $x$ gives the relation $[V(x,z)]+[V(x,w)]=0$.

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