= Solution
Put $R=k[x,y,z]/(xy-z^3)$ and let $X=\operatorname{Spec}R$. The ideal $P=(x,z)$ is a height-one <prime ideal>, because $R/P\cong k[y]$ and $\dim R=2$. It therefore defines a <prime Weil divisor> $D=V(x,z)$.
Localizing at $x$ eliminates $y$, giving
$$
R_x\cong k[x,x^{-1},z],
$$
a <unique factorization domain>. The <Nagata theorem for divisor class groups> says that $\operatorname{Cl}(X)$ is generated by the height-one primes containing $x$. Since
$$
R/(x)\cong k[y,z]/(z^3),
$$
the only such prime is $P$, and hence $[D]$ generates. At the generic point of $D$, $y$ is a <unit in a ring> and $x=z^3/y$, so the order of vanishing is three:
$$
\operatorname{div}(x)=3D.
$$
Thus $3[D]=0$.
This relation has exact order three. Indeed, if a <principal divisor> were supported on $D$, its defining rational function would be a unit on $D(x)$. The units of $R_x=k[z][x,x^{-1}]$ are precisely $c x^m$ with $c\in k^\times$ and $m\in\mathbb Z$, whose divisors are $3mD$. Consequently
$$
\boxed{\operatorname{Cl}\bigl(k[x,y,z]/(xy-z^3)\bigr)\cong\mathbb Z/3\mathbb Z,}
$$
generated by $[V(x,z)]$. This is the $n=3$ case of the <Divisor class group of an A-type surface singularity>.
Back to article page
