Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 113 / 3 / b / Solution

Put $R=k[x,y,z]/(xy-z^3)$ and let $X=\mathrm{Spec}R$. 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 $y$, givinga unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(X)$ is generated by the height-one primes containing $x$. Sincethe 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: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 $3mD$. Consequentlygenerated by $[V(x,z)]$. This is the $n=3$ case of the Divisor class group of an A-type surface singularity.