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

A standard sufficient hypothesis is that $X$ is both a Noetherian scheme and an integral scheme, and is regular in codimension one; in particular, a Noetherian normal scheme qualifies. A Weil divisor is then a finite sumover integral codimension-one closed subschemes $Z$. The local ring at the generic point of each $Z$ is a discrete valuation ring, so every nonzero rational function $g\in k(X{)}^{\times }$ has a principal divisorThe divisor class group is

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.

First prove that $A$ is an integrally closed domain. If $u\in \mathrm{Frac}A$ is integral over $A$, then it is integral over $A_x$. Since the unique factorization domain $A_x$ is integrally closed, write $u=a/x^n$ with $a\in A$ and $n\ge 0$ minimal. If $n>0$, an integral equation for $u$ gives, after multiplication by a suitable power of $x$,Hence $a^m\in (x)$. The ideal $(x)$ is prime, so $a\in (x)$, contradicting minimality of $n$. Therefore $n=0$ and $u\in A$.

Now apply the Nagata theorem for divisor class groups. The class group of $A_x$ vanishes by the divisor-class criterion for unique factorization. Hence $\mathrm{Cl}(A)$ is generated by height-one primes that meet $\{1,x,x^2,\dots \}$. Such a prime contains $x$ and must equal $(x)$, because the Krull principal ideal theorem makes the nonzero prime $(x)$ itself height one. Its divisor class is principal, so $\mathrm{Cl}(A)=0$. A second application of the divisor-class criterion yieldsThis argument is the Nagata criterion for unique factorization domains.