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For $n\ge 2$, the normal affine surfacehas divisor class group $\mathbb{Z}/n\mathbb{Z}$. Inverting $x$ gives $A_n[x^{-1}]\cong k[x,x^{-1},z]$, a unique factorization domain. The only prime Weil divisor over $x=0$ is $D=V(x,z)$, so the Nagata theorem for divisor class groups makes $[D]$ a generator.

At the generic point of $D$, $y$ is a unit and $x=z^n/y$. The local discrete valuation ring has uniformizer $z$, giving $\mathrm{div}(x)=nD$. This proves that the order divides $n$; exactness requires a further step. If $mD=\mathrm{div}(q)$, then $q$ is a unit on $X_n\setminus D$. The units of $k[x,x^{-1},z]$ are precisely $c{x}^r$, so $mD=rnD$. Thus every relation is a multiple of $nD$, proving exact order $n$.

Let $A=k[x,y,z,w]/(xy-zw)$ over a field $k$. Its divisor class group is infinite cyclic. Inverting $x$ givesa unique factorization domain. The Nagata theorem for divisor class groups therefore generates $\mathrm{Cl}(A)$ by the two prime Weil divisors $D_1=V(x,z)$ and $D_2=V(x,w)$ above $x=0$.

At either generic point, the defining equation makes $x$ a uniformizer times a unit, so $\mathrm{div}(x)=D_1+D_2$. To prove that this is the only relation, a rational function with divisor supported on $D_1\cup D_2$ restricts to a unit of $A_x$. Every such unit is $c{x}^r$ with $c\in k^{\ast }$ and $r\in \mathbb{Z}$. Its divisor is $r(D_1+D_2)$, givingThus either ruling plane generates, with the other representing its negative. Computing the localized unit group rules out an additional torsion relation.

Let $A$ be an integral domain that is a Noetherian ring, and let $x\in A$ generate a prime ideal. If the localization of a ring $A_x$ is a unique factorization domain, then $A$ is a unique factorization domain. Indeed, integrality of $a/x^n$ over $A$, with $n$ minimal, forces $x\mid a$ and hence proves that $A$ is an integrally closed domain. The Nagata theorem for divisor class groups then says that $\mathrm{Cl}(A)$ is generated by the height-one primes containing $x$; the only one is the principal prime $(x)$, so $\mathrm{Cl}(A)=0$.

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.

PutSinceis 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 givesa unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(R)$ is generated by the height-one primes containing $x$. Sincethese are $D=V(x,z)$ and $E=V(x,w)$. Both occur with multiplicity one in the principal divisorThe units of $R_x\cong k[z,w][x,x^{-1}]$ are exactly $c{x}^n$ with $c\in k^{\ast }$ and $n\in \mathbb{Z}$. Consequently the only relation supplied by localization is $[D]+[E]=0$, andThis is the divisor class group of the three-dimensional affine quadric cone.