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

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.