= Solution
First prove that $A$ is an <integrally closed domain>. If $u\in\operatorname{Frac}A$ 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 $n\geq0$ minimal. If $n>0$, an integral equation for $u$ gives, after multiplication by a suitable power of $x$,
$$
a^m+c_1a^{m-1}x^n+\cdots+c_mx^{nm}=0.
$$
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 $\operatorname{Cl}(A)$ is generated by height-one primes that meet $\{1,x,x^2,\ldots\}$. 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 $\operatorname{Cl}(A)=0$. A second application of the divisor-class criterion yields
$$
\boxed{A\text{ is a unique factorization domain}.}
$$
This argument is the <Nagata criterion for unique factorization domains>.
Back to article page
