= Solution
Suppose first that $A$ is a <unique factorization domain>, and let $\mathfrak p$ be a minimal nonzero prime ideal. Choose $0\ne a\in\mathfrak p$ and factor it into <irreducible element>[irreducibles]. In a UFD each irreducible is a <prime element>, so one factor $\pi$ belongs to $\mathfrak p$. The nonzero prime ideal $(\pi)\subseteq\mathfrak p$ must equal $\mathfrak p$ by minimality.
Conversely, the <ascending chain condition> in the Noetherian domain implies that every nonzero nonunit factors into irreducibles. Let $\pi$ be irreducible and choose a prime $\mathfrak p$ minimal over $(\pi)$. The <Krull principal ideal theorem> gives $\operatorname{ht}\mathfrak p\leq1$. Since $A$ is a domain and $\mathfrak p\ne(0)$, it is a minimal nonzero prime and hence is principal, say $\mathfrak p=(q)$. The divisibility $q\mid\pi$ and irreducibility of $\pi$ force $q$ to be associate to $\pi$, so $(\pi)=\mathfrak p$ is prime. Thus every irreducible is prime, proving that $A$ is a UFD. This is the <Minimal-prime criterion for a Noetherian unique factorization domain>.
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