 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 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 $\mathrm{ht}\mathfrak{p}\le 1$. 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.