## PIDs are precisely those UFDs which are also Bezout domains

Prove that $R$ is a principal ideal domain if and only if it is a unique factorization domain and a Bezout domain.

We already know that every principal ideal domain is both a unique factorization domain and a Bezout domain.

Suppose conversely that $R$ is a unique factorization domain and a Bezout domain. Let $I \subseteq R$ be an ideal, and choose $\alpha \in I$ with the minimal number of irreducible factors. Let $\beta \in I$ and consider $(\alpha,\beta)$; since $R$ is a Bezout domain, we have $(\alpha,\beta) = (\gamma)$ for some $\gamma \in I$. Now $\alpha = \gamma\delta$ for some $\delta$. Since $R$ is a unique factorization domain, the number of irreducible factors of $\gamma$ is at most that of $\alpha$; since $\alpha$ has the minimal number of irreducible factors of the elements in $I$, in fact $\delta$ is a unit. Thus $(\alpha,\beta) = (\alpha)$, and we have $\beta \in (\alpha)$. Thus $I = (\alpha)$ is principal, and so $R$ is a principal ideal domain.