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.

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