Let be an integral domain, and suppose that every prime ideal in is principal.
- Suppose contains a nonprincipal ideal. Show that contains an inclusion-maximal nonprincipal ideal
- Let be an inclusion-maximal nonprincipal ideal, and let such that while . (These exist because is not prime by hypothesis.) Let and , and define . Prove that and are principal, with and .
- If , show that for some . Deduce that is principal, a contradiction.
- Let denote the set of nonprincipal ideals of . Suppose is nonempty. Note that is partially ordered by inclusion. Let be a chain in , and let . We know that is an ideal of . Suppose now that is principal; say, . We have for some , so that , and thus . But then is principal, a contradiction. Thus is not principal, and so the chain has an upper bound in . By Zorn’s lemma, contains an inclusion-maximal element; let be such an ideal.
- Since , we have properly contained in . Similarly, is properly contained in . Since is maximal with respect to nonprincipalness, both and are principal ideals. Note moreover that , so that . Note that by the definition of , using the fact that is commutative. Note also that is principal, being the product of principal ideals.
- Let . Since , we have for some . However, note that , so that in fact . Thus , and in particular is principal. This contradicts the maximal nonprincipalness of , so that our original assumption that is nonempty was faulty.
Thus contains no nonprincipal ideals and is (by definition) a principal ideal domain.