An integral domain whose every prime ideal is principal is a principal ideal domain

Let R be an integral domain, and suppose that every prime ideal in R is principal.

  1. Suppose R contains a nonprincipal ideal. Show that R contains an inclusion-maximal nonprincipal ideal
  2. Let M be an inclusion-maximal nonprincipal ideal, and let a,b \in R such that ab \in M while a,b \notin M. (These exist because M is not prime by hypothesis.) Let M_a = (M,a) and M_b = (M,b), and define N = \{r \in R \ |\ rM_a \subseteq M \}. Prove that M_a = (\alpha) and M_b = (\beta) are principal, with M \subsetneq M_b \subseteq N and M_aN = (\alpha\beta) \subseteq M.
  3. If x \in M, show that x = s\alpha for some s \in N. Deduce that M = M_aN is principal, a contradiction.

  1. Let \mathcal{A} denote the set of nonprincipal ideals of R. Suppose \mathcal{A} is nonempty. Note that \mathcal{A} is partially ordered by inclusion. Let \{C_i\}_\mathbb{N} be a chain in \mathcal{A}, and let C = \bigcup C_i. We know that C is an ideal of R. Suppose now that C is principal; say, C = (\gamma). We have \gamma \in C_k for some k, so that C \subseteq C_k, and thus C = C_k. But then C_k \in \mathcal{A} is principal, a contradiction. Thus C is not principal, and so the chain \{C_i\}_\mathbb{N} has an upper bound in \mathcal{A}. By Zorn’s lemma, \mathcal{A} contains an inclusion-maximal element; let M be such an ideal.
  2. Since a \notin M, we have M properly contained in M_a = (M,a). Similarly, M is properly contained in M_b. Since M is maximal with respect to nonprincipalness, both M_a = (\alpha) and M_b = (\beta) are principal ideals. Note moreover that M_bM_a = (M,b)(M,a) = (M^2,Ma,Mb,ab) \subseteq M, so that M_b \subseteq N. Note that M_aN \subseteq M by the definition of N, using the fact that R is commutative. Note also that M_aN is principal, being the product of principal ideals.
  3. Let x \in M. Since M \subseteq M_a = (\alpha), we have x = s\alpha for some s \in R. However, note that sM_a = s(\alpha) = (s\alpha) = (x) \subseteq M, so that in fact s \in N. Thus M \subseteq M_aN, and in particular M is principal. This contradicts the maximal nonprincipalness of M, so that our original assumption that \mathcal{A} is nonempty was faulty.

    Thus R contains no nonprincipal ideals and is (by definition) a principal ideal domain.

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