Lemma 10.27.11. Let $R$ be a ring.

1. An ideal $I \subset R$ maximal with respect to not being principal is prime.

2. If every prime ideal of $R$ is principal, then every ideal of $R$ is principal.

Proof. The first part follows from Example 10.27.5 and Proposition 10.27.8. For the second, suppose that there exists an ideal $I \subset R$ which is not principal. The union of a totally ordered chain $\left\{ I_\alpha \right\}$ of ideals that not principal is not principal; indeed, if $I = \bigcup I_\alpha$ were generated by $a$, then $a$ would belong to some $I_\alpha$ and $a$ would generate it. By Zorn's lemma, there is an ideal maximal with respect to not being principal. This ideal is necessarily prime by the first part. $\square$

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