Lemma 10.28.11. 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 10.28.5 and Proposition 10.28.8. 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. $\square$

## Comments (0)

There are also:

• 6 comment(s) on Section 10.28: A meta-observation about prime ideals

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05KH. Beware of the difference between the letter 'O' and the digit '0'.