If $R$ is a principal ideal domain, then every proper ideal of $R$ is contained in a maximal ideal of $R$.

Question

Assume a proper ideal $I_{1}$ of $R$ is not contained in any maximal ideal of $R$. Then $I_{1}$ is not maximal since $I_{1}$ contains itself. Then there is an ideal $I_{2}$ such that $I_{1}\subset I_{2}\subset R$. By assumption, $I_{2}$ is not maximal, so there exists an ideal $I_{3}$ such that $I_{1}\subset I_{2}\subset I_{3}\subset R$. By assumption, $I_{3}$ is not maximal. We continue the process of choosing ideals so that we obtain a strictly ascending infinite chain $I_{1}\subset I_{2}\subset \dotsc$ of ideals, which is a contradiction since $R$ is PID, so every strictly ascending chain of ideals of $R$ should be of a finite length.

As you can see here, I did not use the assumption that $R$ is a domain. $R$ being an arbitrary principal ideal ring should suffice to imply a contradiction. It's either I missed the part where I used the "domain", or there is a flaw in my proof, or the assumption that $R$ is a domain is not necessary at all. Can someone enlighten me?

Answer 1

Your proof is correct. Some possibly-enlightening remarks:

1. You have essentially just used that every ascending chain of ideals in a PID must stabilise.
2. A ring in which every ascending chain stabilises is called Noetherian.
3. An equivalent condition for being Noetherian is: Every ideal is finitely generated.
4. In particular, any principal-ideal-ring is Noetherian and thus, every ascending chain stabilises. So you are correct in noting that that is all that you needed.

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Assuming the axiom of choice, your statement is true for any non-zero commutative ring with unity.