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

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?