I am trying to show that if $R$ is an integral domain such that every prime ideal of $R$ is principal then every ideal of $R$ is principal.
To start this, suppose that $P$ is the set of ideals of $R$ that are not principal. Suppose that $P$ is nonempty. I am trying to apply Zorn's Lemma to show that $P$ has a maximal element and to do this, I want to show that every chain in $P$ is bounded above by $R$ itself.
This leads to my question of whether a principal ideal can contain a non-principal ideal.
Claim: Suppose that $R$ is a principal. Then any ideal $I$ of $R$ is a principal ideal.
Proof: $R$ is principal so for some element $\alpha\in R$, $R=(\alpha)$. Suppose that $I$ is any ideal of $R$. Furthermore, suppose that $I$ is generated by two elements, say $a_{1}$ and $a_{2}$ so $I=(a_{1},a_{2})$. Since $I\subseteq R$, $a_{1},a_{2}\in R$ so since $R=(\alpha)$ there exist elements $\beta_{1},\beta_{2}\in R$ such that $\alpha\beta_{1}=a_{1},\alpha\beta_{2}=a_{2}$. Then since any element of $I$ is of the form $a_{1}b_{1}+a_{2}b_{2}$ where $b_{1},b_{2}\in R$, it follows that the elements of $I$ are of the form $\alpha(\beta_{1}b_{1}+\beta_{2}b_{2})$ where $\beta_{1}b_{1}+\beta_{2}b_{2}$. Since any element of $I$ is of this form, $I=(a_{1},a_{2})=(\alpha)$ so $I$ is a principal ideal.
I then move onto the next part of the exercise which says to show that if $I$ is a non-principal ideal in $R$, then $I_{a}=(I,a)$ for some $a\in R$ is a principal ideal. $I\subseteq I_{a}$ so principal ideals can contain non-principal ideals. Where did I go wrong in the proof of my claim?
Best Answer
The ideal $R = (1)$ is always a principal ideal of the ring $R$, but $R$ is not necessarily a PID.