# $R$ integral domain where every prime ideal is principal then set of non-principal ideals has maximal element

Let $R$ be an integral domain and suppose that every prime ideal in $R$ is principal. Assume that the set of ideals of $R$ that are not principal is nonempty and prove that this set has a maximal element under inclusion. (Use Zorn's Lemma.)

Let $A$ denote the set of all nonprincipal ideals of $R$. I know that $A$ is a collection of sets, so it is partially ordered by set inclusion. I want to say that $A$ has a chain $\{C_\alpha\}_{\alpha\in B}$, where $B$ is an index set, but I want to know why I can say that a chain exists in this case.

I know that every ideal $I\in A$ is nonprincipal, so $I$ is generated by at least two elements. Also, $I$ cannot be prime. Thus, if $ab \in I$, where $a,b \in R$, then $a \notin I$ and $b \notin I$. Would this help me form a chain?

To apply Zorn's lemma, you assume you have such a chain, and show that it has an upper bound in $A$. The burden is for you to prove the upper bound exists. Once you do, Zorn's lemma takes over and gives you the existence of maximal elements in the poset.
Incidentally, the maximal element you find in $A$ is necessarily prime, hence principal (by the assumption of this problem) providing a contradiction that tells you that $A$ is empty. This is where you're headed, right?