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

Question

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?

Thank you in advance.

Answer 1

You don't need to form any chain or prove any such chain exists.

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.

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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?