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?


The ideal $R = (1)$ is always a principal ideal of the ring $R$, but $R$ is not necessarily a PID.

  • $\begingroup$ Yes. This example has crossed my mind. Any chance you could point out where I messed up in my proof? $\endgroup$ – A. Wong Nov 7 '14 at 9:57
  • $\begingroup$ @A.Wong When you write "the elements of $I$ are of the form $\alpha(...)$", all you've proven is that $I$ is included in $(\alpha)$. You cannot conclude from that that $(a_1, a_2) = (\alpha)$; for that you would need to prove that all the elements of the form $\alpha x$ are in $I$. $\endgroup$ – Najib Idrissi Nov 7 '14 at 9:58
  • $\begingroup$ Ahhhh. I see. Thanks. Not sure how that escaped me.. $\endgroup$ – A. Wong Nov 7 '14 at 9:59

A principal ideal can certainly contain a non-principal ideal. Consider, for example, the ring of polynomials with integer coefficients. The ideal generated by $14$ and $7x$ is not principal, but it is contained in the principal ideal generated by $7$.


Try to show that if there exist an ideal which is not principal then there will exist an prime ideal which is not principal. So take the collection of ideal which are not principal,by our assumption this collection would be non-empty.Now take a chain in this collection and apply zorn's lemma.Suppose P is maximal element in that collection.So P will be non-principal.Now try to show that this P is also prime.For this you can use the concept of colon ideal.So P is aprime ideal which is not principal,this is a contradiction to the given condition.

  • $\begingroup$ I have the basic idea on how to solve the exercise. My question pertains to the proof of my claim. $\endgroup$ – A. Wong Nov 7 '14 at 8:27

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