Explicit proof of the fact that a domain which is not a UFD is not a PID

In the same spirit as this question, I would like to prove explicitely that if $$R$$ is a domain which is not a UFD, then it is not a PID.

I am interested in the case where there is an element $$a\in R$$ which has at least two non equivalent decompositions.

Without any loss of generality, we may assume that $$a=u \pi_1\cdots,\pi_r=u' \pi'_1\cdots\pi's,$$ where $$r,s\geq 1$$, and for all $$i,j$$ ,$$\pi$$ and $$\pi'_j$$ are not associate (the case $$r=0$$ or $$s=0$$ is not possible since we want two distinct decompositions)

What I can prove. I can prove that there exists a $$j$$ such that $$(\pi_1,\pi'_j)$$ is not a principal ideal .

Sketch. Assume to the contrary that $$(\pi_1,\pi_j')$$ is generated by some $$\alpha_j$$, for all $$j$$. Then $$\alpha_j$$ is a common divisor of $$\pi_1$$ and $$\pi_j'$$, so it is invertible since these irreducible elements are non associate. Thus $$(\pi_1,\pi'_j)=R$$ for all $$j$$. Hence there is a Bézout relation between $$\pi_1$$ and $$\pi'_j$$, and multipliying all these relations and rearranging yield a Bézout relation between $$\pi_1$$ and $$a$$. Since $$\pi_1$$ divides $$a$$, this yields the contradiction that $$\pi_1$$ is invertible.

My question is: is this result the best resut we may expect? More explicitely:

Question. Assume $$a=u \pi_1\cdots,\pi_r=u' \pi'_1\cdots\pi's,$$ where $$r,s\geq 1$$, and for all $$i,j$$ ,$$\pi$$ and $$\pi'_j$$ are not associate

• Is true that for all $$j$$, the ideal $$(\pi_1,\pi_j')$$ is not a principal ideal ?

or

• can we produce an example where $$(\pi_1,\pi'_k)=R$$ for some $$k$$ and $$(\pi_1,\pi'_j)\neq R$$ for some $$j$$ ?

In this case, it would be nice to have an example with $$R=\mathbb{Z}[\sqrt{d}]$$. I would prefer $$d\not\equiv 1 \ [4]$$ (because if $$d\equiv 1 \ [4]$$, i know another method to produce an explicit non principal ideal).

• Why complicate when the contrapositive is so much clear and has a standard proof?
– lhf
Feb 12 at 13:16
• As already explained in the link provided in the post, the point is not to prove that PID $\Rightarrow$ UFD. The point is to make the proof of the contrapositive non UFD $\Rightarrow$ non PID explicit, by exhibiting a concrete non principal ideal. In other words, I am looking for a systematic way to produce explicit counter examples. Feb 12 at 16:21
• – lhf
Feb 12 at 16:43
• $K[X^2,X^3]$ is not a UFD and its ideal $\langle X^2,X^3 \rangle$ is not principal.
– lhf
Feb 12 at 23:20

I finally found an explicit example, proving that the principality of $$(\pi_1,\pi'_j)$$ depends on $$j$$.
Let $$R=\mathbb{Z}[\sqrt{-21}]$$. Then $$2,3, 7,11,1\pm \sqrt{-21}$$ are irreducible, pairwise non associate, and $$2\cdot 3\cdot 7\cdot 11=-(\sqrt{-21})^2(1-\sqrt{-21})(1+\sqrt{-21})$$.
Then, there is no Bézout relation $$2z+z'(1-\sqrt{-21})=1$$ (otherwise, multiply by $$1+\sqrt{-21}$$ to get a contradiction) , so $$(2,1-\sqrt{-21})$$ is not principal.
However, we have $$2\cdot 11+(\sqrt{-21})^2=1$$, so $$(2,\sqrt{-21})$$ is a principal ideal.