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 ?


  • 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).

  • 2
    $\begingroup$ Why complicate when the contrapositive is so much clear and has a standard proof? $\endgroup$
    – lhf
    Feb 12 at 13:16
  • $\begingroup$ 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. $\endgroup$
    – GreginGre
    Feb 12 at 16:21
  • $\begingroup$ See math.stackexchange.com/questions/3002390/… $\endgroup$
    – lhf
    Feb 12 at 16:43
  • $\begingroup$ $K[X^2,X^3]$ is not a UFD and its ideal $\langle X^2,X^3 \rangle$ is not principal. $\endgroup$
    – lhf
    Feb 12 at 23:20

1 Answer 1


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.


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