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According to wikipedia, all principal ideal domains are also unique factorization domains -- conversely, if a ring $R$ is not a UFD then it is not a PID.

If we suppose that $R$ is a ring which is not a UFD, can we immediately construct an ideal which is not principal in $R$ without any additional information?

EDIT: For a look at the ring(s) in question see this paper, keeping in mind that I was incorrect in the abstract and these rings are not Euclidean domains, nor are they Noetherian. To see this, consider that $$(X^{\omega-n})_{n<\omega}$$ is a countable ascending chain of ideals in the naked polynomial ring $\mathbb{Q}_\infty[X^{\mathbb{Z}_\infty^+}]$, and more generally for any $\gamma$-number $\lambda$ we have that $$(X^{\lambda-n})_{n<\lambda}$$ is an ascending chain of ideals of length $\lambda$.

Alec Rhea
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  • Depends what you mean by "immediately", and by "additional information". Often it works, yes. In general, see this question. – Dietrich Burde Jan 09 '18 at 14:05
  • @DietrichBurde I'm currently trying to understand a ring which I can show is not a Dedekind domain because it is not Noetherian, so it can't be a PID, however I am having difficulty constructing any non-principal ideals. What I'm really wondering is: if I specify that $R$ is a ring which is not a UFD and is not a Dedekind domain, can we abstractly construct a non-principal ideal that I could realize in the ring I'm curious about to better understand it? – Alec Rhea Jan 09 '18 at 14:23
  • You could edit your question and show us your non-Noetherian ring you have. The you could ask a more precise question. – Dietrich Burde Jan 09 '18 at 14:26
  • @DietrichBurde At that point I suspect the question would fit better over on MO (I thought this might be an easy question as I am relatively new to commutative ring theory), but I will edit as you suggest. For a reference on the rings in question, they are the naked polynomial rings from this paper: https://arxiv.org/abs/1712.00662 . Note that I was incorrect in the abstract as these rings are not Euclidean domains, but being Euclidean was a nuke for the problem at hand so I am now trying to establish the structure I need more carefully. – Alec Rhea Jan 09 '18 at 14:29
  • Interesting paper! – Dietrich Burde Jan 09 '18 at 14:34
  • @DietrichBurde Thank you! – Alec Rhea Jan 09 '18 at 14:36

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