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Let $G$ be a finitely generated group. Let $R = \mathbb Z [G]$ be the group ring of $G$ over $\mathbb Z$. Note $R$ is generally not commutative.

Question: Are there any sufficient conditions when $R$ is a UFD?

I looked up on Wikipedia, and there are not many examples of non-commutative UFDs. I'm wondering if there is much work done on UFD and group rings. I'm particularly interested in the case when $G$ is a finitely generated nilpotent group. (In this case, $R = \mathbb Z [G]$ is a Noetherian.)

Any insight will be really appreciated.

ghc1997
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    What definition of "noncommutative UFD" do you want to use? – rschwieb Jun 09 '25 at 13:06
  • I realize that the linked duplicate is not a verbatim duplicate, but it does suffice to show that even the simpler problem of determining when it has nonzero zero divisors is already difficult, and so one cannot hope to get a good answer for this harder question anytime soon. – rschwieb Jun 09 '25 at 13:11
  • Still need to resolve what you mean by "noncommutative UFD." just saying 'there are not many examples of noncommutative UFDs" does not resolve this. – rschwieb Jun 09 '25 at 13:21
  • Maybe this could be helpful. It mentions that ordered groups, supersolvable groups, polycyclic-by-finite groups, and unique product groups have been determined to make $F[G]$ a domain. I'm not a group theorist so I can't immediately tell if any of those encompass f.g. solvable groups (you can tell me.) Anyhow, this information is not too helpful for sufficient conditions. So now hopefully you can see the difficulty, even when $R$ is a field. – rschwieb Jun 09 '25 at 13:27
  • Thank you, I thought a "noncommutative UFD" meant a noncommutative ring with the unique factorisation property? – ghc1997 Jun 09 '25 at 13:51
  • Can you prove any example of a noncommutative UFD? Since a UFD is, by definition, commutative, presumably "noncommutative" here is a modifier word, rather than an a simple condition. So you have to come up with a modified definition of UFD first. – Thomas Andrews Jun 09 '25 at 13:52
  • It is unclear what that would mean if you allowed zero divisors, for example. And , since you can't reorganize the factorization, it means that you can't have any prime in the center of your ring. If there are any rings with this property, you'd have to show them to me. – Thomas Andrews Jun 09 '25 at 13:56
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    thanks for pointing that out. I think this was discussed by @rschwieb in another post https://math.stackexchange.com/a/3593012/976076 – ghc1997 Jun 09 '25 at 13:59
  • For example, in commutative UFDs, you get , for prime $p,$ $p\mid ab\iff p\mid a$ or $p\mid b.$ In the non-commutative case, divisibility comes in three flavors - left-divisor, right-divisor, or middle-divisor. There is just no use for the concept that I can see for a non-commutative UFD. – Thomas Andrews Jun 09 '25 at 14:00
  • @ThomasAndrews Well, inferring that it is useless seems a bit overkill, but certainly how one discusses factorizations becomes significantly messier, as P. M. Cohn's classic paper on the subject indicates. – rschwieb Jun 09 '25 at 14:13
  • I didn't y he word "useless." I said "no use that I can see" for a reason. @rschwieb – Thomas Andrews Jun 09 '25 at 14:23
  • @ThomasAndrews I'll take your word for it but I'd wager the lion's share of readers would read "If there are any rings with this property, you'd have to show them to me. [...] There is just no use for the concept that I can see" to be of significantly more critical tone than you intended. I guess now they will know either way. Regards – rschwieb Jun 09 '25 at 18:01

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