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Assume $R$ is a ring and $(a)$ and $(b)$ are principal ideals in $R$. Under what conditions is the product of the ideals equal to the principal ideal generated by the product, i.e. when is $(a)(b)=(ab)$?

My guess is that they're guaranteed to be equal if $R$ is a U.F.D. (or maybe just an integral domain?), but I do not know how to prove this for an arbitrary ring. I can't think of any counterexamples that would suggest my intuition to be wrong though. Any thoughts?

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    $\begingroup$ Are you assuming your ring is commutative? $\endgroup$ Commented Feb 27, 2018 at 21:34
  • $\begingroup$ I suppose. The example that I'm trying to apply it to is $(x^3-2x) = (x)(x^2-2) \subseteq R = \mathbb{Q}[x]$. I just want to know when we can break $(ab)$ up as $(a)(b)$, since it seems to be a pretty useful fact. $\endgroup$ Commented Feb 27, 2018 at 21:45

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For any commutative ring, it is elementary to show $(a)(b)=(ab)$.

$ab\in (a)(b)$, so $(ab)\subseteq (a)(b)$.

Everything in $(a)(b)$ is of the form $\sum_{i=1}^nar_ibs_i=ab(\sum_{i=1}^nr_is_i)\in (ab)$, so $(a)(b)\subseteq (ab)$.

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    $\begingroup$ Ah I see where that matches up with my proof for the example I'm working on. All we need is commutativity to guarantee $(a)(b) \subseteq (ab)$. $\endgroup$ Commented Feb 27, 2018 at 21:48
  • $\begingroup$ Is the commutative property enough? Do we need to assume the ring has a multiplicative identity to get the result? Thanks! $\endgroup$
    – Sky subO
    Commented May 1, 2021 at 23:00
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    $\begingroup$ @SkysubO most people assume identity by default. But even without, and by using the “right” definition of principal ideal, the result still holds. My expression of “every element looks like” would need to be adapted to a more complicated statement. $\endgroup$
    – rschwieb
    Commented May 1, 2021 at 23:14
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    $\begingroup$ @rschwieb Aha! I understand what you mean by "the right definition". I believe things, as you say, just get a little more complicated. It is surprising that this simple and beautiful result requires so few conditions, even without a identity. Thank you very much! $\endgroup$
    – Sky subO
    Commented May 1, 2021 at 23:47

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