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Consider the principal ideal formed by the product of two elements: $(x\cdot y)$.

Under what circumstances is this:

  • Equal to the product of the two principal ideals $(x)$ and $(y)$?
  • Equal to the intersection of the two principal ideals $(x)$ and $(y)$?

I'm interested in something that holds for commutative rings in general, especially when there are zero divisors.

Mike Battaglia
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  • Where are your thoughts on this matter, Mike? Which sources and/or references have you considered and read that deal with your questions? Answers to my two questions belong under the umbrella of "context", which we expect from all askers. Where did you encounter these questions, and/or, how did these questions arise for you? etc...etc...etc... – amWhy May 22 '17 at 00:43

1 Answers1

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Always $(xy)=(x)(y)$ in a commutative ring $R$.

If $x$ and $y$ are coprime, that is $(x)+(y)=R$ then $(xy)=(x)\cap(y)$. One way is clear. For the other direction, there are $a$ and $b$ with $1=ax+by$. If $c\in (x)\cap (y)$ then $c=axc+byc$. As $c\in (y)$, $xc\in (xy)$ and $axc\in(xy)$. Similarly $byc\in (xy)$. So $c\in(xy)$.

Angina Seng
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  • Don't certain conditions need to hold for $x,y$ coprime implying $(x), (y)$ coprime? This answer says this doesn't generally hold outside of PIDs, but I may be confusing some things here. – Mike Battaglia May 03 '17 at 20:05
  • Just a small comment: In the question, the OP speaks about ideals being isomorphic. Note that in an integral domain, all principal ideals are isomorphic, i.e. $(2) \cap (4) \cong (8)$ would be true, while of course $(2) \cap (4) = (8)$ is false. Why did I say this? To remind the OP that there is a huge difference between submodules/ideals being equal or isomorphic. – MooS May 04 '17 at 06:17
  • Thanks, edited. I'll accept this as the answer once I understand if the restriction I said about PIDs is true. – Mike Battaglia May 04 '17 at 18:39
  • @LordSharktheUnknown How did you get (xy) = (x)(y) – Vinny Chase Dec 08 '17 at 20:05