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I'm working through this question

If the intersection of $x\mathbb{Z}$ and $y\mathbb{Z}$ equals $w\mathbb{Z}$ for positive integers $x$, $y$, and $w$ (ideals in ring $\mathbb{Z}$), express the integer $w$ in terms of $x$ and $y$.

I'm just trying to back track and check if this could be properly expressed with the relationship: $\text{LCM}(x,y) = w$.

Any help would be appreciated.

Ted
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Se yaJ Me
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  • Can you explain why $w = \operatorname{lcm} (x,y)$? – Daniel Fischer Feb 04 '14 at 22:26
  • Because anything that is in both xZ and yZ will be an integer multiple of both x and y. Therefore, if w is the least common multiple of the two, w times any integer will be a multiple of both x and y. I'm sure if this only works for sets or if this line of thinking carries with ideals in rings – Se yaJ Me Feb 04 '14 at 22:30
  • You need to work on the formulation a bit, but the idea is right. So you know that $w = \operatorname{lcm} (x,y)$, and why. This line of reasoning works as is in unique factorisation domains, not in general rings, since the $\operatorname{lcm}$ of two elements generally need not exist. – Daniel Fischer Feb 04 '14 at 22:38

1 Answers1

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Hint $\,\ z \in x{\rm Z}\cap y{\rm Z}\iff z\in x{\rm Z},\, y{\rm Z}\iff x,\,y\mid z\color{#c00}\iff {\rm lcm}(x,y)\mid z\iff z\in {\rm lcm}(x,y){\rm Z}$

Remark $\ $ I presume that $\, Z\,$ means $\,\Bbb Z,\,$ or a domain where lcms exist, e.g. a UFD or gcd-domain. The equivalence $\color{#c00}\iff$ is the universal property of lcm (which is the definition of lcm in general domains).

Bill Dubuque
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