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Question: Let $D$ be a PID. Show that $d\in D$ is a least common multiple of $a$ and $b$ iff $(a)\cap(b)=(d)$.

Attempt/Thoughts: First, let's define a least common multiple: Suppose $a,b,c,d\in D$. $c$ is a least common multiple of $a$ and $b$ if $a\mid c$ and $b\mid c$, and if $a\mid d$ and $b\mid d$ then $c\mid d$.
Next, in ideals, $a\mid b\iff b=ua\iff b\in(a)\iff (b)\subseteq(a)$.

For the forward direction, let $d\in D$ be a LCM of $a$ and $b$. Then $a|d$ and $b|d$ thus $(d)\subseteq (a)$ and $(d)\subseteq (b)$, so $(d)\subseteq (a)\cap(b)$. $\textbf{but how do I show $(a)\cap(b)\subseteq (d)$?}$

For the backward direction, suppose $(a)\cap(b)=(d)$. Then, $(d)\subseteq (a)$ and $(d)\subseteq (b)$. So, $b\mid d$ and $b\mid a$. If there exists a $c\in D$ such that $a\mid c$ and $b\mid c$, then $c\in (a)\cap(b)$, $\textbf{but can I then just say that $d\mid c$ and be done?}$

Bonnaduck
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User7238
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1 Answers1

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$\textbf{but how do I show $(a)\cap(b)\subseteq (d)$?}$

To show that $(a)\cap (b) \subseteq (d)$, suppose $x\in (a)\cap (b)$. Then $a\mid x$ and $b\mid x$, hence $d\mid x$ by definition of $d$. Thus $x\in (d)$, as desired.

If there exists a $c\in D$ such that $a|c$ and $b|c$, then $c\in (a)\cap(b)$,$\textbf{but can I then just say that $d|c$ and be done?}$

You should first note that $c\in (d)$ because $(a)\cap (b) = (d)$. Then, claim $d\mid c$ from the fact that $c\in (d)$.

kobe
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