Let $a_1,a_2,b_1$ and $b_2$ be elements in an integral domain $D$. Show that $a_1|b_1$ if and only if $a_2\mid b_2$

Asked Jun 14 '21 at 10:30

Active Jun 14 '21 at 18:10

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We may assume that $a_1$ and $a_2$ are associated, and that $b_1$ and $b_2$ are associated.

So here is what I've got so far:

Lets assume that $a_1|b_1$. By definition, this is: $b_1=a_1\cdot c$ for some $c\in D$.

By associativity, $a_1=a_2\cdot u_1$ and $b_1=b_2\cdot u_2$, we can rewrite this as:

$b_2\cdot u_2=(a_2\cdot u_1)\cdot c$

Where $u_1$ and $u_2$ are units in $D$.

Now here is where I am stuck. I might be missing something related to units of groups, but if I could only get them to magically disappear from this equation I would have $a_2|b_2$. Any pointers?

edited Jun 14 '21 at 18:10

Bill Dubuque

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asked Jun 14 '21 at 10:30

Max

Note that $u_2^{-1}\in D$. – Jun 14 '21 at 16:10
1

Simply scale your final equation by $,u_2^{-1}.,$ Generally $, c\mid ud\iff c\mid d,$ when $u$ is a unit. – Bill Dubuque Jun 14 '21 at 18:12
@BillDubuque So if I express it as $b_2=(a_2\cdot u_1)\cdot c\cdot u_2^{-1}$ this means that I can simply neglect the units? Sorry, divisibility is not my strong suit apparently. :( – Max Jun 15 '21 at 08:21
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You have $,b_2 = a_2 (u_1 c, u_2^{-1}),$ so $,a_2\mid b_2,,$ since $,u_1 c, u_2^{-1}\in D.\ \ $ – Bill Dubuque Jun 15 '21 at 08:36
@BillDubuque Oh, that makes sense! Thank you :) – Max Jun 15 '21 at 08:48

Let $a_1,a_2,b_1$ and $b_2$ be elements in an integral domain $D$. Show that $a_1|b_1$ if and only if $a_2\mid b_2$

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