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This is an exercise in a text I am reading.

Let $a$, $b$, and $c$ be elements of a commutative ring, and suppose that $a$ is a unit. Prove that $b$ divides $c$ if and only if $ab$ divides $c$.

If $ab$ divides $c$ then there is no problem proving that $b$ divides $c$. However I do not believe the converse is true. Consider the commutative ring $(\mathbb{Z}_7, +_7, \cdot_7)$. Letting $b=2$, $c=4$, and $a = 6$, we have $2 \mid 4$ but $2 \cdot_7 6 = 5 \not\mid 4$.

Am I missing something or is there an error in my textbook?

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    $\begingroup$ The ring you give in your example is actually a field, so every element is divisible by every other nonzero element. $\endgroup$ – Matt Samuel Jun 11 '15 at 20:19
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    $\begingroup$ $5\cdot 5=4$ ${}{}{}$ $\endgroup$ – Zircht Jun 11 '15 at 20:19
  • $\begingroup$ OK I think I see this now. So if b|c there is an x such that bx=c and since a has an inverse we can say a(a^-1*bx) =c. Does that look right? $\endgroup$ – Geoffrey Critzer Jun 11 '15 at 20:23
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    $\begingroup$ Almost but not quite - this is where commutativity comes in. To conclude $ab\vert c$ we need some $y$ such that $aby=c$. Taking $y=a^{-1}x$ works if the ring is commutative, since then we have $aby=aba^{-1}x=aa^{-1}bx=bx=c,$ but without commutativity this doesn't necessarily work. $\endgroup$ – Noah Schweber Jun 11 '15 at 20:27
  • $\begingroup$ Yes. thanks for this point. I see that Matt Samuel has the proof now. $\endgroup$ – Geoffrey Critzer Jun 11 '15 at 20:30
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If $b\mid c$ then $$c=bd$$ for some $d$. Furthermore, $$ac=abd$$ so $$c=ab(a^{-1}d)$$ Hence $ab\mid c$.

The ring you give in your example is a field, so every nonzero element divides every other element.

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    $\begingroup$ A good exercise for the OP: show that (1) the assumption that $a$ is a unit can't be removed, and (2) the property "For all $b, c$, $b\vert c\iff ab\vert c$" is actually equivalent to "$a$ is a unit." $\endgroup$ – Noah Schweber Jun 11 '15 at 20:25
  • $\begingroup$ For(1) would you agree that a valid counter example could be R=({0,1,2,3,4,5,6,7}, mod 8 addition and multiplication). Letting b=3, c=6 and a =4 (not a unit). Then 3|6 but 3*4==4 does not divide 6. $\endgroup$ – Geoffrey Critzer Jun 11 '15 at 20:47
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    $\begingroup$ @GeoffreyCritzer Yes that's fine. $\endgroup$ – Matt Samuel Jun 11 '15 at 20:50
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5 divides 4 so there is no error.

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Hint $\,\ ab\mid \underbrace{a^{-1}ab}_{\large b}\mid c\,\Rightarrow\,ab\mid c\,\ $ by transitivity of divisibility.

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