# Prove that if $b \mid c$ then $ab \mid c$?

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?

• The ring you give in your example is actually a field, so every element is divisible by every other nonzero element. – Matt Samuel Jun 11 '15 at 20:19
• $5\cdot 5=4$ ${}{}{}$ – Zircht Jun 11 '15 at 20:19
• 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? – Geoffrey Critzer Jun 11 '15 at 20:23
• 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. – Noah Schweber Jun 11 '15 at 20:27
• Yes. thanks for this point. I see that Matt Samuel has the proof now. – Geoffrey Critzer Jun 11 '15 at 20:30

If $b\mid c$ then $$c=bd$$ for some $d$. Furthermore, $$ac=abd$$ so $$c=ab(a^{-1}d)$$ Hence $ab\mid c$.
• 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." – Noah Schweber Jun 11 '15 at 20:25
Hint $\,\ ab\mid \underbrace{a^{-1}ab}_{\large b}\mid c\,\Rightarrow\,ab\mid c\,\$ by transitivity of divisibility.