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

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

Okay so here is the proof I came up with. Please be as critical as possible of the proof I need to have a very rigid understanding of this concept.

($\Rightarrow$) Since $b\mid c$ then there exists some $n\in R$ such that $bn = c$.

Since $a$ is a unit we can rewrite this as $abna^{-1} = c$.

Thus we know that $na^{-1}$ is an element of $R$, so rename it as $x$.

So $abx = c$. Therefore $(ab)x =c$. Thus $ab\mid c$.

($\Leftarrow$) Assume $ab\mid c$, then there exists some integer $x\in R$ such that $abx = c$.

Now since $a$ is a unit we can conclude that $abx = bx$, therefore $bx = c$, so $b\mid c$.

• In your last lines: how do you come to $abc=bx$ on base of $abx=c$? Btw, the equality $abx=c$ tells us immediately that $b\mid c$. If $r_1\cdots r_k=c$ then every $r_i$ divides $c$. You don't need the fact that $a$ is a unit there. Also you should mention where it is used that $R$ is commutative (e.g. when you are saying that $abna^{-1}=bn=c$). – drhab Apr 18 '15 at 6:09
• There it was a typo I have corrected my mistake, okay thank you for the input....I will post a revised version of the proof shortly – B ry Apr 18 '15 at 14:48
• --> Since b/c then there exists some n that is an element of the ring R, such that bn = c. Since a is a unit and the ring R is commutative we can rewrite this as abna^-1 = c. Thus we know that na^-1 is an element of R, so rename it as x. So abx = c. Therefore, by commutativity, (ab)x =c. Thus ab/c. <-- Assume ab/c, then there exists some integer x in R such that abx = c. Now we can conclude that abx = b(ax), and we know that axis an element of R, therefore b/c. – B ry Apr 18 '15 at 14:53
• "some integer $x\in R$...." Elements of $\mathbb Z$? – drhab Apr 18 '15 at 14:55

The $(\Rightarrow)$ part is good.
The $(\Leftarrow)$ part is wrong, unfortunately. You can't say that $abx=bx$.
Assume $ab\mid c$; then, for some $x\in R$, $(ab)x=c$ and so $a(bx)=c$. Then $b(ax)=c$ and therefore $b\mid c$.
Note that this part doesn't require $a$ is a unit.