Does $a$ divide $c$ if $a$ divides $b$ and $b$ divides $c$?

Question

Suppose R is a commutative ring with identity. Let $a$, $b$, $c\in R$. If $a\mid b$ and $b\mid c$, does $a\mid c$?

I am able to prove this statement using integers as the set $\mathbb{Z}$ of integers is closed under multiplication but I am not too sure if this statement holds when dealing with Rings. Does this have to do with the associates or units in $R$?

You can write down a proof for $\mathbb Z$ and replace all instances of $\mathbb Z$ by $R$ to get the proof in the general setting. It works exactly the same. — Suzet, Nov 19 '19 at 05:09
As @LordSharktheUnknown says, you should start from the definition of the notation $a\mid b$. — justadzr, Nov 19 '19 at 05:15

score 0 · Answer 1 · answered Nov 19 '19 at 20:31

By definition of divisibility in a commutative ring: $x \mid y$ if and only if $\exists a$ so that $ax = y$.

Thus: $a \mid b$ implies that $ax = b$ for some $x$. Similarly, $b \mid c \Rightarrow by = c$ for some $y$. Plug everything in to get:

$$axy = c$$

Thus there does exists some $z$ (to be precise, $z = xy$, which is in $R$ since rings are closed under multiplication) so that $az = c$, and $a \mid c$ by definition.

Does $a$ divide $c$ if $a$ divides $b$ and $b$ divides $c$?

1 Answers1