This question already has an answer here:

Here is the problem If $c\ |\ ab$ and $\text{gcd}(a,c)=1$ then $c\ |\ b$ Here's my approach. There exists $x,y$ such that $ax+cy=1$, so $c\ |\ axb+cyb=b$ I'm pretty sure my first step is wrong. Any help would be appreciated!


marked as duplicate by Bill Dubuque abstract-algebra Jul 26 at 14:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Your proof is fine. $\endgroup$ – Wuestenfux Jul 26 at 6:10
  • $\begingroup$ Note $c|axb+cyb \Leftrightarrow c|(ab)x$ $\endgroup$ – Kai Jul 26 at 6:17
  • 3
    $\begingroup$ @Kai Bézout's Identity i.e. $ax + cd = 1$ fails in arbitrary UFDs. You cannot use that result. $\endgroup$ – 0XLR Jul 26 at 6:19

It is not true that $\gcd(a, c) = 1$ implies that there exist $x, y \in R$ with $ax + cy = 1$. (For example when $R = \mathbb Z[X]$, $a = 2$, $c = X$.)

Instead work with the factorizations of $a, b, c$ into irreducibles.

  • $\begingroup$ Oh! Yeah, so all irreducible factors of $c$ doesn't divide $a$, so by definition of UFD, they divide $b$. Is that right? $\endgroup$ – Kai Jul 26 at 6:19
  • $\begingroup$ Indeed. All prime powers dividing $c$ divide $b$. $\endgroup$ – punctured dusk Jul 26 at 6:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.