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!
2026-03-27 11:49:07.1774612147
Let $R$ be a UFD and $a,b,c \in R$ be nonzero. If $c \mid ab$ and $\gcd(a, c) = 1$, then $c \mid b$.
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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.