Exercise A0.V.2.02

Proposition

Let $R$ be an UFD and let $a,b,c \in R$ such that $a \mid bc$ and $\gcd(a,b) = 1.$ Then $a \mid c.$

Proof

Since $a \mid bc$ there exists $v \in R$ such that $va = bc.$ Hence $\mathsf{F}(v) + \mathsf{F}(a) = \mathsf{F}(b) + \mathsf{F}(c).$ Since $\gcd(a,b) = 1,$ $\mathsf{F}(a) \cap \mathsf{F}(b) = \emptyset.$ Thus $\mathsf{F}(b) \subseteq \mathsf{F}(v).$

Let $u$ be such that $\mathsf{F}(u) + \mathsf{F}(b) = \mathsf{F}(v).$ Then $\mathsf{F}(u) + \mathsf{F}(b) + \mathsf{F}(a) = \mathsf{F}(b) + \mathsf{F}(c)$ and so $\mathsf{F}(u) + \mathsf{F}(a) = \mathsf{F}(c)$ which implies that there exists a unit $j$ such that $jua = c.$

This establishes $a \mid c.$

$\blacksquare$