divisibility by product

Theorem.

Let $R$ be a Bézout ring, i.e. a commutative ring with non-zero unity where every finitely generated ideal is a principal ideal. If $a,b,c$ are three elements of $R$ such that $a$ and $b$ divide $c$ and  $\gcd (a,b)=1$,  then also $ab$ divides $c$.

Proof. The divisibility assumptions that  $c=a{a}_1=b{b}_1$  where $a_1$ and $b_1$ are some elements of $R$.  Because $R$ is a Bézout ring, there exist such elements $x$ and $y$ of $R$ that  $\gcd (a,b)=1=xa+yb$. This implies the equation  $a_1=xa{a}_1+yb{a}_1=xb{b}_1+yb{a}_1$  which shows that $a_1$ is divisible by $b$, i.e.  $a_1=b{b}_2$,  $b_2\in R$. Consequently,  $c=a{a}_1=ab{b}_2$,  or  $ab\mid c$  Q.E.D.

Note 1. The theorem may by induction be generalized for several factors (http://planetmath.org/Divisibility) of $c$.

Note 2. The theorem holds e.g. in all Bézout domains, especially in principal ideal domains, such as $\mathbb{Z}$ and polynomial rings over a field.