# 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  $c=aa_{1}=bb_{1}$  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$a_{1}=xaa_{1}+yba_{1}=xbb_{1}+yba_{1}$  which shows that $a_{1}$ is divisible by $b$, i.e.  $a_{1}=bb_{2}$,  $b_{2}\in R$. Consequently,  $c=aa_{1}=abb_{2}$,  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.

Title divisibility by product DivisibilityByProduct 2013-03-22 14:50:37 2013-03-22 14:50:37 pahio (2872) pahio (2872) 13 pahio (2872) Theorem msc 11A51 msc 13A05 BezoutDomain ProductDivisibleButFactorCoprime CorollaryOfBezoutsLemma Bézout ring