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.

Title	divisibility by product
Canonical name	DivisibilityByProduct
Date of creation	2013-03-22 14:50:37
Last modified on	2013-03-22 14:50:37
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	13
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11A51
Classification	msc 13A05
Related topic	BezoutDomain
Related topic	ProductDivisibleButFactorCoprime
Related topic	CorollaryOfBezoutsLemma
Defines	Bézout ring