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 $a b$ 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.

单词	DivisibilityByProduct
释义	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 $a b$ 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.
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