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Homedivisibility by product

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# 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 mean 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 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.

## Mathematics Subject Classification

11A51*no label found*13A05

*no label found*

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