# 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 $\mathrm{gcd}(a,b)=1$, 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 $\mathrm{gcd}(a,b)=1=xa+yb$. 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 |