# corollary of Bézout’s lemma

###### Theorem.

If $\mathrm{gcd}(a,c)=1$ and $c|ab$, then $c|b$.

Proof. Bézout’s lemma (http://planetmath.org/BezoutsLemma) gives the integers $x$ and $y$ such that $xa+yc=1$. This implies that $xab+ybc=b$, and because here the both summands are divisible by $c$, so also the sum, i.e. $b$, is divisible by $c$ .

Note. A similar theorem holds in all Bézout domains (http://planetmath.org/BezoutDomain), also in Bézout rings.

Title | corollary of Bézout’s lemma |
---|---|

Canonical name | CorollaryOfBezoutsLemma |

Date of creation | 2013-03-22 14:48:16 |

Last modified on | 2013-03-22 14:48:16 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 15 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11A05 |

Synonym | Euclid’s lemma |

Synonym | product divisible but factor coprime |

Related topic | GreatestCommonDivisor |

Related topic | DivisibilityInRings |

Related topic | DivisibilityByProduct |