# divisibility by prime number

###### Theorem.

Let $a$ and $b$ be integers and $p$ any prime number^{}. Then we have:

$p\mid ab\mathit{\hspace{1em}}\iff \mathit{\hspace{1em}}p\mid a\vee p\mid b$ | (1) |

Proof. Suppose that $p\mid ab$. Then either $p\mid a$ or $p\nmid a$. In the latter case we have $\mathrm{gcd}(a,p)=1$, and therefore the corollary of Bézout’s lemma gives the result $p\mid b$. Conversely, if $p\mid a$ or $p\mid b$, then for example $a=mp$ for some integer $m$; this implies that $ab=mb\cdot p$, i.e. $p\mid ab$.

Remark 1. The theorem means, that if a product is divisible by a prime number, then at least one of the factor is divisibe by the prime number. Also conversely.

Remark 2. The condition (1) is expressed in of principal ideals^{} as

$(ab)\subseteq (p)\mathit{\hspace{1em}}\iff \mathit{\hspace{1em}}(a)\subseteq (p)\vee (b)\subseteq (p).$ | (2) |

Here, $(p)$ is a prime ideal^{} of $\mathbb{Z}$.

Title | divisibility by prime number |
---|---|

Canonical name | DivisibilityByPrimeNumber |

Date of creation | 2013-03-22 14:48:18 |

Last modified on | 2013-03-22 14:48:18 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 18 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11A05 |

Synonym | divisibility by prime |

Related topic | PrimeElement |

Related topic | DivisibilityInRings |

Related topic | EulerPhiAtAProduct |

Related topic | RepresentantsOfQuadraticResidues |