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  $\gcd (a,p)=1$,  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