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 $\mathbb{Z}$.