Proving $a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}$ The question is 

Prove that $a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}$.
  Hint:First prove $b\mid a \Leftrightarrow a\mathbb Z\le b\mathbb Z$ and then prove $a\mathbb Z+b\mathbb Z=(a,b)\mathbb Z$

I managed to prove $b\mid a\Leftrightarrow a\mathbb Z\le b\mathbb Z$. About the next claim: From the first statement ($b\mid a...$) it follows that since $(a,b)\mid a$ and $(a,b)\mid b, a\mathbb Z+b\mathbb Z\le(a,b)\mathbb Z$. My problem is proving the opposite side (a.k.a $a\mathbb Z+b\mathbb Z\ge(a,b)\mathbb Z$). How can I do so and how can I proceed from $(a,b)$ to $[a,b]$?

An idea:
Denoting by $\,p_n\,$ a prime that divides an integer $\,n\,$ , we have
$$m\in a\Bbb Z\cap b\Bbb Z\implies m=ar=bs\;,\;r,s\in\Bbb Z\implies \begin{cases}p_a\mid bs\;\forall\,p_a\\{}\\p_b\mid ar\;\;\forall\,p_b\end{cases}\;\implies$$
$$\implies m=[a,b]t\;,\;t\in\Bbb Z\implies m\in [a,b]\Bbb Z$$
since we know that $\;[a,b]=\frac{ab}{(a,b)}\;$ (and thus every prime common to both $\,a,b\,$ appears both in $\,ar\,$ and in $\;bs\;$).
The other direction is trivial
