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 $[a,b]$?
 A: 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
A: $\newcommand{\lcm}{\operatorname{lcm}}$

$$a\mid b\iff a\Bbb Z\supset b\Bbb Z$$

P First, suppose $b=ak$. Pick $x\in b\Bbb Z$. Then $x=by=a(ky)\in a\Bbb Z$. Conversely, if $a\Bbb Z\geq b\Bbb Z$ we have $b\in a\Bbb Z$ so $b=ak$, $a\mid b$.

$$a\Bbb Z\cap b\Bbb Z=\ell\Bbb Z\;,\; \ell=\lcm(a,b)$$

P Since $a\mid \ell$, $a\Bbb Z \supset \ell \Bbb Z$. Similarily,  $b\Bbb Z \supset \ell \Bbb Z$, so $a\Bbb Z\cap b\Bbb Z\supset \ell \Bbb Z$. Now pick $x\in a\Bbb Z\cap b\Bbb Z$. Then $x=ak=bj$ for some $k,j$. Thus $a,b\mid x$. So $x$ is a common multiple, whence $ \ell\mid x$, that is $\ell =xm$ for some $m$ and $x\in\ell\Bbb Z$

$$a\Bbb Z+b\Bbb Z=d\Bbb Z\;,\;d=\gcd(a,b)$$

P Pick $y\in a\Bbb Z+b\Bbb Z$. Then $y=am+bn$. But then $d\mid y$, since $d\mid a,b$, and thus $y=kd\in d\Bbb Z$. Now pick $y\in d\Bbb Z$. Then $y=dk$ for some $k$. Bezout tells us we can write $d=an+bm$ so $y=adn+bdm\in a\Bbb Z+b\Bbb Z$.
A: Well firstly, how do you define $[a,b]$ aka $\text{lcm}(a,b)$? I'll define it by the converse of a proposition in Artin Algebra (Prop 2.3.8)



Namely, we'll prove the converse of Prop 2.3.8 where $m:=\text{lcm}(a,b)$ is defined by the integer s.t.
(a) $m$ is divisible by both $a$ and $b$
(b) If $n$ is divisible by $a$ and $b$, then $n$ is divisible by $m$.
Pf: $(\subseteq)$
Let $n \in \mathbb Z m$. Then there is an integer $n_m$ s.t. $n_m = \frac n m$. Observe that $n_m = \frac{n_a}{m_a} = \frac{n_b}{m_b}$ where we define $n_a := \frac n a, m_a := \frac m a, m_b := \frac m b, n_b := \frac n b$. Observe that $m_a, m_b$ are integers by assumption (a) while we want to show that $n_a, n_b$ are integers because showing such is equivalent to showing $n \in \mathbb Za \cap \mathbb Zb$.
Now, $n_m m_a = n_a$ is a product of integers and hence an integer. The same is true for $n_m m_b = n_b$. Therefore, $n_a, n_b$ are integers and thus, $n \in \mathbb Za \cap \mathbb Zb$
$(\supseteq)$
This one is easier. Let $n \in \mathbb Za \cap \mathbb Zb$. Then $n_a, n_b$ as defined earlier are integers, i.e. $n$ is divisible by both $a$ and $by$. By assumption (b), $n$ is divisible by $m$.
QED
