Proving that $a \mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb Z$ 
Given $m \in \mathbb{Z}$, let $m\mathbb{Z}$ denote the set of integer multiples of $m$, i.e. $m\mathbb{Z} := \{mk\mid k \in \mathbb{Z}\}$.
  Now let $a,b \in \mathbb{Z}$ with $a,b$ not both $0$. Prove that $a\mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb{Z}$.

I am trying to write a proof for this, but I am unsure of what method to use. Also I am confused by $mk\mid k$, because wouldn't $m=1$ for this to be true. 
 A: Hint: 
Step 1: Can you prove that $\operatorname{lcm}(a,b)\in a\mathbb{Z}\cap b\mathbb{Z}$?
Step 2: Can you prove that if $a$ and $b$ both divide $c$, then $c\in a\mathbb{Z}\cap b\mathbb{Z}$?
Step 3: How do steps 1 and 2, together, imply your result?
A: First of all, to clear up your confusion:
$$m\mathbb Z=\{mk\mid k\in\mathbb Z\}$$
does not mean that $mk$ divides $k$. The vertical line can be read as "so that" or "where". It means that the set $m\mathbb Z$ is the set of numbers in the form of $mk$, where $k$ is any element of $\mathbb Z$.
That said, to prove that $a\mathbb Z\cap b\mathbb Z=\operatorname{lcm}(a,b)\mathbb Z$, you need to prove that 


*

*if $x\in a\mathbb Z\cap b\mathbb Z$, then $x\in \operatorname{lcm}(a,b)\mathbb Z$.

*if $x\in\operatorname{lcm}(a,b)\mathbb Z$, then $x\in a\mathbb Z\cap b\mathbb Z$.

A: I give another demonstration you might find useful, which allows you to reach the result without double implication, using only the definition of $lcm(a,b)$.

Definition: $d \in \Bbb Z$ is defined least common multiple of $a$ and $b$, in symbols $d := lcm(a,b)$ if:


*

*$\exists m,n \in \Bbb Z$ such that $am = d = bn$, otherwise written as $a|d$ and $b|d$

*If $c \in \Bbb Z$ is such that $a|c$ and $b|c$, then $d|c$



So we can write
$$a\Bbb Z \cap b\Bbb Z =_{(1)} \left \{ x \in \Bbb Z : a|x, b|x \right \} =_{(2)} \left \{ x \in \Bbb Z : lcm(a,b)|x \right \} =_{(3)} lcm(a,b)\Bbb Z$$
Where:


*

*$=_{(1)}, =_{(3)}$ follows from the definition of $a\Bbb Z \cap b\Bbb Z$ and $lcm(a,b)\Bbb Z$

*$=_{(2)}$ follows from the point $2$ of the definition of $lcm(a,b)$

A: Well firstly, how do you define $\text{lcm}(a,b)$? 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
