Let $n, m ∈ Z$ (integer set) , $(n, m) = 1$. Suppose that $d$ is a positive divisor of $mn$. Show that there exist positive integers $d_1$ and $d_2$ such that $d =$ $d_1$$d_2$ where $d_2$ divides $n$ and $d_2$ divides $m$.
Could someone explain how to approach this proof? I would greatly appreciate it!
My Attempt:
Since it is assumed that d is a positive divisor of $mn$ we get $d_1$ and $d_2$, but I don't understand how $d_2$ divides $n$ and $m$, this is where I get confused.
For every prime power $p^a$ that divides $d$, we have $p^a \mid mn$. If $p \mid m$, then $p \nmid n$, and vice versa, because $(m,n) = 1$; so all the $p$s must be in only one of $m$ and $n$. We get that $p^a$ divides either $m$ or $n$. Since we're doing this for every $p^a$ that divides $d$, we will eventually assign each factor to either $m$ or $n$, and we will get the $d_1$ and $d_2$ required.