When $m,n$ are co-prime integers and $d_1d_2|mn$, show that $\gcd(\frac {m} {d_1} , \frac {n} {d_2})=1$ In my number theory notes, I come across a theorem that if $m,n$ are co-prime integers, then the divisors of $mn$ are of the form $d_1d_2$, where $d_1 | m$ and $d_2 | n$. 
Later in the notes, the author states that for $d_1d_2|mn$ that $\gcd(\frac {m} {d_1} , \frac {n} {d_2})=1$. I'm trying to figure out why this is the case. 
If $p$ is a prime number dividing $m/d_1$ and $n/d_2$, then I should find some contradiction, hints appreciated.
 A: If $(m,n)=1$, there are integers $x, y$ such that $mx+ny =1$.  Then
$$\frac{m}{d_1}(d_1x) + \frac{n}{d_2}(d_2y) =1.$$
Since $d_1x$ and $d_2y$ are integers, you have your result.
A: The OP essentially asserts that divisors of coprimes remain coprime - which is the  special case below where $\,(m,n) = 1,\,$ with the divisors $\,\bar m  = m/d_1,\,\ \bar n  = n/d_2,\, $  of $\,m,n$.
Lemma $\ \begin{align} \bar m &\mid m\\ \bar n &\mid n\end{align}$ $\Rightarrow\, (\bar m ,\bar n )\mid (m,n),\ \ $  where $\ (x,y) :=\gcd(x,y)$
Proof $\ \ \ \begin{align}&(\bar m ,\bar n )\mid \bar m \mid m\\ &(\bar m ,\bar n )\mid \bar n \,\mid n\end{align}\Rightarrow\, (\bar m ,\bar n )\mid (m,n)\ $ by the  GCD Universal Property, i.e.
$$ c\mid a,b\iff c\mid (a,b)\qquad$$
Remark $ $ This universal property is the definition of GCD in more general rings where the Bezout identity need not hold true,  e.g. in the polynomial UFD rings  $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y]\,$ where the gcds $\,(x,2) = 1 = (x,y)\,$ cannot be written as Bezout linear combinations. There the above proof still works, but Bezout-based proofs do not.

The theorem that you mention follows immediately from a more general fundamental property $$\ a\mid BC\:\Rightarrow a=bc,\ b\mid B,\ c\mid C$$  This goes by many names, e.g. Euler's four number theorem (Vierzahlensatz), Schreier refinement, Riesz interpolation, etc. For much further discussion of this and related factorization concepts see this answer.
