Let m and n be positive integers with (m,n) = 1. Prove that each divisor d > 0 of m*n can be written uniquely as d1*d2, where d1,d2 > 0, d1 | m, d2 | n, and (d1,d2) = 1 and each such prouduct d1*d2 corresponds to a divisor d of m*n
I'm really struggling with this one. All I have is the fact that m cannot equal n and that their prime factorization's were different so I figured that since this were the case and they were relatively prime then I could say there must be two d's that divide both of these
Let $m = \prod p_i^{a_i}$ be the prime factorization of $m$ and $n = \prod q_i^{b_i}$ be the prime factorization of $n$.
As $\gcd(m,n) = 1$ we know the primes $\{p_i\}$ and $\{q_i\}$ distinct and disjoint.
So $mn = \prod p_i^{a_i}\prod q_i^{b_i}$ is the unique prime factorization of $mn$.
If $d$ is a factor then $d = \prod p_i^{c_i}\prod q_i^{e_i}$ for some $0 \le c_i \le a_i; 0\le e_i < b_i$.
The $d_1 = \prod p_i^{c_i}$ and $d_2 = \prod q_i^{e_i}$ are precisely the factors asked for. They are unique as $d = \prod p_i^{c_i}\prod q_i^{e_i}$ is a unique prime factorization of $d$.