Divisors of the product of two coprime integers can be written as the product of two coprimes

Question

In my lecture notes:

Let $m,n\in \mathbb{N}$ be relatively prime. The fundamental theorem of arithmetic implies that each divisor of $mn$ is the product of two unique positive relatively prime integers $d_1|m$ and $d_2|n$.

Please could someone help me understand how this is implied? I have no idea

thanks!

Answer 1

Simply write

$$m=\prod_{i=1}^rp_i^{e_i},\quad n=\prod_{j=1}^sq_j^{f_j}$$

Since they are coprime, no $p_i=q_j$.

A divisor of $mn$ is determined by taking each $p_i$ to some power $0\le n_{p_i}\le e_i$ and $q_j$ to power $0\le n_{q_j}\le f_j$. Grouping the primes $\{p:p|m\}$ and $\{q: q|n\}$ together, we get

$$d_1=\prod_{i=1}^rp_i^{n_{p_i}},\quad d_2=\prod_{j=1}^sq_i^{n_{q_j}}$$

Answer 2

$d_1$ and $d_2$ are not, in general, prime numbers. They are relatively prime (to each other); that means that the highest common factor of $d_1$ and $d_2$ is $1$.