Divisors of coprime numbers

Setup. Let $d_1=\gcd(d,a)$ and $d_2=d/d_1$. Then $d_1\mid a$.

Show $d_2\mid b$. Since $d\mid ab$ and $d_1\mid a$, we have $d_2=d/d_1\mid (a/d_1)b$. Since $\gcd(d_1,d_2)=1$ and $\gcd(a/d_1,b)=1$ (as $\gcd(a,b)=1$), Gauss's lemma gives $d_2\mid b$.

Coprimality. $\gcd(d_1,d_2)=1$ by construction.

Conclusion. Every divisor of $ab$ splits uniquely as a product of coprime divisors of $a$ and $b$.