divisor as factor of principal divisor

Let an integral domain $𝒪$ have a divisor theory  ${𝒪}^{*}\to 𝔇$.  The definition of divisor theory (http://planetmath.org/DivisorTheory) implies that for any divisor $𝔞$, there exists an element $\omega$ of $𝒪$ such that $𝔞$ divides the principal divisor $(\omega )$, i.e. that  $𝔞𝔠=(\omega )$  with $𝔠$ a divisor.  The following theorem states that $𝔠$ may always be chosen such that it is coprime with any beforehand given divisor.

Theorem.  For any two divisors $𝔞$ and $𝔟$, there is a principal divisor $(\omega )$ such that

$$𝔞𝔠=(\omega )$$

and

$$\gcd (𝔟,𝔠)=(1).$$

Proof.  Let  ${𝔭}_1,\mathrm{\dots },{𝔭}_s$  all distinct prime divisors, which divide the product $𝔞𝔟$, and let the divisor $𝔞$ be exactly divisible (http://planetmath.org/ExactlyDivides) by the powers  ${𝔭}_1^{a_1},\mathrm{\dots },{𝔭}_s^{a_s}$ (the cases  $a_i=0$  are not excluded).  For each  $i=1,\mathrm{\dots },s$,  we choose a nonzero element ${\alpha }_i$ of $𝒪$ being exactly divisible by the power ${𝔭}_i^{a_i}$; the choosing is possible, since any nonzero element of the ideal determined by the divisor ${𝔭}_i^{a_i}$, not belonging to the sub-ideal determined by the divisor ${𝔭}_i^{a_i+1}$, will do.  According to the Chinese remainder theorem (http://planetmath.org/ChineseRemainderTheoremInTermsOfDivisorTheory), there exists a nonzero element $\omega$ of the ring $𝒪$ such that

$\omega \equiv {\alpha }_{i}mod{𝔭}_i^{a_i+1}\mathit{\hspace{1em}}(i=\mathrm{\hspace{0.17em}1},\mathrm{\dots },s).$(1)

Because ${\alpha }_i$ is divisible by ${𝔭}_i^{a_i}$, the element $\omega$ is divisible by  ${𝔭}_1^{a_1}\mathrm{\cdots }{𝔭}_s^{a_s}=𝔞$,  i.e.  $(\omega )=𝔞𝔠$.  If one of the divisors ${𝔭}_i$ would divide $𝔠$, then $(\omega )$ would be divisible by ${𝔭}_i^{a_i+1}$ and thus by (1), also ${\alpha }_i$ were divisible by ${𝔭}_i^{a_i+1}$.  Therefore, no one of the prime divisors  ${𝔭}_1,\mathrm{\dots },{𝔭}_s$  divides $𝔠$.  On the other hand, every prime divisor dividing the divisor $𝔟$ divides $𝔞𝔟$ and thus is one of  ${𝔭}_1,\mathrm{\dots },{𝔭}_s$.  Accordingly, the divisors $𝔟$ and $𝔠$ have no common prime divisor, i.e.  $\gcd (𝔟,𝔠)=(1)$.