divisor as factor of principal divisor

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

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

\\mathfrak{ac}\\;=\\;(\\omega)

and

\\gcd(\\mathfrak{b},\\,\\mathfrak{c})\\;=\\;(1).

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

\\displaystyle\\omega\\;\\equiv\\;\\alpha_{i}\\mod{\\mathfrak{p}_{i}^{a_{i}+1}}\\quad(i%\\,=\\,1,\\,\\ldots,\\,s).

(1)

Because $\\alpha_{i}$ is divisible by $\\mathfrak{p}_{i}^{a_{i}}$ , the element $\\omega$ is divisible by $\\mathfrak{p}_{1}^{a_{1}}\\cdots\\mathfrak{p}_{s}^{a_{s}}=\\mathfrak{a}$ , i.e. $(\\omega)=\\mathfrak{ac}$ . If one of the divisors $\\mathfrak{p}_{i}$ would divide $\\mathfrak{c}$ , then $(\\omega)$ would be divisible by $\\mathfrak{p}_{i}^{a_{i}+1}$ and thus by (1), also $\\alpha_{i}$ were divisible by $\\mathfrak{p}_{i}^{a_{i}+1}$ . Therefore, no one of the prime divisors $\\mathfrak{p}_{1},\\,\\ldots,\\,\\mathfrak{p}_{s}$ divides $\\mathfrak{c}$ . On the other hand, every prime divisor dividing the divisor $\\mathfrak{b}$ divides $\\mathfrak{ab}$ and thus is one of $\\mathfrak{p}_{1},\\,\\ldots,\\,\\mathfrak{p}_{s}$ . Accordingly, the divisors $\\mathfrak{b}$ and $\\mathfrak{c}$ have no common prime divisor, i.e. $\\gcd(\\mathfrak{b},\\,\\mathfrak{c})=(1)$ .

References

1 М. М. Постников:Введение в теорию алгебраических чисел. Издательство ‘‘Наука’’. Москва (1982).