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).

Title	divisor as factor of principal divisor
Canonical name	DivisorAsFactorOfPrincipalDivisor
Date of creation	2013-03-22 18:02:09
Last modified on	2013-03-22 18:02:09
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	9
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13A05
Classification	msc 11A51
Related topic	EveryIdealInADedekindDomainIsAFactorOfAPrincipalIdeal