# 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 DivisorAsFactorOfPrincipalDivisor 2013-03-22 18:02:09 2013-03-22 18:02:09 pahio (2872) pahio (2872) 9 pahio (2872) Theorem msc 13A05 msc 11A51 EveryIdealInADedekindDomainIsAFactorOfAPrincipalIdeal