# divisor as factor of principal divisor

Let an integral domain $\mathcal{O}$ have a divisor theory β$\mathcal{O}^{*}\to\mathfrak{D}$.β 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. thatβ $\mathfrak{ac}=(\omega)$β with $\mathfrak{c}$ a divisor.β The following theorem states that $\mathfrak{c}$ may always be chosen such that it is coprime with any beforehand given divisor.

Theorem.β 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.β Letβ $\mathfrak{p}_{1},\,\ldots,\,\mathfrak{p}_{s}$β 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 powersβ $\mathfrak{p}_{1}^{a_{1}},\,\ldots,\,\mathfrak{p}_{s}^{a_{s}}$ (the casesβ $a_{i}=0$β are not excluded).β For eachβ $i=1,\,\ldots,\,s$,β 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.β 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 byβ $\mathfrak{p}_{1}^{a_{1}}\cdots\mathfrak{p}_{s}^{a_{s}}=\mathfrak{a}$,β i.e.β $(\omega)=\mathfrak{ac}$.β 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 $\mathfrak{p}_{i}^{a_{i}+1}$.β Therefore, no one of the prime divisorsβ $\mathfrak{p}_{1},\,\ldots,\,\mathfrak{p}_{s}$β divides $\mathfrak{c}$.β On the other hand, every prime divisor dividing the divisor $\mathfrak{b}$ divides $\mathfrak{ab}$ and thus is one ofβ $\mathfrak{p}_{1},\,\ldots,\,\mathfrak{p}_{s}$.β Accordingly, the divisors $\mathfrak{b}$ and $\mathfrak{c}$ have no common prime divisor, i.e.β $\gcd(\mathfrak{b},\,\mathfrak{c})=(1)$.

## References

• 1 Π. Π. ΠΠΎΡΡΠ½ΠΈΠΊΠΎΠ²: ΠΠ²Π΅Π΄Π΅Π½ΠΈΠ΅β Π²β ΡΠ΅ΠΎΡΠΈΡβ Π°Π»Π³Π΅Π±ΡΠ°ΠΈΡΠ΅ΡΠΊΠΈΡβ ΡΠΈΡΠ΅Π». βΠΠ·Π΄Π°ΡΠ΅Π»ΡΡΡΠ²ΠΎβ ββΠΠ°ΡΠΊΠ°ββ. ΠΠΎΡΠΊΠ²Π°β(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