# any divisor is gcd of two principal divisors

Using the exponent valuations, one can easily prove the

Proof.  Let  $\mathcal{O}^{*}\to\mathfrak{D}$  be a divisor theory and $\mathfrak{d}$ an arbitrary divisor in $\mathfrak{D}$.  We may suppose that $\mathfrak{d}$ is not a principal divisor (if $\mathfrak{D}$ contains exclusively principal divisors, then  $\mathfrak{d}=\gcd(\mathfrak{d},\,\mathfrak{d})$  and the proof is ready).  Let

 $\mathfrak{d}=\prod_{i=1}^{r}\mathfrak{p}_{i}^{k_{i}}$

where the $\mathfrak{p}_{i}$’s are pairwise distinct prime divisors and every $k_{i}>0$.  Then third condition in the theorem concerning divisors and exponents allows to choose an element $\alpha$ of the ring $\mathcal{O}$ such that

 $\nu_{\mathfrak{p}_{1}}(\alpha)=k_{1},\;\;\ldots,\;\;\nu_{\mathfrak{p}_{r}}(% \alpha)=k_{r}.$

Let the principal divisor corresponding to $\alpha$ be

 $(\alpha)=\prod_{i=1}^{r}\mathfrak{p}_{i}^{k_{i}}\prod_{j=1}^{s}\mathfrak{q}_{j% }^{l_{j}}=\mathfrak{dd}^{\prime},$

where the prime divisors $\mathfrak{q}_{j}$ are pairwise different among themselves and with the divisors $\mathfrak{p}_{i}$.  We can then choose another element $\beta$ of $\mathcal{O}$ such that

 $\nu_{\mathfrak{p}_{1}}(\beta)=k_{1},\;\;\ldots,\;\;\nu_{\mathfrak{p}_{r}}(% \beta)=k_{r},\;\;\nu_{\mathfrak{q}_{1}}(\beta)=\ldots=\nu_{\mathfrak{q}_{s}}(% \beta)=0.$

Then we have  $(\beta)=\mathfrak{dd}^{\prime\prime}$,  where  $\mathfrak{d}^{\prime\prime}\in\mathfrak{D}$  and

 $\gcd(\mathfrak{d}^{\prime},\,\mathfrak{d}^{\prime\prime})=\mathfrak{q}^{0}% \cdots\mathfrak{q}^{0}=\mathfrak{e}=(1).$

The gcd of the principal divisors $(\alpha)$ and $(\beta)$ is apparently $\mathfrak{d}$, whence the proof is settled.

Title any divisor is gcd of two principal divisors AnyDivisorIsGcdOfTwoPrincipalDivisors 2013-03-22 17:59:37 2013-03-22 17:59:37 pahio (2872) pahio (2872) 5 pahio (2872) Theorem msc 13A05 msc 13A18 msc 12J20 TwoGeneratorProperty SumOfIdeals