any divisor is gcd of two principal divisors

Using the exponent valuations, one can easily prove the

Theorem.  In any divisor theory, each divisor is the greatest common divisor of two principal divisors.

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

$$𝔡=\prod _{i=1}^r{𝔭}_i^{k_i}$$

where the ${𝔭}_i$’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 $𝒪$ such that

$${\nu }_{{𝔭}_1}(\alpha )=k_1,\mathrm{\dots },{\nu }_{{𝔭}_r}(\alpha )=k_r.$$

Let the principal divisor corresponding to $\alpha$ be

$$(\alpha )=\prod _{i=1}^r{𝔭}_i^{k_i}\prod _{j=1}^s{𝔮}_j^{l_j}=𝔡{𝔡}^{\prime },$$

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

$${\nu }_{{𝔭}_1}(\beta )=k_1,\mathrm{\dots },{\nu }_{{𝔭}_r}(\beta )=k_r,{\nu }_{{𝔮}_1}(\beta )=\mathrm{\dots }={\nu }_{{𝔮}_s}(\beta )=0.$$

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

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

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

Title	any divisor is gcd of two principal divisors
Canonical name	AnyDivisorIsGcdOfTwoPrincipalDivisors
Date of creation	2013-03-22 17:59:37
Last modified on	2013-03-22 17:59:37
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	5
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13A05
Classification	msc 13A18
Classification	msc 12J20
Related topic	TwoGeneratorProperty
Related topic	SumOfIdeals