any divisor is gcd of two principal divisors


Using the exponent valuations, one can easily prove the

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

Proof.  Let  𝒪*→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=∏i=1r𝔭iki

where the 𝔭i’s are pairwise distinct prime divisors and every ki>0.  Then third condition in the theorem concerning divisors and exponents allows to choose an element α of the ring 𝒪 such that

ν𝔭1⁢(α)=k1,…,ν𝔭r⁢(α)=kr.

Let the principal divisor corresponding to α be

(α)=∏i=1r𝔭iki⁢∏j=1s𝔮jlj=𝔡⁢𝔡′,

where the prime divisors 𝔮j are pairwise different among themselves and with the divisors 𝔭i.  We can then choose another element β of 𝒪 such that

ν𝔭1⁢(β)=k1,…,ν𝔭r⁢(β)=kr,ν𝔮1⁢(β)=…=ν𝔮s⁢(β)=0.

Then we have  (β)=𝔡⁢𝔡′′,  where  𝔡′′∈𝔇  and

gcd⁡(𝔡′,𝔡′′)=𝔮0⁢⋯⁢𝔮0=𝔢=(1).

The gcd of the principal divisors (α) and (β) 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