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*→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=r∏i=1𝔭kii |
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)=r∏i=1𝔭kiis∏j=1𝔮ljj=𝔡𝔡′, |
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)=𝔡𝔡′′, 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 |