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[parent] divisibility by product (Theorem)
Theorem 1   Let $ R$ be a Bézout ring, i.e. a commutative ring with non-zero unity where every finitely generated ideal is a principal ideal. If $ a,\,b,\,c$ are three elements of $ R$ such that $ a$ and $ b$ divide $ c$ and $ \gcd(a,\,b) = 1$, then also $ ab$ divides $ c$.

Proof. The divisibility assumptions mean that $ c = aa_1 = bb_1$ where $ a_1$ and $ b_1$ are some elements of $ R$. Because $ R$ is a Bézout ring, there exist such elements $ x$ and $ y$ of $ R$ that $ \gcd(a,\,b) = 1 = xa+yb$. This implies the equation $ a_1 = xaa_1+yba_1 = xbb_1+yba_1$ which shows that $ a_1$ is divisible by $ b$, i.e. $ a_1 = bb_2$, $ b_2\in R$. Consequently, $ c = aa_1 = abb_2$, or $ ab \mid c$ Q.E.D.

Note 1. The theorem may by induction be generalized for several factors of $ c$.

Note 2. The theorem holds e.g. in all Bézout domains, especially in principal ideal domains, such as $ \mathbb{Z}$ and polynomial rings over a field.



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See Also: Bezout domain, product divisible but factor coprime, corollary of Bézout's lemma

Also defines:  Bézout ring
Keywords:  Bézout ring

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Cross-references: field, polynomial rings, principal ideal domains, Bézout domains, induction, divisible, equation, implies, divisibility, proof, divide, principal ideal, ideal, finitely generated, non-zero unity, commutative ring
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This is version 10 of divisibility by product, born on 2004-11-22, modified 2008-03-18.
Object id is 6513, canonical name is DivisibilityByProduct.
Accessed 2183 times total.

Classification:
AMS MSC: 11A51 (Number theory :: Elementary number theory :: Factorization; primality)
 13A05 (Commutative rings and algebras :: General commutative ring theory :: Divisibility)
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