What is the necessary and sufficient condition for an integral domain to have gcd for every pair of elements and why?
Existence of GCD in an integral domain
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$\begingroup$
abstract-algebra
ring-theory
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1Well, it has to be a "gcd-domain", which is defined *precisely* as an integral domain in which every pair of elements has a gcd. Unique factorization is sufficient, but is not necessary (e.g., the ring of all algebraic integers). Every finitely generated ideal being principal (Bezout domain) is sufficient, but again not necessary (e.g., $\mathbb{Z}[x]$). In other words, you are asking for characterizations of GCD-domains. – 2011-11-13
1 Answers
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For a silly condition: every pair of elements in a domain $D$ has a gcd if and only if every pair of elements in $D$ has an lcm.
More seriously, a good survey is:
GCD domains, Gauss' Lemma, and content of polynomials by D.D. Anderson. In Non-Noetherian commutative ring theory, pages 1-13. Math. Appl. 520, Kluwer Acad, Publ., Dordrecht, 2000. MR 1858155 (2002g:13039).
Though, in general, the equivalent conditions are probably not what you are looking for (for example, one of the equivalent conditions is that a domain $D$ is a GCD-domain if and only if the group of divisibility of $D$ is a complete lattice ordered group).
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0@ Arturo Magidin, I have small comment on the last paragraph of your answer. I think $D$ is a GCD domain if and only if group of divisibility is a lattice ordered group. This is the theorem by Krull, Kaplansky and Jaffard. – 2013-05-26