# Existence of GCD in an integral domain

What is the necessary and sufficient condition for an integral domain to have gcd for every pair of elements and why?

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Well, 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. –  Arturo Magidin Nov 13 '11 at 6:51

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.
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).
@ 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. –  Rajesh May 26 '13 at 20:01