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

Arturo Magidin
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Mohan
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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 06:51

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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).

Arturo Magidin
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    Note that that silly condition is nontrivial! I'm going my memory here, but IIRC, if a pair of elements has an LCM, then it has a GCD, but the reverse doesn't hold. You need that every pair of elements has a GCD in order to automatically get LCMs as well. (Perhaps you can do it with less? No idea.) – Harry Altman Nov 13 '11 at 10:23
  • @Harry: By "silly" I did not mean "trivial"; but it's like the fact that every nonempty set that is bounded above has a supremum if and only if every nonempty set that is bounded below has an infinimum; in fact, it's essentially the same argument (the gcd of a and b is the lcm of all common divisors; the lcm of a and b is the gcd of all common multiples). It's not really a good way to "characterize" GCD-domains, since it's a closely-related conditions that is usually about as hard to check. – Arturo Magidin Nov 13 '11 at 21:56
  • @ 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
  • This "silly condition" has since been proven at https://math.stackexchange.com/a/2148964/ . – darij grinberg Aug 15 '19 at 22:35
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    @darijgrinberg: No, it’s not quite the same. The “silly condition” I give is that lcms for every pair exist is equivalent to gcds for every pair exist. That is not what is done in that question, where it is proven that the lcm for a particular pair $a$ and $b$ exists is equivalent to gcds for all pairs of the form $ra,rb$ exist. It’s not same assertion. – Arturo Magidin Aug 16 '19 at 02:09
  • @darijgrinberg: As I note in the comment, you can prove that if all lcms exist, then gcds exist because the gcd of $a$ and $b$ is the lcm of all common divisors; and if all gcds exist, then the lcm is the gcd of all common multiples. You can see that it’s a different argument from that given in that link, which is a related but more restricted assertion. – Arturo Magidin Aug 16 '19 at 02:10
  • Yeah, to be more precise it's a more general fact that is proven over there. I'm a bit unsure about your argument, though -- there are infinitely many common multiples, and we don't really want to assume that gcds of infinite sets exist, do we? – darij grinberg Aug 16 '19 at 07:14
  • @darijgrinberg: Okay, you may need to mod out by units, and bound above, say, "all common multiples that divide $ab$". Or extend to "all gcds exist" rather than just pairs. – Arturo Magidin Aug 16 '19 at 14:49