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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?

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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
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up vote 2 down vote accepted

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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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
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