an integral domain is lcm iff it is gcd

Proposition 1.

Let D be an integral domain. Then D is a lcm domain iff it is a gcd domain.

This is an immediate consequence of the following

Proposition 2.

Let D be an integral domain and a,bD. Then the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath:

  1. 1.

    a,b have an lcm,

  2. 2.

    for any rD, ra,rb have a gcd.


For arbitrary x,yD, denote LCM(x,y) and GCD(x,y) the sets of all lcm’s and all gcd’s of x and y, respectively.

(12). Let cLCM(a,b). Then c=ax=by, for some x,yD. For any rD, since rab is a multiple of a and b, there is a dD such that rab=cd. We claim that dGCD(ra,rb). There are two steps: showing that d is a common divisor of ra and rb, and that any common divisor of ra and rb is a divisor of d.

  1. 1.

    Since c=ax, the equation rab=cd=axd reduces to rb=xd, so d divides rb. Similarly, ra=yd, so d is a common divisor of ra and rb.

  2. 2.

    Next, let t be any common divisor of ra and rb, say ra=ut and rb=vt for some u,vD. Then uvt=rav=rbu, so that z:=av=bu is a multiple of both a and b, and hence is a multiple of c, say z=cw for some wD. Then the equation axw=cw=z=av reduces to xw=v. Multiplying both sides by t gives xwt=vt. Since vt=rb=xd, we have xd=xwt, or d=wt, so that d is a multiple of t.

As a result, dGCD(ra,rb).

(21). Suppose kGCD(a,b). Write ki=a, kj=b for some i,jD. Set =kij, so that ab=k. We want to show that LCM(a,b). First, notice that =aj=bi, so that a and b. Now, suppose at and bt, we want to show that t as well. Write t=ax=by. Then ta=aby and tb=abx, so that abta and abtb. Since GCD(ta,tb), we have tkGCD(ta,tb) (see proof of this here (, implying abtk. In other words tk=abz for some zD. As a result, tk=abz=kz, or t=z. In other words, t, as desired. ∎

Since the first statement is equivalent to D being an lcm domain, and the second statement is equivalent to D being a gcd domain, PropositionPlanetmathPlanetmath 1 follows.

Another way of stating Proposition 1 is the following: let L be the set of equivalence classesMathworldPlanetmath on the integral domain D, where ab iff a and b are associates. Partial orderMathworldPlanetmath L so that [a][b] iff ac=b for some cD. Then L is a semilattice (upper or lower) implies that L is a latticeMathworldPlanetmath.

Title an integral domain is lcm iff it is gcd
Canonical name AnIntegralDomainIsLcmIffItIsGcd
Date of creation 2013-03-22 18:19:38
Last modified on 2013-03-22 18:19:38
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 10
Author CWoo (3771)
Entry type DerivationPlanetmathPlanetmath
Classification msc 13G05