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 $a\mathrm{,}b\mathrm{\in}D$. Then the following are equivalent^{}:

1.
$a,b$ have an lcm,

2.
for any $r\in D$, $ra,rb$ have a gcd.
Proof.
For arbitrary $x,y\in D$, denote $\mathrm{LCM}(x,y)$ and $\mathrm{GCD}(x,y)$ the sets of all lcm’s and all gcd’s of $x$ and $y$, respectively.
$(1\Rightarrow 2)$. Let $c\in \mathrm{LCM}(a,b)$. Then $c=ax=by$, for some $x,y\in D$. For any $r\in D$, since $rab$ is a multiple of $a$ and $b$, there is a $d\in D$ such that $rab=cd$. We claim that $d\in \mathrm{GCD}(ra,rb)$. 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.
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.
Next, let $t$ be any common divisor of $ra$ and $rb$, say $ra=ut$ and $rb=vt$ for some $u,v\in D$. 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 $w\in D$. 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, $d\in GCD(ra,rb)$.
$(2\Rightarrow 1)$. Suppose $k\in \mathrm{GCD}(a,b)$. Write $ki=a$, $kj=b$ for some $i,j\in D$. Set $\mathrm{\ell}=kij$, so that $ab=k\mathrm{\ell}$. We want to show that $\mathrm{\ell}\in \mathrm{LCM}(a,b)$. First, notice that $\mathrm{\ell}=aj=bi$, so that $a\mid \mathrm{\ell}$ and $b\mid \mathrm{\ell}$. Now, suppose $a\mid t$ and $b\mid t$, we want to show that $\mathrm{\ell}\mid t$ as well. Write $t=ax=by$. Then $ta=aby$ and $tb=abx$, so that $ab\mid ta$ and $ab\mid tb$. Since $\mathrm{GCD}(ta,tb)\ne \mathrm{\varnothing}$, we have $tk\in \mathrm{GCD}(ta,tb)$ (see proof of this here (http://planetmath.org/PropertiesOfAGCDDomain)), implying $ab\mid tk$. In other words $tk=abz$ for some $z\in D$. As a result, $tk=abz=k\mathrm{\ell}z$, or $t=\mathrm{\ell}z$. In other words, $\mathrm{\ell}\mid t$, 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, Proposition^{} 1 follows.
Another way of stating Proposition 1 is the following: let $L$ be the set of equivalence classes^{} on the integral domain $D$, where $a\sim b$ iff $a$ and $b$ are associates. Partial order^{} $L$ so that $[a]\le [b]$ iff $ac=b$ for some $c\in D$. Then $L$ is a semilattice (upper or lower) implies that $L$ is a lattice^{}.
Title  an integral domain is lcm iff it is gcd 

Canonical name  AnIntegralDomainIsLcmIffItIsGcd 
Date of creation  20130322 18:19:38 
Last modified on  20130322 18:19:38 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Derivation^{} 
Classification  msc 13G05 