an integral domain is lcm iff it is gcd
This is an immediate consequence of the following
Let be an integral domain and . Then the following are equivalent:
have an lcm,
for any , have a gcd.
For arbitrary , denote and the sets of all lcm’s and all gcd’s of and , respectively.
. Let . Then , for some . For any , since is a multiple of and , there is a such that . We claim that . There are two steps: showing that is a common divisor of and , and that any common divisor of and is a divisor of .
Since , the equation reduces to , so divides . Similarly, , so is a common divisor of and .
Next, let be any common divisor of and , say and for some . Then , so that is a multiple of both and , and hence is a multiple of , say for some . Then the equation reduces to . Multiplying both sides by gives . Since , we have , or , so that is a multiple of .
As a result, .
. Suppose . Write , for some . Set , so that . We want to show that . First, notice that , so that and . Now, suppose and , we want to show that as well. Write . Then and , so that and . Since , we have (see proof of this here (http://planetmath.org/PropertiesOfAGCDDomain)), implying . In other words for some . As a result, , or . In other words, , as desired. ∎
Since the first statement is equivalent to being an lcm domain, and the second statement is equivalent to being a gcd domain, Proposition 1 follows.
Another way of stating Proposition 1 is the following: let be the set of equivalence classes on the integral domain , where iff and are associates. Partial order so that iff for some . Then is a semilattice (upper or lower) implies that is a lattice.