Let be a unique factorization domain and let be nonzero elements. Prove that and have a least common multiple, and describe one such multiple in terms of the factorizations of and .
We begin with some remarks on the elements of a UFD. Let be a unique factorization domain, and let be a set of irreducibles such that (1) every irreducible in is a multiple of some element in and (2) no two elements of are multiples of each other. Then we can write every nonzero element of in the form , where and all but finitely many are zero. The fact that this product is taken over the (possibly infinite) set is okay because our exponent function comes from the “direct sum” of . With this in mind, we will typically not mention and simply refer to factorizations of nonzero elements in as , with the understanding that ranges over a set of representatives of irreducible classes and all but finitely many are zero. Using this notation, for factorizations and , we have using the usual laws of exponents (because is commutative).
Now we prove a lemma.
Lemma: Let be a UFD and let be nonzero, where and have the factorizations and . Then divides if and only if for all . Proof: Suppose divides . Then we can write for some , where . Then , so that for each . Since these exponents are natural numbers, we have for all . Conversely, suppose for all ; then there exists a natural number such that . Note that all but finitely many are zero, so that all but finitely many are zero. Let . Certainly then , so that divides .
Now to the main result. Let be nonzero elements with factorizations and . Note that all but finitely many of are zero. Define . Certainly and , using the lemma. Suppose now that is an element of which is divisible by both and . Again using the lemma, we have and . Thus , and thus divides . Hence is a least common multiple of and .