Describe least common multiples in a UFD

Let R be a unique factorization domain and let a,b \in R be nonzero elements. Prove that a and b have a least common multiple, and describe one such multiple in terms of the factorizations of a and b.


We begin with some remarks on the elements of a UFD. Let R be a unique factorization domain, and let S \subseteq R be a set of irreducibles such that (1) every irreducible in R is a multiple of some element in S and (2) no two elements of S are multiples of each other. Then we can write every nonzero element of R in the form \prod_S s^{k_s}, where k_s \in \mathbb{N} and all but finitely many k_s are zero. The fact that this product is taken over the (possibly infinite) set S is okay because our exponent function k comes from the “direct sum” of \mathbb{N}. With this in mind, we will typically not mention S and simply refer to factorizations of nonzero elements in R as \prod s^{k_s}, with the understanding that s ranges over a set of representatives of irreducible classes and all but finitely many k_s are zero. Using this notation, for factorizations \prod s^{k_s} and \prod s^{\ell_s}, we have (\prod s^{k_s})(\prod s^{\ell_s}) = \prod s^{k_s + \ell_s} using the usual laws of exponents (because R is commutative).

Now we prove a lemma.

Lemma: Let R be a UFD and let a,b \in R be nonzero, where a and b have the factorizations a = \prod s^{k_s} and b = \prod s^{\ell_s}. Then a divides b if and only if k_s \leq \ell_s for all s. Proof: Suppose a divides b. Then we can write ac = b for some c, where c = \prod s^{m_s}. Then \prod s^{k_s + m_s} = ac = b = \prod s^{\ell_s}, so that k_s + m_s = \ell_s for each s. Since these exponents are natural numbers, we have k_s \leq \ell_s for all s. Conversely, suppose k_s \leq \ell_s for all s; then there exists a natural number m_s such that k_s + m_s = \ell_s. Note that all but finitely many \ell_s are zero, so that all but finitely many m_s are zero. Let c = \prod s^{m_s}. Certainly then ac = (\prod s^{k_s})(\prod s^{m_s}) = \prod s^{k_s + m_s} = \prod s^{\ell_s} = b, so that a divides b. \square

Now to the main result. Let a,b \in R be nonzero elements with factorizations a = \prod s^{k_s} and b = \prod s^{\ell_s}. Note that all but finitely many of \max(k_s,\ell_s) are zero. Define m = \prod s^{\max(k_s,\ell_s)}. Certainly a|c and b|c, using the lemma. Suppose now that c = \prod s^{n_s} is an element of R which is divisible by both a and b. Again using the lemma, we have k_s \leq n_s and \ell_s \leq n_s. Thus \max(k_s,\ell_s) \leq n_s, and thus m divides c. Hence m is a least common multiple of a and b.

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