Definition and basic properties of least common multiples

Let R be a commutative ring with 1. Given nonzero elements a,b \in R, a least common multiple of a and b is an element \ell \in R such that a|\ell and b|\ell and if m \in R such that a|m and b|m, then \ell|m.

  1. Prove that a least common multiple of a and b, if it exists, is precisely a generator for the \subseteq-largest principal ideal contained in (a) \cap (b).
  2. Deduce that any two nonzero elements in a Euclidean domain have a least common multiple which is unique up to multiplication by a unit.
  3. Prove that in a Euclidean domain, the least common multiple of a and b is \dfrac{ab}{\mathsf{gcd}(a,b)}.

  1. Suppose first that a least common multiple \ell of a and b exists. Then by definition, we have a|\ell and b|\ell, so that (\ell) \subseteq (a) and (\ell) \subseteq (b). Thus (\ell) \subseteq (a) \cap (b). Suppose now that (m) \subseteq (a) \cap (b). Then (m) \subseteq (a) and (m) \subseteq (b), so that a|m and b|m. Thus \ell|m, and we have (m) \subseteq (\ell). Thus a least common multiple of a and b, provided it exists, generates a \subseteq-largest principal ideal in (a) \cap (b).

    Suppose conversely that (a) \cap (b) contains a \subseteq-largest principal ideal, say (\ell). Then (\ell) \subseteq (a) and (\ell) subseteq (b), so that a|\ell and b|\ell. Moreover, if a|m and b|m, then we have (m) \subseteq (a) \cap (b), so that (m) \subseteq (\ell), and thus \ell|m. Thus a \subseteq-largest principal ideal in (a) \cap (b), if it exists, is a least common multiple of a and b.

  2. Suppose R is a Euclidean domain, and let a,b \in R be nonzero. Now (a) \cap (b) is an ideal in R, and thus is principal (since all ideals in a Euclidean domain are principal). Say (a) \cap (b) = (\ell). Now (\ell) is necessarily the \subseteq-largest principal ideal contained in (a) \cap (b), and so (by the previous part) \ell is a least common multiple of a and b. Moreover, if u is a unit in R, then (u\ell) = (\ell). Thus the least common multiple of a and b is unique up to multiplication by a unit.
  3. Let d be a greatest common divisor of a and b. Note that \dfrac{ab}{d} = a \cdot \dfrac{b}{d}, so that (\frac{ab}{d}) \subseteq (a). Similarly, (\frac{ab}{d}) \subseteq (b), so that (\frac{ab}{d}) \subseteq (a) \cap (b). By the previous part, we have (\frac{ab}{d}) \subseteq (\ell), where \ell is a least common multiple of a and b. Now note that ax = \ell for some x; then abx = b\ell, so that \frac{ab}{\ell}x = b. In particular, \frac{ab}{\ell} divides b. Similarly, \frac{ab}{\ell} divides a. Thus \frac{ab}{\ell} divides d; say \frac{ab}{\ell} \cdot k = d. Then \frac{ab}{d} \cdot k = \ell, so that \frac{ab}{d} divides \ell, and in fact we have (\ell) \subseteq (\frac{ab}{d}). Thus \frac{ab}{d} is a least common multiple of a and b, as desired.
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Comments

  • Jamie Sorrosa  On July 25, 2011 at 8:00 pm

    On the first line of your proof of #3, I think you meant to say gcd instead of lcm.

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