## Definition and basic properties of least common multiples

Let be a commutative ring with 1. Given nonzero elements , a *least common multiple* of and is an element such that and and if such that and , then .

- Prove that a least common multiple of and , if it exists, is precisely a generator for the -largest principal ideal contained in .
- Deduce that any two nonzero elements in a Euclidean domain have a least common multiple which is unique up to multiplication by a unit.
- Prove that in a Euclidean domain, the least common multiple of and is .

- Suppose first that a least common multiple of and exists. Then by definition, we have and , so that and . Thus . Suppose now that . Then and , so that and . Thus , and we have . Thus a least common multiple of and , provided it exists, generates a -largest principal ideal in .
Suppose conversely that contains a -largest principal ideal, say . Then and , so that and . Moreover, if and , then we have , so that , and thus . Thus a -largest principal ideal in , if it exists, is a least common multiple of and .

- Suppose is a Euclidean domain, and let be nonzero. Now is an ideal in , and thus is principal (since all ideals in a Euclidean domain are principal). Say . Now is necessarily the -largest principal ideal contained in , and so (by the previous part) is a least common multiple of and . Moreover, if is a unit in , then . Thus the least common multiple of and is unique up to multiplication by a unit.
- Let be a greatest common divisor of and . Note that , so that . Similarly, , so that . By the previous part, we have , where is a least common multiple of and . Now note that for some ; then , so that . In particular, divides . Similarly, divides . Thus divides ; say . Then , so that divides , and in fact we have . Thus is a least common multiple of and , as desired.

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## Comments

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

Thanks!