## 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.

• 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.

• nbloomf  On August 5, 2011 at 8:39 am

Thanks!