Exercise 1.19

Proof. As $\mathit{a\mathbb{Z}}\cap \mathit{b\mathbb{Z}}$ is an ideal of $\mathbb{Z}$, and $\mathbb{Z}$ is a principal ideal domain, there exists an unique $m\ge 0$ such that $\mathit{a\mathbb{Z}}\cap \mathit{b\mathbb{Z}}=\mathit{m\mathbb{Z}}$. So by definition,

We may note also $[a,b]=a\vee b$.

Characterization of lcm :

$\text{(⇒)}$ By definition, $m\ge 0$. $m\in \mathit{m\mathbb{Z}}=\mathit{a\mathbb{Z}}\cap \mathit{b\mathbb{Z}}$, so $a\mid m$ and $b\mid m$ : (ii) is verified. If $\mu \in \mathbb{Z}$ is such that $a\mid \mu ,b\mid \mu$, then $\mu \in \mathit{a\mathbb{Z}}\cap \mathit{b\mathbb{Z}}=\mathit{m\mathbb{Z}}$, so $m\mid \mu$ : (iii) is true.

$\text{(⇐)}$ Suppose that $m$ verifies (i),(ii),(iii). Let $m^{\prime }$ such that $\mathit{a\mathbb{Z}}\cap \mathit{b\mathbb{Z}}=m^{\prime }\mathbb{Z},m^{\prime }\ge 0$. We show that $m=m^{\prime }$.

As $m^{\prime }\in \mathit{a\mathbb{Z}}\cap \mathit{b\mathbb{Z}},a\mid m^{\prime },b\mid m^{\prime }$, so we see from (iii) that $m\mid m^{\prime }$. From (ii), we obtain that $m\in \mathit{a\mathbb{Z}}\cap \mathit{b\mathbb{Z}}=m^{\prime }\mathbb{Z}$, thus $m^{\prime }\mid m,m\ge 0,m^{\prime }\ge 0$. The conclusion is $m=m^{\prime }$ and $\mathit{a\mathbb{Z}}\cap \mathit{b\mathbb{Z}}=\mathit{m\mathbb{Z}},m\ge 0$, so $m=a\vee b$. □