Exercise 1.13

Proof. Let $n_1,n_2,\dots ,n_s\in \mathbb{Z}$. The ideal of $\mathbb{Z}$, $(n_1,\dots ,n_s)=n_1\mathbb{Z}+\cdots +n_s\mathbb{Z}$ is principal, so there exists an unique $d\in \mathbb{Z},d\ge 0$ such that

We define

The characterization of the gcd is

$\text{(⇒)}$ Indeed, if we suppose (1), then $d\ge 0$, and $n_1=n_1.1+n_2.0+\cdots +n_s.0\in n_1\mathbb{Z}+\cdots +n_s\mathbb{Z}=d\mathbb{Z}$, so $d\mid n_1$. Similarly $d\mid n_i,1\le i\le s$ so (i)(ii) are true. if $\delta \mid n_i,1\le i\le s$, as $d=n_1{m}_1+\cdots +n_s{m}_s,m_1,\dots ,m_s\in \mathbb{Z}$, then $\delta \mid d$.

$\text{(⇐)}$ Suppose that $d$ verify (i)(ii)(iii). From (ii), we see that $n_i\mathbb{Z}\subset \mathit{d\mathbb{Z}},i=1,\dots ,s$, so $n_1\mathbb{Z}+\cdots +n_s\mathbb{Z}\subset \mathit{d\mathbb{Z}}$.

As $\mathbb{Z}$ is a principal ring, there exists $\delta \ge 0$ such that $n_1\mathbb{Z}+\cdots +n_s\mathbb{Z}=\delta \mathbb{Z}$. $n_i\in n_1\mathbb{Z}+\cdots +n_s\mathbb{Z}$ so $n_i\in \mathit{\delta \mathbb{Z}},i=1,\dots ,s$ : $\delta \mid n_1,\dots ,\delta \mid n_s$. From (iii), we deduce $\delta \mid d$. As $\mathit{\delta \mathbb{Z}}\subset \mathit{d\mathbb{Z}}$, $d\mid \delta$, with $d\ge 0,\delta \ge 0$. Consequently, $d=\delta$ and $n_1\mathbb{Z}+\cdots +n_s\mathbb{Z}=d\mathbb{Z},d\ge 0$, so $d=\gcd (n_1,\dots ,n_s)$.

At last, as $n_1\mathbb{Z}+\cdots +n_s\mathbb{Z}=d\mathbb{Z}$, there exist integers $m_1,m_2,\dots ,m_s$ such that $n_1{m}_1+n_2{m}_2+\cdots +n_s{m}_s=d$. □