Ideals in $\mathbb{Z}$

Setup. If $I=\{0\}$, then $I=0\mathbb{Z}$. Otherwise, $I$ contains positive elements (if $x\in I$, $-x\in I$, so $|x|\in I$). Let $n$ be the smallest positive element of $I$.

$n\mathbb{Z}\subseteq I$. Since $n\in I$ and $I$ is closed under multiplication by $\mathbb{Z}$, every multiple of $n$ is in $I$.

$I\subseteq n\mathbb{Z}$. Let $a\in I$. Euclidean division: $a=qn+r$ with $0\le r<n$. Then $r=a-qn\in I$. By minimality of $n$, $r=0$, so $a=qn\in n\mathbb{Z}$.

Conclusion. $I=n\mathbb{Z}$. $\mathbb{Z}$ is a principal ideal domain.