> [!NOTE] Theorem (Ideals of Integers) > Let $(I,+)$ be an [[Ideal of Ring|ideal]] of the [[Integers|integers]] $\mathbb{Z}.$ Then $I=n\mathbb{Z}=\{ nm \mid m \in \mathbb{Z} \}$ for some $n\in\mathbb{Z}.$ > >**Equivalently**, every ideal of the integers $(I,+)$ is cyclic (or a [[Principal Ideal of Ring|principal ideal]]) and so the **ring of integers is a [[Principal Ideal Domain|principal ideal domain]]**. *Proof*. If $I=\{ 0 \}$ then let $n=0.$ Now suppose $I\neq \{ 0 \}.$ If $x\in I$ then $-x\in I$ and so $I$ will contain a positive integer. Let $n$ be the smallest positive integer contained in $I$ which exists by [[Well-Ordering Principle|well-ordering]]. Since $I$ is an ideal and $n\in I,$ for all $m\in \mathbb{Z},$ $nm\in I$ hence $n\mathbb{Z}\subset I.$ Now let $x\in I.$ By [[Division with remainder for integers|division with remainder]], there exists $q,r\in \mathbb{Z}$ with $0\leq r<n$ such that $x=nq+r.$ But $r=x-nq \in I$ since $x,nq\in I$ and $(I,+)$ is a group. By minimality of $n$ this means that $r=0$ and $x=nq\in \mathbb{Z}.$ Therefore $I \subset n\mathbb{Z}$ and $I=n\mathbb{Z}.$ # Applications Generalisations: [[Ring of Polynomial Forms over Field is a Principal Ideal Domain]].