[proofplan] We first separate the zero ideal, whose normalized generator is $0$. For a nonzero ideal, we use closure under additive inverses to produce positive elements and then choose the least positive element $d$ using the well-ordering property of $\mathbb{Z}$. The ideal property gives $(d) \subseteq I$, while the [division algorithm](/theorems/725) forces every element of $I$ to have remainder $0$ upon division by $d$, giving $I \subseteq (d)$. Finally, equality of two principal ideals with nonnegative generators implies mutual divisibility, and nonnegativity forces the generators to be equal. [/proofplan] [step:Handle the zero ideal by choosing the normalized generator $0$] Suppose first that \begin{align*} I = \{0\}. \end{align*} By the definition of the principal ideal generated by $0$, \begin{align*} (0)=0\mathbb{Z}=\{0n:n\in\mathbb{Z}\}=\{0\}. \end{align*} Thus $I=(0)$, and the required generator in this case is $d=0$, which satisfies $d \ge 0$. [/step] [step:Choose the least positive element of a nonzero ideal] Assume now that $I \ne \{0\}$. Choose an element $a \in I$ with $a \ne 0$. Since $I$ is an ideal of $\mathbb{Z}$, it is an additive subgroup of $\mathbb{Z}$, so $-a \in I$. Hence either $a \in I$ is positive or $-a \in I$ is positive. Define \begin{align*} S=\{n\in I:n>0\}. \end{align*} The preceding paragraph shows that $S$ is nonempty. Since every element of $S$ is bounded below by $1$, [citetheorem:9702] gives a least element of $S$. Let $d$ denote this least element. Then \begin{align*} d\in I \end{align*} and \begin{align*} d>0. \end{align*} [guided] Because $I$ is assumed nonzero, there is some nonzero integer $a \in I$. This element may be negative, but ideals are additive subgroups, so they are closed under additive inverses. Therefore $-a \in I$ as well. Exactly one of $a$ and $-a$ is positive, so the ideal contains at least one positive integer. We collect all positive integers in $I$ into the set \begin{align*} S=\{n\in I:n>0\}. \end{align*} This set is nonempty by the previous argument. It is also bounded below, for every $n \in S$ satisfies $n \ge 1$. The ordered-ring property of the integers in [citetheorem:9702] says that every nonempty subset of $\mathbb{Z}$ that is bounded below has a least element. Applying that result to $S$, we obtain a least element. We call it $d$. Thus $d$ is not an arbitrary nonzero element of $I$; it is the smallest positive element of $I$. This minimality is the key point that will force all remainders in the next step to vanish. We have \begin{align*} d\in I \end{align*} and \begin{align*} d>0. \end{align*} [/guided] [/step] [step:Show that the principal ideal generated by $d$ is contained in $I$] We prove \begin{align*} (d)\subseteq I. \end{align*} Let $x\in(d)$. By definition of $(d)$, there exists $n\in\mathbb{Z}$ such that \begin{align*} x=dn. \end{align*} Since $d\in I$ and $I$ is an ideal of $\mathbb{Z}$, absorption under multiplication by elements of $\mathbb{Z}$ gives \begin{align*} nd\in I. \end{align*} Because multiplication in $\mathbb{Z}$ is commutative, $dn=nd$, and therefore $x\in I$. Hence $(d)\subseteq I$. [/step] [step:Use division with remainder to prove that every element of $I$ is a multiple of $d$] We prove \begin{align*} I\subseteq(d). \end{align*} Let $a\in I$. Since $d>0$, the [division algorithm for integers](/theorems/9333) gives integers $q,r\in\mathbb{Z}$ such that \begin{align*} a=qd+r \end{align*} and \begin{align*} 0\le r<d. \end{align*} Because $d\in I$ and $I$ is an ideal, $qd\in I$. Since $a\in I$ and $I$ is closed under subtraction, \begin{align*} r=a-qd\in I. \end{align*} If $r>0$, then $r\in S$ and $r<d$, contradicting the definition of $d$ as the least element of $S$. Therefore $r=0$. Hence \begin{align*} a=qd\in(d). \end{align*} Since $a\in I$ was arbitrary, $I\subseteq(d)$. Combining this inclusion with $(d)\subseteq I$, we obtain \begin{align*} I=(d). \end{align*} [guided] Take an arbitrary element $a\in I$. Our goal is to prove that $a$ is a multiple of $d$, because that is exactly what it means to prove $a\in(d)$. Since $d>0$, the division algorithm for integers applies to the pair $(a,d)$. It gives integers $q,r\in\mathbb{Z}$ satisfying \begin{align*} a=qd+r \end{align*} and \begin{align*} 0\le r<d. \end{align*} The quotient $q$ is not important by itself; the important object is the remainder $r$. We will show that the ideal condition forces this remainder to belong to $I$, and then the minimality of $d$ forces the remainder to be zero. Because $d\in I$ and $I$ is an ideal of $\mathbb{Z}$, multiplying $d$ by the integer $q$ keeps us inside $I$: \begin{align*} qd\in I. \end{align*} We also have $a\in I$ by choice. Since ideals are additive subgroups, they are closed under subtraction, so \begin{align*} r=a-qd\in I. \end{align*} Now compare this with the inequalities from the division algorithm: \begin{align*} 0\le r<d. \end{align*} If $r>0$, then $r$ is a positive element of $I$, so $r\in S$. But $r<d$, contradicting the choice of $d$ as the least positive element of $I$. Therefore the only possible value is \begin{align*} r=0. \end{align*} Substituting back into the division equation gives \begin{align*} a=qd. \end{align*} Thus $a\in(d)$. Since the element $a\in I$ was arbitrary, this proves \begin{align*} I\subseteq(d). \end{align*} Together with the previous inclusion $(d)\subseteq I$, this gives \begin{align*} I=(d). \end{align*} [/guided] [/step] [step:Prove uniqueness of the nonnegative generator] Let $d,e\in\mathbb{Z}$ satisfy $d\ge0$, $e\ge0$, and \begin{align*} (d)=(e). \end{align*} We prove $d=e$. First suppose $d=0$. Then \begin{align*} (e)=(d)=(0)=\{0\}. \end{align*} Since $e\in(e)$, it follows that $e=0$, so $d=e$. The same argument with $d$ and $e$ interchanged handles the case $e=0$. It remains to consider the case \begin{align*} d>0 \end{align*} and \begin{align*} e>0. \end{align*} Since $d\in(d)=(e)$, there exists $m\in\mathbb{Z}$ such that \begin{align*} d=me. \end{align*} Since $d>0$ and $e>0$, this forces $m>0$, hence $m\ge1$, and therefore \begin{align*} d=me\ge e. \end{align*} Similarly, since $e\in(e)=(d)$, there exists $n\in\mathbb{Z}$ such that \begin{align*} e=nd. \end{align*} Because $e>0$ and $d>0$, we have $n\ge1$, and hence \begin{align*} e=nd\ge d. \end{align*} Thus $d\ge e$ and $e\ge d$, so $d=e$. Therefore the integer $d\ge0$ with $I=(d)$ is unique. This completes the proof. [/step]