Prove that there exists $n_0 \in \mathbb{Z}$ such that $I = \{kn_0 | k \in \mathbb{Z}\}$ given that $I$ satisfies the following properties [duplicate]

Question

Assume that the set $I \subseteq \mathbb{Z}$ satisfies the following properties:

(a) There exists $n \in I$ such that $n \not = 0$

(b) If $m, n \in I$, then $m + n \in I$

(c) If $m \in I$ and $a \in \mathbb{Z}$, then $am \in I$

Prove that there exists $n_0 \in \mathbb{Z}$ such that $I = \{kn_0 | k \in \mathbb{Z}\}$

Not too sure if I am correct but I am thinking if $n_0$ is actually $gcd(m, n)$, then $I$ would be the set of all integer combinations of $m$ and $n$. So essentially this question is to relate the set of all integer linear combinations to the set of all integer multiples of the $gcd(m, n)$.

Answer 1

You are right in guessing that whenever $m, n \in I$, then $\gcd(m,n) \in I$ by the Lemma of Bézout. We will use this in the following proof.

Let $n_0$ be the smallest non-zero positive integer in $I$ (if no such $n_0$ exists, then $I = \{0\}$ and $n_0 = 0$). Then, by (b), we get that $\{kn_0 : k \in \mathbb Z\} \subset I$. For the other inclusion, suppose that $m \in I$. If $m = 0$ then $m = 0\cdot n_0$. If not, assume $m$ is positive by replacing $m$ by $-m$ otherwise. Then $0<\gcd(m,n_0) \leq n_0$, and by the Lemma of Bézout, $\gcd(m,n_0) \in I$. But now, since $n_0$ is minimal, we have that $\gcd(m, n_0) = n_0$. In other words, $m$ is a multiple of zero. This yields $I \subset \{kn_0:k \in \mathbb Z\}$.

For additional context, you are asking for a proof of the fact that $\mathbb Z$ is a so-called Principal Ideal Domain (a ring in which any ideal is principal). This is actually true for any euclidean domain.