A Fascinating Multiples Problem

Problem 1. Let $I \subseteq \mathbb Z$ be a subset with the following properties:

$I \backslash \{0\} \neq \emptyset$ .
For any $x,y \in I$ , $x+y \in I$ .
For any $x \in I$ and $\alpha \in \mathbb Z$ , $\alpha x \in I$ .

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

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Solution. By the first property, there exists some nonzero $n \in I$ . By the third property, $-n = (-1)n \in I$ . Thus, without loss of generality, suppose $n > 0$ .

By the well-ordering principle, there exists some unique $n_0 > 0$ such that

$\displaystyle n_0 = \min \{m \in I : m > 0\}.$

Now we prove that $I = n_0 \mathbb Z$ .

For $(\supseteq)$ , fix $u \in n_0 \mathbb Z$ . Then there exists $k \in \mathbb Z$ such that $u = k n_0 \in I$ .

For $(\subseteq)$ , fix $v \in I$ . By the division algorithm, find integers $q, r$ with $0 \leq r < n_0$ such that $v = qn_0 + r$ .

We claim that $r = 0$ by way of contradiction. Suppose otherwise that $1 \leq r < n_0$ . Then by the third and second properties respectively, $r = v + (-q)n_0 \in I$ .