Principal Ideals of Integers

Theorem
Let $J$ be a non-zero ideal of $\Z$.

Then $J = \left({b}\right)$ where $b$ is the smallest strictly positive integer belonging to $J$.

Proof
It follows from Ring of Integers is a Principal Ideal Domain‎ that $J = \left({b}\right)$ is actually a principal ideal.


 * Let $c \in J, c \ne 0$.

Then $-c \in J$ and by Natural Numbers are Non-Negative Integers, exactly one of them is strictly positive.

Thus $J$ does actually contain strictly positive elements, so that's a start.


 * Let $b$ be the smallest strictly positive element of $J$.

This exists because Natural Numbers are Non-Negative Integers and the Natural Numbers are a Naturally Ordered Semigroup which is well-ordered by definition.

It is clear that $\left({b}\right) \subseteq J$.

We need to show that $J \subseteq \left({b}\right)$.

So, let $a \in J$.

By the Division Theorem, $\exists q, r: a = b q + r, 0 \le r < b$.

As $a, b \in J$, then so does $r = a - b q$.

So, by the definition of $b$, it follows that $r = 0$.

Thus $a = b q \in \left({b}\right)$.