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