# Principal Ideals of Integers

## Theorem

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

Then $J = \ideal b$ where $b$ is the smallest strictly positive integer belonging to $J$.

## Proof

It follows from Ring of Integers is Principal Ideal Domain that $J$ is 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 Well-Ordering Principle.

By definition of a principal ideal, we have $\ideal b \subseteq J$ as $b \in J$.

We need to show that $J \subseteq \ideal b$.

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 $a = b q \in \ideal b$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 24$: Theorem $24.3$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 59.2$ Principal ideals in a commutative ring with a one