Ring of Integers is Principal Ideal Domain/Proof 3

Proof
Let $U$ be an arbitrary ideal of $\Z$.

Let $c$ be a non-zero element of $U$.

Then both $c$ and $-c$ belong to $\left({a}\right)$ and one of them is positive.

Thus $U$ contains strictly positive elements.

Let $b$ be the smallest strictly positive element of $U$.

By the Set of Integers Bounded Below by Integer has Smallest Element, $b$ is guaranteed to exist.

If $\left({b}\right)$ denotes the ideal generated by $b$, then $\left({b}\right) \subseteq U$ because $b\in U$ and $U$ is an ideal.

Let $a \in U$.

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

As $a, b \in U$ it follows that so does $r = a - bq$.

By definition of $b$ it follows that $r = 0$.

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

and so:
 * $U \subseteq \left({b}\right)$

From the above:
 * $U = \left({b}\right)$

It follows by definition that $U$ is a principal ideal of $\Z$.

Recall that $U$ was an arbitrary ideal of $\Z$.

Hence by definition $\Z$ is a principal ideal domain.