Ring of Integers is Principal Ideal Domain/Proof 3
Let $U$ be an arbitrary ideal of $\Z$.
Then both $c$ and $-c$ belong to $\ideal a$ and one of them is positive.
Thus $U$ contains strictly positive elements.
By the Set of Integers Bounded Below by Integer has Smallest Element, $b$ is guaranteed to exist.
If $\ideal b$ denotes the ideal generated by $b$, then $\ideal b \subseteq U$ 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 $r = a - b q$.
By definition of $b$ it follows that $r = 0$.
- $a = b q \in \ideal b$
- $U \subseteq \ideal b$
From the above:
- $U = \ideal b$
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.