Ring of Integers is Principal Ideal Domain
Theorem
The integers $\Z$ form a principal ideal domain.
Proof 1
Let $J$ be an ideal of $\Z$.
Then $J$ is a subring of $\Z$, and so $\left({J, +}\right)$ is a subgroup of $\left({\Z, +}\right)$.
But by Integers under Addition form Infinite Cyclic Group, the group $\left({\Z, +}\right)$ is cyclic, generated by $1$.
Thus by Subgroup of Cyclic Group is Cyclic, $\left({J, +}\right)$ is cyclic, generated by some $m \in \Z$.
Therefore from the definition of principal ideal, $J = \left\{{k m: k \in \Z}\right\} = \left({m}\right)$, and is thus a principal ideal.
$\blacksquare$
Proof 2
We have that Integers are Euclidean Domain.
Then we have that Euclidean Domain is Principal Ideal Domain.
Hence the result.
$\blacksquare$
Proof 3
Let $U$ be an arbitrary ideal of $\Z$.
Let $c$ be a non-zero element of $U$.
Then both $c$ and $-c$ belong to $\ideal a$ 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 $\ideal b$ 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 $r = 0$.
Thus:
- $a = b q \in \ideal b$
and so:
- $U \subseteq \ideal b$
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.
$\blacksquare$