# Ring of Integers is Principal Ideal Domain/Proof 3

## Theorem

The integers $\Z$ form a principal ideal domain.

## 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 $\ideal a$ 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.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.3$ - 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $5$: Rings: $\S 21$. Ideals: Theorem $38$