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 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, $J = \left\{{k m: m \in \Z}\right\} = \left({m}\right)$, 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.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 24$: Theorem $24.3$
• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 5.21$: Theorem $38$