Ring of Integers is Principal Ideal Domain/Proof 1

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

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.

$\blacksquare$