Ring of Integers is Principal Ideal Domain

From ProofWiki
Jump to navigation Jump to search


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.


Proof 2

We have that Integers are Euclidean Domain.

Then we have that Euclidean Domain is Principal Ideal Domain.

Hence the result.


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 $r = 0$.


$a = b q \in \ideal b$

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.