# Principal Ideals and Principal Ideal Domains (PIDs)

Recall from the Ideals of Rings that if $(R, +, *)$ is a ring then an ideal if a subring $(I, +, *)$ such that for all $r \in R$ and for all $i \in I$ we have that $r * i \in I$ and $i * r \in I$.

We now define a special type of ideal called a principal ideal.

Definition: Let $(R, +, *)$ be a commutative ring. An ideal of the form $aR = \{ a * r : r \in R \}$ is called a Principal Ideal generated by $a$. |

It is easy to verify that if $R$ is a commutative ring then for every $a \in R$, $aR$ is indeed an ideal of $R$. To show this, let $q \in R$ and let $ar \in aR$. Then $qar = a(qr) \in aR$. Similarly, $(ar)q = a(qr) \in aR$.

For example, consider the ring of integers $\mathbb{Z}$. Then $2\mathbb{Z} = \{ 0, \pm 2, \pm 4, ... \}$ is a principal ideal and is generated by $2$.

In fact, the principal ideal generated by $k \in \mathbb{Z}$ is:

(1)Definition: Let $(R, +, *)$ be an integral domain. Then $R$ is said to be a Principal Ideal Domain (PID) if every ideal in $(R, +, *)$ is a principal ideal. |

For example, consider the set of integers $\mathbb{Z}$. We will prove that $\mathbb{Z}$ is a principal ideal domain. Let $I$ be an ideal of $R$.

First suppose that $I = \{ 0 \}$. Then $I = 0R$, so $I$ is a principal ideal.

Now instead suppose that $I \neq \{ 0 \}$. Then there exists a smallest positive integer $a \in I$ such that $a > 0$. Now let $b \in I$ and suppose that $b > a$. By the division algorithm there exists $q, r \in R$ such that $b = aq + r$ with $0 \leq r < a$. So $r = b - aq$. Since $a \in I$ and $q \in R$ we have that $aq \in I$. So $b - aq \in I$. This shows that $r \in I$. Since $a$ is the smallest positive integer in $I$, we must have that $r = 0$. So $b = aq$. So every element in $I$ is of the form $b = aq$, so $I = aR$. Hence $I$ is a principal ideal.

So every ideal of $\mathbb{Z}$ is a principal ideal, so $\mathbb{Z}$ is a principal ideal domain.