# Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication

## Theorem

Let $\struct {D, +, \circ}$ be a principal ideal domain.

Let $p$ be an irreducible element of $D$.

Let $\ideal p$ be the principal ideal of $D$ generated by $p$.

Then $\ideal p$ is a maximal ideal of $D$.

## Proof

Let $p$ be irreducible in $D$.

Let $U_D$ be the group of units of $D$.

By definition, an irreducible element is not a unit.

So from Principal Ideals in Integral Domain:

- $\ideal p \subset D$

Suppose the principal ideal $\ideal p$ is not maximal.

Then there exists an ideal $K$ of $D$ such that:

- $\ideal p \subset K \subset R$

Because $D$ is a principal ideal domain:

- $\exists x \in R: K = \ideal x$

Thus:

- $\ideal p \subset \ideal x \subset D$

Because $\ideal p \subset \ideal x$:

- $x \divides p$

by Principal Ideals in Integral Domain.

That is:

- $\exists t \in D: p = t \circ x$

But $p$ is irreducible in $D$, so $x \in U_D$ or $t \in U_D$.

That is, either $x$ is a unit or $x$ is an associate of $p$.

But since $K \subset D$:

- $\ideal x \ne D$ so $x \notin U_D$

by Principal Ideals in Integral Domain.

Also, since $\ideal p \subset \ideal x$:

- $\ideal p \ne \ideal x$

so $x$ is not an associate of $p$, by Principal Ideals in Integral Domain.

This contradiction shows that $\ideal p$ is a maximal ideal of $D$.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 63.2$ Construction of fields as factor rings