Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal

Theorem
Let $\left({D, +, \circ}\right)$ be a principal ideal domain.

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

Let $\left({p}\right)$ be the principal ideal of $D$ generated by $p$.

Then $\left({p}\right)$ is a maximal ideal of $D$.

Proof
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, $\left({p}\right) \subset D$.


 * Suppose the principal ideal $\left({p}\right)$ is not maximal.

Then there is an ideal $K$ of $D$ such that $\left({p}\right) \subset K \subset R$.

Because $D$ is a principal ideal domain, $\exists x \in R: K = \left({x}\right)$.

Thus $\left({p}\right) \subset \left({x}\right) \subset D$.

Because $\left({p}\right) \subset \left({x}\right)$, $x \backslash p$ 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 $K \subset D$, $\left({x}\right) \ne D$ so $x \notin U_D$ by Principal Ideals in Integral Domain.

Also, since $\left({p}\right) \subset \left({x}\right)$, $\left({p}\right) \ne \left({x}\right)$ so $x$ is not an associate of $p$, by Principal Ideals in Integral Domain.

The result follows from this contradiction.