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

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 $p$ 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:
 * $\left({p}\right) \subset D$

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

Then there exists 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 $\left({p}\right) \subset \left({x}\right)$:
 * $x \mathop \backslash p$

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.

This contradiction shows that $\left({p}\right)$ is a maximal ideal of $D$.