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.