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

## Theorem

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

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

Then $p$ is irreducible if and only if $\ideal p$ is a maximal ideal of $D$.

## Proof

### Necessary Condition

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.

$\ideal p \subset D$

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 $\ideal p \subset \ideal x$:

$x \divides p$

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 $x \notin U_D$

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 $D$.

$\blacksquare$

### Sufficient Condition

Let $\left({p}\right)$ be a maximal ideal of $D$.

Let $p = f g$ be any factorization of $p$.

We must show that one of $f, g$ is a unit.

Suppose that neither of $f, g$ is a unit.

First it will be shown that:

$\left({p}\right) \subsetneqq \left({f}\right)$

Let $x \in \left({p}\right)$.

That is:

$\exists q \in D: x = p q$

Then:

$x = f g q \in \left({f}\right)$

so:

$\left({p}\right) \subseteq \left({f}\right)$

Now suppose $f \in \left({p}\right)$.

Then:

$\exists r \in D: f = r p$

and so from $p = f g$ above:

$f = r g f$

Therefore:

$r g = 1$

and so $g$ is a unit.

This is a contradiction.

Thus:

$f \notin \left({p}\right)$

and clearly:

$f \in \left({f}\right)$

so:

$\left({p}\right) \subsetneqq \left({f}\right)$

as claimed.

Therefore, since $\left({p}\right)$ is maximal, we must have:

$\left({f}\right) = D$

But we assumed that $f$ is not a unit.

So there is no $h \in D$ such that $f h = 1$.

Therefore:

$1 \notin \left({f}\right) = \left\{{f h: h \in D}\right\}$

and:

$\left({f}\right) \subsetneqq D$

This is a contradiction.

Therefore at least one of $f, g$ must be a unit.

This completes the proof.

$\blacksquare$