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