Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Reverse Implication
Theorem
Let $\left({D, +, \circ}\right)$ be a principal ideal domain.
Let $\left({p}\right)$ be the principal ideal of $D$ generated by $p$.
Let $\left({p}\right)$ be a maximal ideal of $D$.
Then $p$ is irreducible.
Proof
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.
This article contains statements that are justified by handwavery. In particular: "Clearly" needs to be replaced by a link to the definition which specifies this fact. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding precise reasons why such statements hold. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Handwaving}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
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.
$\blacksquare$