Proof that in a PID a nonzero element is irreducible iff it generates a maximal ideal

Proof

Let $R$ be a principal ideal domain and let $p \in R$ with $p \neq 0$.

1. For $b \in R$, the set $b R = \set{b t : t \in R}$ is an ideal that contains $b$. Every ideal containing $b$ contains $b R$. So the ideal generated by $b$ is $(b) = b R$.
2. For $b \in R$, $(b) = R$ exactly when $b$ is a unit. If $(b) = R$, then $1 = b t$ for some $t \in R$, so $b$ is a unit. If $b t = 1$, then each $r \in R$ equals $b (t r) \in b R$.

First suppose that $p$ is irreducible.

3. Since $p$ is not a unit, step 2 gives $(p) \neq R$.
4. Let $I$ be an ideal with $(p) \subseteq I$. Since $R$ is a principal ideal domain, $I = (a)$ for some $a \in R$, a principal ideal.
5. Since $p \in (a) = a R$, there is $b \in R$ with $p = a b$. Since $p$ is irreducible, $a$ is a unit or $b$ is a unit.
6. If $a$ is a unit, then $I = (a) = R$ by step 2.
7. If $b$ is a unit, then $a = p b^{-1} \in (p)$. So $(a) \subseteq (p)$, and $I = (p)$.
8. By steps 3 to 7, $(p)$ is a maximal ideal.

Now suppose that $(p)$ is a maximal ideal.

9. We have $p \neq 0$ by hypothesis. Since $(p) \neq R$, step 2 shows that $p$ is not a unit.
10. Let $p = a b$ with $a, b \in R$. Then $p \in (a)$, so $(p) \subseteq (a)$. By maximality, $(a) = (p)$ or $(a) = R$.
11. If $(a) = R$, then $a$ is a unit by step 2.
12. If $(a) = (p)$, then $a = p t$ for some $t \in R$. So $p \cdot 1 = p = a b = p (t b)$.
13. Since $p \neq 0$, cancellation gives $t b = 1$, so $b$ is a unit.
14. So in every factorization $p = a b$, one factor is a unit. With step 9, $p$ is irreducible.