Prime implies Irreducible in integral domains

Proposition 8.3.10:
In an integral domain, a prime element is always irreducible.

First recall some definitions:
Definition: An integral domain is a commutative ring with no zero divisors.
The lack of zero divisors gives integral domains a useful cancellation property:
$$ab=ac \Longrightarrow b=c \quad or \quad a=0$$ A nonzero elemnent $p$ is prime in the ring $R$ if the ideal generated by $p$, $(p)$ is a prime ideal.
An ideal $P$ is prime if $P \neq R$ and $ab\in{P} \Longrightarrow a\in{P}$ or $b\in{P}$.
An element $r\in{R}-\{0\}-R^\times$ is irreducible if $r=xy \Longrightarrow x\in{R^\times}$ or $y\in{R^\times}$ where $R^\times$ is the group of units of $R$.

$\textit{Proof}$ : Let $R$ be an integral domain and let $p\in{R}$ be a prime element. So, $(p)$ is a nonzero prime ideal generated by $p$ and $p=ab$. We want to show that $a\in{R^\times}$ or $b\in{R^\times}$.
So, $ab=p\in{(p)}$. Since $(p)$ is a prime ideal this means that $(p) \neq R$ and $ab\in{(p)} \Longrightarrow a\in{(p)}$ or $b\in{(p)}$. Say $a\in{(p)}$. Then, $\exists r\in{R}$ s.t. $a=pr$. So, we have:
$$p=ab=pr \cdot b$$ $$p=p \cdot rb$$ $$p \cdot 1 = p \cdot rb$$ Since $R$ is an integral domain, this imples
$$p=0 \quad or \quad 1 = rb$$ But $p \neq 0$ since prime elements are nonzero. Thus, $1=rb \Longrightarrow b\in{R^\times}$.
Swapping the roles of $a$ and $b$, if $b\in{(p)}$ then $a\in{R^\times}$. In either case, $p$ is irreducible. $\blacksquare$