Proof: By Induction
(related to Proposition: Principal Ideals being Prime Ideals)
By assumption, $(R, + ,\cdot)$ be an integral domain, $a\in R,$ $a\neq 0,$ and $(a)\lhd R$ a principal ideal.
"$\Rightarrow$"
 Let $(a)$ be a prime ideal.
 By definition of prime ideals, $(a)\neq R.$
 By principal ideal generated by a unit, $a\not\in R^\ast$ (i.e. $a$ is not a unit)
 Let $r,s\in R$ with $a\mid rs$ (i.e. $a$ is a divisor of $rs.$)
 Since $(a)$ is a principal ideal $rs\in (a).$
 Since $(a)$ is a prime ideal, we have $r\in (a)$ or $s\in (a).$
 Therefore, $a\mid r$ or $a\mid s.$
 We have already shown that $a$ is not a unit and, by hypothesis, $a\neq 0.$
 Altogether, by definition of prime elements, it follows that $a$ is a prime element of $R.$
"$\Leftarrow$"
 Let $a\in R$ be a prime element.
 By definition, $a\neq 0$ and $a$ is not a unit.
 Therefore, $(a)\neq (0).$
 Let $rs\in (a)$ for some $r,s\in R.$
 Since $a\mid rs$ and $a$ prime, it follows $a\mid r$ or $a\mid s.$
 Therefore, $r\in (a)$ or $s\in (a).$
 Since $a$ is not a unit, by principal ideal generated by a unit, $(a)\neq R.$
 By definition of prime ideals, $(a)$ is a prime ideal.
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