Let $c$ be prime and $d$ be an associate of $c$.
Claim: $d$ is prime.
$c$ prime $\Rightarrow c\neq 0$, nonunit
Suppose that $c\vert ab$. Since $c$ is prime, $c\vert a$ or $c\vert b$
$c\sim d \Rightarrow c\vert d$ and $d\vert c$
$R$ ID $\Rightarrow d\neq 0$
$d\vert c$ and $c\vert a \Rightarrow c=dx_1$ and $a=cx_2, \;x_1,x_2\in R$
$\Rightarrow a=d(x_1x_2), \; x_1x_2\in R$
$\Rightarrow d\vert a$
Also,
$d\vert c$ and $c\vert b\Rightarrow c=dx_1$ and $b=cx_2, x_1,x_2\in R$
$\Rightarrow b=d(x_1x_2), \; x_1x_2\in R$
$\Rightarrow d\vert b$
I'm having trouble trying to show that $d$ is non-unit.
Definition: Let $R$ be a commutative ring with 1. An element $p\in R$ is said to be prime provided that
1) $p\neq 0$ and nonunit.
2) $p\vert ab \Rightarrow p\vert a$ or $p\vert b$
