Proving that that an element is prime via association. The question I'm stuck on is
"Let R be an integral domain and a, b ∈ R so that a and b are associated.
Show that if a is prime element then b is a prime element."
I don't know how to solve this, like I don't know how to show that an element is a prime element.
 A: Hint $ $ Divisibility is preserved by association: $ $ if $\,\bar p\sim p\,$ then $\,\bar p\mid a\color{#c00}\iff  p\mid a,\,$ therefore any property that is defined purely by divisibility is also preserved, including primality, e.g.
$$\begin{align} \bar p\mid ab \color{#c00}\iff &\,p\mid ab\\ 
\iff &\,p\mid a\,\ {\rm or}\,\ p\mid b,\ \ \text{by $\,p\,$ prime}\\
\color{#c00}\iff &\,\bar p\mid a\,\ {\rm or}\,\ \bar p\mid b\\[.1em] \text{thus $\,p\,$ prime}  \Longrightarrow\,&\, \bar p\ {\rm prime}\end{align}\qquad$$
Remark $ $ In fact when studying divisibility theory in domains it is often convenient to ignore units by  working modulo the unit group, i.e. we consider elements congruent if they are associate. The quotient monoid is known as the reduced monoid and it is the standard place to begin study of factorization in general domains (at least for those properties that are monoid-theoretic). See this answer for further discussion, including examples and literature references.
A: Take $p,q \in R$ associated. Then $p \mid q$ and $q \mid p.$ 
Suppose $p$ prime. 
You have to prove that $(\forall a,b \in R.\ q \mid ab \implies q  \mid a \lor q \mid b).$
Take $a,b\in R$ such that $q \mid ab.$
Since $p \mid q$, you know that $p \mid ab$.
Since $p$ is prime, you know that $p\mid a \lor p \mid b.$
Since $q \mid p$, you know that $q\mid a \lor q \mid b$.
Do the case with $q$ prime and conclude.
