My textbook on algebra states these definitions:
$R$ - Commutative integral domain with unity, where integral domain means $xy = 0 \Rightarrow x= 0 $ or $y=0$
Two elements $a,b \in R$ are called associates if there exists a unit $u \in R$ such that $a = bu$.
An element $b \in R$ is called an improper divisor of an elemenent $a \in R$ if $b$ is either a unit or an associate of $a$.
Later it says for irreducible $a$: "Every divisor of $a$ is improper, i.e. $a=bc$ implies either $b$ or $c$ is a unit.
For this to make sense it would seem that if $a = bc$, where $b$ is an associate of $a$, then $c$ must be a unit. Is this obvious from the definition?
If $a = bc$ where $b$ is an associate of $a$, then $a = bu$ for some unit $u$. So $bc = bu$, so $bcu^{-1} = b$, so $b(1 - cu^{-1}) = 0$. So if $b$ is nonzero (and the last statement is obviously not true if $a = b = 0$) then since $R$ is a domain, $1 - cu^{-1} = 0$, and so $c = u$.