Associates in Domains Let D be a domain and $a, b \in D^*$. Show that $a$ is a proper divisor of $b$ if and only if $b=ax$ for some nonzero nonunit $x$. 
I'm just really not sure how to start this. Any advice would be great!
 A: Unpacking the first hypothesis, you're showing this:

For $a,b\in D^\ast$, $(b)\subsetneq (a)$ if and only if $ax=b$ for a nonzero nonunit $x\in D$.

For $\Rightarrow$: from $(b)\subseteq (a)$ conclude that $b=ax$ for some $x$. Show that if $x$ is zero or a unit, you have a contradiction with $(b)\neq (a)$, therefore $x$ is a nonzero nonunit.
For $\Leftarrow$: conclude that $(b)\subseteq (a)$, and work to show that $(b)=(a)$ is not possible. You will be able to show that if $(a)\subseteq (b)$ and $b=ax$, then $x$ is actually a unit. 
A: It's not clear which definition of "proper divisor" you use. In any case, this should work:
Theorem $\ $ The following are equivalent for $\,a,b\in D^*$
$(1)\ \ \ ax = b,\quad\ x\nmid 1,\ $ for some $\, x\in D$
$(2)\qquad\ a\mid b,\quad b\nmid a$
$(3)\ \ \ (a)\supseteq (b),\,\ (b)\not\supseteq (a)$
Proof  $\ $ (sketch)  $\ \ \ \exists x\!:\, b=ax\iff a\mid b\iff (a)\supseteq (b)$
If  so, by $\  \dfrac{a}b = \dfrac{1}x\, $ we infer $\, x\mid1 \!\iff\! b\mid a\!\iff\! (b)\supseteq(a)$
Remark $\ $ Above we used $ $  Divides $\!\iff\!$ Contains for principal ideals.
