An exercise in number theory: associates elements I have a question for you about associates elements in an integral domain.
Let $R$ be an integral domain and define $aR := \{ ar \; | \; r \in R\}$. In the following, for $unity$ (denoted with $u$) I mean an element in R that divides $1$. So, here's the question: $a,b \in R$ are associates ($a=ub$) if and only if $aR=bR$. Why??
Thank you very much for the support, I appreciate!
 A: The following applies inside any commutative monoid $R$.

Definition. Write $a \mid b$ iff $b \in aR$, equivalently $a \mid b$ iff $b=ra$ for some $r$.
Proposition. $a \mid b$ iff $bR \subseteq aR$.
Suppose $a \mid b$. Then $b \in aR$. So $bR \subseteq aRR \subseteq aR.$ So $bR \subseteq aR$.
Conversely, suppose $bR \subseteq aR$. Then $b1 \in aR,$ so $b \in aR$. In other words, $a \mid b$. Hence $aR = bR$ iff $a \mid b$ and $b \mid a$.
Definition. A unit is an element $u$ such that there exists $v$ with $uv=1$.
Proposition. Suppose $a=ub$ where $u$ is a unit. Then $a \mid b$ and $b \mid a$.
Proof. Its clear that $b \mid a$. Now let $v$ denote multiplicative inverse of $u$. Then $va=vub$, so $va=1b$, so $b=va$. Hence $a \mid b$.
Corollary. If $a=ub$ for some unit $u$, then $aR = bR.$

To get the backward direction, we need some further assumptions. Suppose our commutative monoid $R$ also has a distinguished element $0$ that is absorptive:
$$x0 = 0, \quad 0x = 0$$
Suppose also that the "integral domain" law holds.
$$(ax=ay) \rightarrow (a=0) \vee (x=y)$$
Then we have the following.

Proposition. Suppose $a \mid b$ and $b \mid a$. Then $a=ub$ for some unit $b$.
Proof. We know that $ar=b$ and $bs=a$ for some $r,s$. Hence $bsr=b$. So either $b=0$, or $sr=1$. In the first case, we also have that $a=0$, so $a=1b$, hence $1$ is our required unit $u$ such that $a=ub$. In the second case, we that $s$ is a unit, so $s$ is our required unit $u$ such that $a=ub$.
Corollary. $aR = bR$ iff $a=ub$ for some unit $u$.

With a bit more care, similar definitions/arguments work in the non-commutative setting.
A: Hint: if $u$ is a unit, then $\exists k \in R$ such that $uk = 1$. So $1 \in uR$, so $uR = R$.
