Show that the relation defined on an integral domain by "$x\sim y$" ("$x$ is associated with $y$") is a relation of equivalence. I did the following demonstration: Would it be possible to give me a feedbak if i'm to wrong.
Be $A$ an integral domain and $x,y \in A$, then: 
$x\sim y \Rightarrow$ $\exists u \in \mathfrak{U}(A)$ such that $y=xu \Rightarrow x \mid y$ and $y\mid x$. So there's $a,b \in A$ such that $xa=y$ and $yb=x$.
$1º$ case: $x=0.$
$x=0$ and $xa=y \Rightarrow y=0.$
Take $u=1 \in \mathfrak{U}(A)$ and then,
$0=0\cdot 1$, or $y=xu$.
$2º$ case: $a\neq 0$.
$xa=y$ and $yb=x$.
$\Rightarrow xab=x$
$\Rightarrow x(ab-1)=0$
$\Rightarrow ab=1$
$\Rightarrow a$ is invertible.
Taking $u=a \in \mathfrak{U}(A)$ and then $xu=y$.
By hypothesis, $y=xu$, $u \in \mathfrak{U}(A)$. It follows that $x\mid y.$ In addition, from $y=xu$, we have $y \cdot u^{-1}=x$ and then $ y\mid x$.
So, $x\sim y$.
$\square$
 A: I'm not sure exactly what you're trying to show in your exposition. To answer the question in the title, i.e. show that a relation ~ is an equivalence relation, we need to show three things, reflexivity, symmetry, and transitivity.
Reflexivity is clear, every element is associate to itself since $x = 1 \cdot x \forall x \in R$.
Symmetry is also easy since units are invertible, so if $x$~$y$ then there exists some unit $u$ such that $x = uy$ so $u^{-1}x=y$ i.e. $y$ ~ $x$.
Lastly we consider transitivity. Suppose $x$ ~ $y$ and $y$~$z$. Then there exist units $u,v$ such that $x = uy$ and $y = vz$. So we have that $x = uvz$. Since the units are closed under multiplication, $x$~$z$ as desired.
A: This holds not only for the unit group but for any subgroup $G$ of the multiplicative group, i.e.
suppose that $\,G\subset A^{\times}\,$ satisfies $\,\color{#c00}1\in G\,$ and $\,g,h\in G\,\Rightarrow\, \color{#08f}{gh}\in G,\,$ and $\, \color{#0a0}{g^{-1}}\in G$.
Note $\ x\approx y \!\overset{\rm def\!\!}\iff\! x = g\, y\ \ {\rm for\ some}\ \ g\in G\ $ is an equivalence relation, $ $  since
$\qquad\ \ \ \  \approx\,$ is $\ \ \rm\color{#c00}{reflexive}\,\ \ $ by $\  x = \color{#c00}1x\,\Rightarrow\, x\approx x$
$\qquad\ \ \ \  \approx\,$ is $\,\rm\color{#0a0}{symmetric}\,$ by $\ x\approx y\,\Rightarrow\, x = g y\,\Rightarrow\, y = \color{#0a0}{g^{-1}}x\,\Rightarrow\,y\approx x$
$\qquad\ \ \ \ \approx\,$ is $\ \rm\color{#08f}{transitive}\,\ $  by $\,x\approx y\approx z\,\Rightarrow\, y = hz,\,  x = g y = g(hz) = (\color{#08f}{gh})z\,\Rightarrow\, x\approx z$
Remark $\ G x = \{ gx\, :\, g\in G\}\ $ is called the $G$-orbit of $x.\, $ It is a basic concept in group theory.
A quick way to recognize such group structure is by the subgroup test, i.e. a nonempty $G\subset H$ of a group $H$ forms a group $\iff$ it is closed under division, i.e. $\, g,h\in g\,\Rightarrow g/h = gh^{-1}\in G,\,$ which is clear for units.
When studying divisibility it is often useful to ignore units by factoring out the unit group, e.g. see the divisibility group and  its associated characterizations of UFDs and gcd, valuation and Reisz domains mentioned there.
