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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, $x\sim y$.

$\square$

Eric Wofsey
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Fernando Sousa
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  • I'm not clear on what exactly you're trying to show. You've structured your argument as an if and only if, but you seem to showing that $x $~$y$ if and only if $x,y$ are associate, but this is how you define ~? – Noah Solomon Dec 27 '20 at 05:13
  • @NoahSolomon I corrected my statement. It is not if and only if. – Fernando Sousa Dec 27 '20 at 05:20
  • Perhaps I am a little slow on the pickup today, but from what I can tell you start by assuming that $x$ ~ $y$ and end by concluding $x$ ~ $y$, so again I'm not sure what you're trying to show. I posted an answer explaining how I would answer what I think you're trying to prove. – Noah Solomon Dec 27 '20 at 05:22
  • @NoahSolomon I really want to prove $x\sim y$, but on the ring of integrity. It is more advanced than showing the equivalence relation. – Fernando Sousa Dec 27 '20 at 05:27
  • You want to prove that $x$~$y$ given what initial assumption? – Noah Solomon Dec 27 '20 at 05:32
  • @NoahSolomon, yes... I want to prove that given an integral domain, then there exist $x\sim y$ – Fernando Sousa Dec 27 '20 at 05:40
  • @Noah Usually domains are hypothesized to be commutative rings with $1$ etc (esp. when studying divisibility theory). – Bill Dubuque Dec 27 '20 at 06:33

2 Answers2

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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.

Noah Solomon
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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.

Bill Dubuque
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