Associatehood is Equivalence Relation

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Theorem

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.

Let $\sim$ be the relation defined on $D$ as:

$\forall x, y \in D: x \sim y$ if and only if $x$ is an associate of $y$


Then $\sim$ is an equivalence relation.


Proof

Checking in turn each of the criteria for equivalence:


Reflexive

Clearly $x \divides x$ as $x = 1_D \circ x$, so $x \sim x$.

Thus $\sim$ is reflexive.

$\Box$


Symmetric

By the definition:

$x \sim y \iff x \divides y \land y \divides x \iff y \sim x$

Thus $\sim$ is symmetric.

$\Box$


Transitive

\(\ds \) \(\) \(\ds x \sim y \land y \sim z\)
\(\ds \) \(\leadsto\) \(\ds \paren {x \divides y \land y \divides x} \land \paren {y \divides z \land z \divides y}\) Definition 1 of Associate
\(\ds \) \(\leadsto\) \(\ds \paren {x \divides y \land y \divides z} \land \paren {z \divides y \land y \divides x}\) Rule of Association and Rule of Commutation
\(\ds \) \(\leadsto\) \(\ds x \divides z \land z \divides x\) Divisor Relation in Integral Domain is Transitive
\(\ds \) \(\leadsto\) \(\ds x \sim z\) Definition 1 of Associate

Thus $\sim$ is transitive.

$\Box$


All criteria are fulfilled for $\sim$ to be an equivalence relation.

Hence the result.

$\blacksquare$


Sources

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