# Associatehood is Equivalence Relation

## 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.

$\blacksquare$