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:

$\forall x, y \in D: x \sim y$ $x$ 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.

Symmetric
By the definition:
 * $x \sim y \iff x \divides y \land y \divides x \iff y \sim x$

Thus $\sim$ is symmetric.

Transitive
Thus $\sim$ is transitive.

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

Hence the result.