Associates are Unit Multiples

Theorem
Let $\left({D, +, \circ}\right)$ be an integral domain whose unity is $1_D$.

Let $\left({U_D, \circ}\right)$ be the group of units of $\left({D, +, \circ}\right)$.

Then:
 * $\forall x, y \in D: x \backslash y \land y \backslash x \iff \exists u \in U_D: y = u \circ x$.

That is, if an element of $D$ is an associate of another element of $D$, then one is a unit multiple of the other.

Proof

 * If $y = u \circ x$, then $x = u^{-1} y$, and by the definition of divisor, both $x \backslash y$ and $y \backslash x$.


 * Suppose $x \backslash y$ and $y \backslash x$.

Then $\exists s, t \in D: y = t \circ x, x = s \circ y$.

If either $x = 0_D$ or $y = 0_D$, then so must be the other (as an integral domain has no zero divisors by definition).

So $x = 1_D \circ y$ and $y = 1_D \circ x$, and the result holds.

Otherwise:

So $s \circ t = 1_D$ and both $s \in U_D, t \in U_D$.

The result follows.