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.