Associates are Unit Multiples

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Theorem

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

Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$.


Then:

$\forall x, y \in D: x \divides y \land y \divides x \iff \exists u \in U_D: y = u \circ x$


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 $x \divides y$ and $y \divides x$.


Suppose $x \divides y$ and $y \divides x$.

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:

\(\displaystyle 1_D \circ x\) \(=\) \(\displaystyle x\) $\quad$ Definition of Unity of Ring $\quad$
\(\displaystyle \) \(=\) \(\displaystyle s \circ y\) $\quad$ from above $\quad$
\(\displaystyle \) \(=\) \(\displaystyle s \circ \paren {t \circ x}\) $\quad$ from above $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {s \circ t} \circ x\) $\quad$ Definition of Associative Operation $\quad$


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

The result follows.

$\blacksquare$


Sources