Equivalence of Definitions of Associates

Theorem
Let $\left({D, +, \cdot}\right)$ be an integral domain.

Proof

 * $(1) \implies (3)$:

Suppose that $x, y \in D$ such that $x \mathrel \backslash x$ and $y \mathrel \backslash x$.

Then $x = a y$, $y = b x$.

So $x = a y = a b x$ and so $\left({1 - a b}\right) x = 0$.

Since $D$ is an integral domain, this implies that $1 - a b = 0$, or equivalently $a b = 1$.

In particular, $b$ divides $1$, so by Divisor of Unit is Unit, $b$ is a unit of $D$ with $y = b x$.

$(3) \implies (2)$:

Suppose that $x, y \in D$ such that there exists a unit $u$ of $\left({D, +, \circ}\right)$ such that $u x = y$.

Then by definition of ideals:
 * $\left({y}\right) = \left({u x}\right)$

so it remains to be shown that:
 * $\left({u x}\right) = \left({x}\right)$

Let $z = a x \in \left({x}\right)$.

Then:
 * $z = u^{-1} a \cdot u x \in \left({u x}\right)$

Therefore:
 * $\left({x}\right) \subseteq \left({u x}\right)$

The reverse inclusion is trivial:

Let $z = a u x \in \left({u x}\right)$.

Then:
 * $z = a u \cdot x \in \left({x}\right)$

Therefore:
 * $\left({u x}\right) = \left({x}\right)$

$(2) \implies (1)$:

Suppose that $x, y \in D$ such that $\left({x}\right) = \left({y}\right)$.

In particular since an integral domain has a unity $1$, this implies that:
 * $x = 1 \cdot x \in \left({y}\right)$

and:
 * $y = 1 \cdot y \in \left({x}\right)$

Thus there exist $a, b \in D$ such that $x = a y$ and $y = b x$.

But this states precisely that $x \mathrel \backslash y$ and $y \mathrel \backslash x$.

$(3) \implies (1)$:

Let $\exists u \in R: y = u x$.

Then by definition of unit:
 * $x = u^{-1} y$

By the definition of divisor, both:
 * $x \mathrel \backslash y$

and:
 * $y \mathrel \backslash x$