Equivalence of Definitions of Associates

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

Then the definitions of associate elements in $D$ are equivalent.

That is, the following conditions are equivalent for any elements $x,y \in D$:


 * $(1): \quad$ $x \mathop \backslash y$ and $y \mathop \backslash x$


 * $(2): \quad$ there exists a unit $u$ of $\left({D, +, \circ}\right)$ such that $u x = y$


 * $(3): \quad$ $\left({ x }\right) = \left({ y }\right)$, where $\left({ x }\right)$ and $\left({ y }\right)$ denote the ideals generated by $x$ and $y$ respectively

Proof
$(1) \implies (2)$:

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

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

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

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

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

$(2) \implies (3)$:

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

Then of course we have $\left({ y }\right) = \left({ ux }\right)$ so we want to show that $\left({ ux }\right) = \left({ x }\right)$.

If $z = ax \in \left({ x }\right)$ then we have:
 * $z = u^{-1}a \cdot ux \in \left({ ux }\right)$

Therefore $\left({ x }\right) \subseteq \left({ ux }\right)$.

The converse is trivial: if $z = aux \in \left({ ux }\right)$, then $z = au \cdot x \in \left({ x }\right)$.

Therefore $\left({ ux }\right) = \left({ x }\right)$.

$(3) \implies (1)$:

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

In particular we have that $x \in \left({ y }\right)$ and $y \in \left({ x }\right)$.

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

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