# Equivalence of Definitions of Associate in Integral Domain

## Theorem

The following definitions of the concept of Associate in the context of Integral Domain are equivalent:

Let $\struct {D, +, \circ}$ be an integral domain.

Let $x, y \in D$.

### Definition 1

$x$ is an associate of $y$ (in $D$) if and only if they are both divisors of each other.

That is, $x$ and $y$ are associates (in $D$) if and only if $x \divides y$ and $y \divides x$.

### Definition 2

$x$ and $y$ are associates (in $D$) if and only if:

$\ideal x = \ideal y$

where $\ideal x$ and $\ideal y$ denote the ideals generated by $x$ and $y$ respectively.

### Definition 3

$x$ and $y$ are associates (in $D$) if and only if there exists a unit $u$ of $\struct {D, +, \circ}$ such that:

$y = u \circ x$

and consequently:

$x = u^{-1} \circ y$

That is, if and only if $x$ and $y$ are unit multiples of each other.

## Proof

### $(1)$ is Equivalent to $(2)$

We are to show that:

$x \divides y \text{ and } y \divides x \iff \ideal x = \ideal y$

Thus:

 $\ds$  $\ds x \divides y \text{ and } y \divides x$ Definition 1 of Associate in Integral Domain $\ds$ $\leadstoandfrom$ $\ds \ideal y \subseteq \ideal x \text{ and } \ideal x \subseteq \ideal y$ Element in Integral Domain is Divisor iff Principal Ideal is Superset $\ds$ $\leadstoandfrom$ $\ds \ideal x = \ideal y$ Definition 2 of Set Equality

$\blacksquare$

### $(1)$ is Equivalent to $(3)$

 $\ds y$ $=$ $\ds u \circ x$ $\ds \leadstoandfrom \ \$ $\ds x$ $=$ $\ds u^{-1} \circ y$ Definition of Unit of Ring

By the definition of divisor:

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

$\Box$

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

Then $\exists s, t \in D$ such that:

$(1): \quad y = t \circ x$

and:

$(2): \quad x = s \circ y$

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:

 $\ds 1_D \circ x$ $=$ $\ds x$ Definition of Unity of Ring $\ds$ $=$ $\ds s \circ y$ from $(2)$ $\ds$ $=$ $\ds s \circ \paren {t \circ x}$ from $(1)$ $\ds$ $=$ $\ds \paren {s \circ t} \circ x$ Definition of Associative Operation

So:

$s \circ t = 1_D$

and both $s \in U_D$ and $t \in U_D$.

The result follows.

$\blacksquare$