# 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.

$\blacksquare$