# Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 3

## 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 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

\(\ds y\) | \(=\) | \(\ds u \circ x\) | ||||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(=\) | \(\ds u^{-1} \circ y\) | 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.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Theorem $50$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 62.1$ Factorization in an integral domain: $\text{(iv)}$