Equivalence of Definitions of Associate in Integral Domain

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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.


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

We are to show that:

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


\(\ds \) \(\) \(\ds x \divides y \text{ and } y \divides x\) 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


$(1)$ 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 $y \divides x$


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

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

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


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

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.


\(\ds 1_D \circ x\) \(=\) \(\ds x\) 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


$s \circ t = 1_D$

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

The result follows.