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.
$\blacksquare$