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

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Associate_in_Integral_Domain/Definition_1_Equivalent_to_Definition_3&oldid=401851"