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: $\text{(iv)}$