Divisor Relation in Integral Domain is Transitive
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Theorem
Let $\struct {D, +, \circ}$ be an integral domain.
Let $x, y, z \in D$.
Then:
- $x \divides y \land y \divides z \implies x \divides z$
Corollary
The divisibility relation is a transitive relation on $\Z$, the set of integers.
That is:
- $\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$
Proof
Let $x \divides y \land y \divides z$.
Then from the definition of divisor, we have:
- $x \divides y \iff \exists s \in D: y = s \circ x$
- $y \divides z \iff \exists t \in D: z = t \circ y$
Then:
- $z = t \circ \paren {s \circ x} = \paren {t \circ s} \circ x$
Thus:
- $\exists \paren {t \circ s} \in D: z = \paren {t \circ s} \circ x$
and 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 $49 \ \text{(i)}$