Proof: By Induction
(related to Lemma: A Criterion for Associates)
- Assume, $a,b\in R$ are associates $a\sim b.$
- By definition, $a\mid b$ and $b\mid a.$
- This means that there are $c,d\in R$ with $ac=b$ and $bd=a.$
- Therefore, $a=bd=acd$, which means $0=a(cd-1)$
- Since $(R, + , \cdot)$ is an integral domain, we have either $a=0$ or $cd=1$
- If $a=0$ then $b=0$ and, trivially, $a=cb$ for some $c\in R^\ast,$ ($(R^\ast,\cdot)$ being the group of units).
- If $a\neq 0$ then $cd=1$ but then $c$ is a unit, therefore $c\in R^\ast,$ and again $a=cb.$
- Let $a=cb$ with $c$ being a unit.
- Therefore, $b\mid a.$
- But since $c$ has an multiplicative inverse $d\in R^\ast$ with $cd=1$, it follows that $b=1\cdot b=(cd)b=(cb)d=ad.$
- It follows that $a\mid b.$
- By dfinition, the elements $a,b$ are associates $a\sim b.$
Thank you to the contributors under CC BY-SA 4.0!
- Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013