# Proof: By Induction

(related to Lemma: A Criterion for Associates)

### "$\Rightarrow$"

• 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 $a=cb.$

### "$\Leftarrow$"

• 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 $a\sim b.$

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### References

#### Bibliography

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013