Let $(R,\cdot,+)$ be an integral domain with the multiplicative neutral element $1.$ And element $a\in R$ is called a unit of $R,$ if $$a\mid 1\,$$ i.e. $a$ is a divisor of $1$.
- Unfolding the definition of a divisor in a ring, this means that there exists \(b\in R\) with \(a\cdot b=1\).
- In other words, units in \(R\) are exactly those of its elements, which have inverse elements with respect to the operation "\(\cdot\)".
Table of Contents
- Proposition: Group of Units
Definitions: 1 2 3
Proofs: 4 5 6 7 8 9
Propositions: 10 11
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- Koch, H.; Pieper, H.: "Zahlentheorie - Ausgewählte Methoden und Ergebnisse", Studienbücherei, 1976