Let $(R,\cdot,+)$ be an integral domain with the multiplicative neutral element $1,$ and let $a\in R.$

We call $a$ a **unit** of $R$ if and only if $$a\mid 1\,$$ i.e. $a$ is a divisor of $1$.

- Unfolding the definition of 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 “\(\cdot\)”.

| | | | | created: 2019-06-27 18:36:41 | modified: 2019-06-27 18:48:22 | by: *bookofproofs* | references: [8250]

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[8250] **Koch, H.; Pieper, H.**: “Zahlentheorie – Ausgewählte Methoden und Ergebnisse”, Studienbücherei, 1976

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