(related to Proposition: Group of Units)

- By hypothesis, $(R, + ,\cdot)$ is an integral domain.
- Let $R^\ast\subset R$ be the subset of $R$ containing all units of $R.$
- Obviously, $1\in R^\ast,$ therefore $R^\ast$ is not empty.
- If $a\in R^\ast$ then, by definition of a unit, there exists an $b\in R$ with $a\cdot b=1.$
- Since $R$ is an integral domain, the multiplication "$\cdot$" is commutative.
- Therefore, $b\cdot a=1,$ thus $b$ is a also unit.
- Therefore $b\in R^\ast,$ if $a\in R^\ast.$
- We have shown that $R^\ast$ contains always both, an element $a\in R^\ast$ and also its multiplicative inverse $b\in R^\ast.$
- Since "$\cdot$" is associative, we have shown all properties of a group.
- This proves that the
*group of units*$(R^\ast,\cdot)$ is indeed a group.∎