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group of units

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 \, The set $E$ of units of a ring $R$ forms a group with respect to ring multiplication. 

{\em Proof.}\, If $u$ and $v$ are two units, then there are the elements $r$ and $s$ of $R$ such that\, $ru = ur = 1$\, and\, $sv = vs = 1$.\, Then we get that\,$(sr)(uv) = s(r(uv)) = s((ru)v) = s(1v) = sv = 1$,\, similarly\, $(uv)(sr) = 1$.\, Thus also $uv$ is a unit, which means that $E$ is closed under multiplication.\, Because\, $1 \in E$\, and along with $u$ also its inverse $r$ belongs to $E$, the set $E$ is a group.\\

\textbf{Corollary.}\, In a commutative ring, a ring product is a unit iff all \PMlinkescapetext{factors} are units.\\

The group $E$ of the units of the ring $R$ is called the {\it group of units of the ring}.\, If $R$ is a field, $E$ is said to be the {\it multiplicative group of the field}.\\

 \item When\, $R = \mathbb{Z}$, then\, $E = \{1,\,-1\}$.
 \item When\, $R = \mathbb{Z}[i]$,\, the ring of Gaussian integers, then\, 
$E = \{1,\,i,\,-1,\,-i\}$.
 \item When\, $R = \mathbb{Z}[\sqrt{3}]$, \PMlinkname{then}{UnitsOfQuadraticFields}\, 
$E = \{\pm(2\!+\!\sqrt{3})^n\,\vdots\,\,\, n\in\mathbb{Z}\}$.
 \item When\, $R = K[X]$\, where $K$ is a field, then\, $E = K\!\smallsetminus\!\{0\}$.
 \item When\, 
$R = \{0\!+\!\mathbb{Z},\,1\!+\!\mathbb{Z},\,\ldots,\, 
 m\!-\!1\!+\!\mathbb{Z}\}$\, is the residue class ring modulo $m$, then\, $E$ consists of the prime classes modulo $m$, i.e. the residue classes $l\!+\!\mathbb{Z}$ satisfying\, $\gcd(l, m) = 1$.