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# group of units

###### Theorem.

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

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.

Corollary. In a commutative ring, a ring product is a unit iff all factors are units.

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

Examples

1. When $R=\mathbb{Z}$, then $E=\{1,\,-1\}$.

2. When $R=\mathbb{Z}[i]$, the ring of Gaussian integers, then $E=\{1,\,i,\,-1,\,-i\}$.

3. When $R=\mathbb{Z}[\sqrt{3}]$, then $E=\{\pm(2\!+\!\sqrt{3})^{n}\,\vdots\,\,\,n\in\mathbb{Z}\}$.

4. When $R=K[X]$ where $K$ is a field, then $E=K\!\smallsetminus\!\{0\}$.

5. 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 $\gcd(l,m)=1$.

## Mathematics Subject Classification

16U60*no label found*13A05

*no label found*

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