# 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 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. 1.

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

2. 2.

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

3. 3.

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

4. 4.

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

5. 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  $\gcd(l,m)=1$.

 Title group of units Canonical name GroupOfUnits Date of creation 2013-03-22 14:41:32 Last modified on 2013-03-22 14:41:32 Owner pahio (2872) Last modified by pahio (2872) Numerical id 24 Author pahio (2872) Entry type Theorem Classification msc 16U60 Classification msc 13A05 Synonym unit group Related topic CommutativeRing Related topic DivisibilityInRings Related topic NonZeroDivisorsOfFiniteRing Related topic PrimeResidueClass Defines group of units of ring Defines multiplicative group of field