Editing Unit (ring theory) (section)

== Examples ==
{{anchor|−1}}The multiplicative identity {{math|1}} and its additive inverse {{math|−1}} are always units.  More generally, any [[root of unity]] in a ring {{mvar|R}} is a unit: if {{math|1=''r<sup>n</sup>'' = 1}}, then {{math|1=''r''<sup>''n''−1</sup>}} is a multiplicative inverse of {{mvar|r}}.
In a [[zero ring|nonzero ring]], the [[additive identity|element 0]] is not a unit, so {{math|''R''{{sup|×}}}} is not closed under addition.
A nonzero ring {{mvar|R}} in which every nonzero element is a unit (that is, {{math|1=''R''{{sup|×}} = ''R'' &setminus; {{mset|0}}}}) is called a [[division ring]] (or a skew-field). A commutative division ring is called a [[field (mathematics)|field]]. For example, the unit group of the field of [[real number]]s {{math|'''R'''}} is {{math|'''R''' &setminus; {{mset|0}}}}.

=== Integer ring ===
In the ring of [[integers]] {{math|'''Z'''}}, the only units are {{math|1}} and {{math|−1}}.

In the ring {{math|'''Z'''/''n'''''Z'''}} of [[Modular arithmetic#Integers modulo m|integers modulo {{mvar|n}}]], the units are the congruence classes {{math|(mod ''n'')}} represented by integers [[coprime]] to {{mvar|n}}. They constitute the [[multiplicative group of integers modulo n|multiplicative group of integers modulo {{mvar|n}}]].

=== Ring of integers of a number field ===
In the ring {{math|'''Z'''[{{sqrt|3}}]}} obtained by adjoining the [[quadratic integer]] {{math|{{sqrt|3}}}} to {{math|'''Z'''}}, one has {{math|1= (2 + {{sqrt|3}})(2 − {{sqrt|3}}) = 1}}, so {{math|2 + {{sqrt|3}}}} is a unit, and so are its powers, so {{math|'''Z'''[{{sqrt|3}}]}} has infinitely many units.

More generally, for the [[ring of integers]] {{mvar|R}} in a [[number field]] {{mvar|F}}, [[Dirichlet's unit theorem]] states that {{math|''R''{{sup|×}}}} is isomorphic to the group 
<math display="block">\mathbf Z^n \times \mu_R</math>
where <math>\mu_R</math> is the (finite, cyclic) group of roots of unity in {{mvar|R}} and {{mvar|n}}, the [[rank of a module|rank]] of the unit group, is
<math display="block">n = r_1 + r_2 -1, </math>
where <math>r_1, r_2</math> are the number of real embeddings and the number of pairs of complex embeddings of {{mvar|F}}, respectively.

This recovers the {{math|'''Z'''[{{sqrt|3}}]}} example: The unit group of (the ring of integers of) a [[real quadratic field]] is infinite of rank 1, since <math>r_1=2, r_2=0</math>.

=== Polynomials and power series ===
For a commutative ring {{mvar|R}}, the units of the [[polynomial ring]] {{math|''R''[''x'']}} are the polynomials
<math display="block">p(x) = a_0 + a_1 x + \dots + a_n x^n</math>
such that {{math|''a''<sub>0</sub>}} is a unit in {{mvar|R}} and the remaining coefficients <math>a_1, \dots, a_n</math> are [[nilpotent]], i.e., satisfy <math>a_i^N = 0</math> for some {{math|''N''}}.{{sfn|Watkins|2007|loc=Theorem 11.1|ps=}}
In particular, if {{mvar|R}} is a [[domain (ring theory)|domain]] (or more generally [[reduced ring|reduced]]), then the units of {{math|''R''[''x'']}} are the units of {{mvar|R}}.
The units of the [[power series ring]] <math>R[[x]]</math> are the power series
<math display="block">p(x)=\sum_{i=0}^\infty a_i x^i</math>
such that {{math|''a''<sub>0</sub>}} is a unit in {{mvar|R}}.{{sfn|Watkins|2007|loc=Theorem 12.1|ps=}}

=== Matrix rings ===
The unit group of the ring {{math|M<sub>''n''</sub>(''R'')}} of [[square matrix|{{math|''n'' × ''n''}} matrices]] over a ring {{mvar|R}} is the group {{math|[[general linear group|GL<sub>''n''</sub>(''R'')]]}} of [[invertible matrix|invertible matrices]].  For a commutative ring {{mvar|R}}, an element {{mvar|A}} of {{math|M<sub>''n''</sub>(''R'')}} is invertible if and only if the [[determinant]] of {{mvar|A}} is invertible in {{mvar|R}}.  In that case, {{math|''A''{{sup|−1}}}} can be given explicitly in terms of the [[adjugate matrix]].

=== In general ===
For elements {{mvar|x}} and {{mvar|y}} in a ring {{mvar|R}}, if <math>1 - xy</math> is invertible, then <math>1 - yx</math> is invertible with inverse <math>1 + y(1-xy)^{-1}x</math>;{{sfn|Jacobson|2009|loc=§2.2 Exercise 4|ps=}} this formula can be guessed, but not proved, by the following calculation in a ring of noncommutative power series:
<math display="block">(1-yx)^{-1} = \sum_{n \ge 0} (yx)^n = 1 + y \left(\sum_{n \ge 0} (xy)^n \right)x = 1 + y(1-xy)^{-1}x.</math>
See [[Hua's identity]] for similar results.