Unit (ring theory)

In the ring of integers Z, the only units are +1 and −1.

Rings of integers ${\displaystyle R={\mathfrak {O}}_{F}}$ in a number field F have, in general, more units. For example,

(√5 + 2)(√5 − 2) = 1

in the ring {{math|[[quadratic integer|Z[1 + √5/2]]]}}, and in fact the unit group of this ring is infinite.

In fact, Dirichlet's unit theorem describes the structure of U(R) precisely: it is isomorphic to a group of the form

${\displaystyle \mathbf {Z} ^{n}\oplus \mu _{R}}$

where ${\displaystyle \mu _{R}}$ is the (finite, cyclic) group of roots of unity in R and n, the rank of the unit group is

${\displaystyle n=r_{1}+r_{2}-1,}$

where ${\displaystyle r_{1},r_{2}}$ are the numbers of real embeddings and the number of pairs of complex embeddings of F, respectively.

This recovers the above example: the unit group of (the ring of integers of) a real quadratic field is infinite of rank 1, since ${\displaystyle r_{1}=2,r_{2}=0}$.

In the ring Z/nZ of integers modulo n, the units are the congruence classes (mod n) represented by integers coprime to n. They constitute the multiplicative group of integers modulo n.

For a commutative ring R, the units of the polynomial ring R[x] are precisely those polynomials

${\displaystyle p(x)=a_{0}+a_{1}x+\dots a_{n}x^{n}}$

such that ${\displaystyle a_{0}}$ is a unit in R, and the remaining coefficients ${\displaystyle a_{1},\dots ,a_{n}}$ are nilpotent elements, i.e., satisfy ${\displaystyle a_{i}^{N}=0}$ for some N.[4] In particular, if R is a domain (has no zero divisors), then the units of R[x] agree with the ones of R. The units of the power series ring ${\displaystyle R[[x]]}$ are precisely those power series

${\displaystyle p(x)=\sum _{i=0}^{\infty }a_{i}x^{i}}$

such that ${\displaystyle a_{0}}$ is a unit in R.[5]

The unit group of the ring M_n(R) of n × n matrices over a commutative ring R (for example, a field) is the group GL_n(R) of invertible matrices.

An element of the matrix ring ${\displaystyle \operatorname {M} _{n}(R)}$ is invertible if and only if the determinant of the element is invertible in R, with the inverse explicitly given by Cramer's rule.

Let ${\displaystyle R}$ be a ring. For any ${\displaystyle x,y}$ in ${\displaystyle R}$, if ${\displaystyle 1-xy}$ is invertible, then ${\displaystyle 1-yx}$ is invertible with the inverse ${\displaystyle 1+y(1-xy)^{-1}x}$.[6] The formula for the inverse can be found as follows: thinking formally, suppose ${\displaystyle 1-yx}$ is invertible and that the inverse is given by a geometric series: ${\displaystyle (1-yx)^{-1}=\sum _{0}^{\infty }(yx)^{n}}$. Then, manipulating it formally,

${\displaystyle (1-yx)^{-1}=1+y\left(\sum _{0}^{\infty }(xy)^{n}\right)x=1+y(1-xy)^{-1}x.}$

See also Hua's identity for a similar type of results.