Domain (ring theory)

In algebra, a domain is a nonzero ring in which ab = 0 implies a = 0 or b = 0.^[1] (Sometimes such a ring is said to "have the zero-product property".) Equivalently, a domain is a ring in which 0 is the only left zero divisor (or equivalently, the only right zero divisor). A commutative domain is called an integral domain.^[1][2] Mathematical literature contains multiple variants of the definition of "domain".^[3]

Examples and non-examples

• The ring ${\displaystyle \mathbb {Z} /6\mathbb {Z} }$ is not a domain, because the images of 2 and 3 in this ring are nonzero elements with product 0. More generally, for a positive integer ${\displaystyle n}$, the ring ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$ is a domain if and only if ${\displaystyle n}$ is prime.
• A finite domain is automatically a finite field, by Wedderburn's little theorem.
• The
  invertible
  .
• The set of all
  Lipschitz quaternions
  , that is, quaternions of the form ${\displaystyle a+bi+cj+dk}$ where a, b, c, d are integers, is a noncommutative subring of the quaternions, hence a noncommutative domain.
• Similarly, the set of all Hurwitz quaternions, that is, quaternions of the form ${\displaystyle a+bi+cj+dk}$ where a, b, c, d are either all integers or all
  half-integers
  , is a noncommutative domain.
• A matrix ring M_n(R) for n ≥ 2 is never a domain: if R is nonzero, such a matrix ring has nonzero zero divisors and even nilpotent elements other than 0. For example, the square of the matrix unit E₁₂ is 0.
• The tensor algebra of a vector space, or equivalently, the algebra of polynomials in noncommuting variables over a field, ${\displaystyle \mathbb {K} \langle x_{1},\ldots ,x_{n}\rangle ,}$ is a domain. This may be proved using an ordering on the noncommutative monomials.
• If R is a domain and S is an Ore extension of R then S is a domain.
• The Weyl algebra is a noncommutative domain.
• The universal enveloping algebra of any Lie algebra over a field is a domain. The proof uses the standard filtration on the universal enveloping algebra and the Poincaré–Birkhoff–Witt theorem.

Group rings and the zero divisor problem

Suppose that G is a group and K is a field. Is the group ring R = K[G] a domain? The identity

${\displaystyle (1-g)(1+g+\cdots +g^{n-1})=1-g^{n},}$

shows that an element g of finite order n > 1 induces a zero divisor 1 − g in R. The zero divisor problem asks whether this is the only obstruction; in other words,

Given a

torsion-free group

G, is it true that K[G] contains no zero divisors?

No counterexamples are known, but the problem remains open in general (as of 2017).

For many special classes of groups, the answer is affirmative. Farkas and Snider proved in 1976 that if G is a torsion-free

• Zero divisor
• Zero-product property
• Divisor (ring theory)
• Integral domain

Notes