Domain (ring theory)

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In mathematics, and more specifically in algebra, a domain is a ring such that ab = 0 implies a = 0 or b = 0.[1] That is, it is a ring which has no left or right zero divisors. (Sometimes such a ring is said to "have the zero-product property.") Some authors require the ring to be nontrivial (that is, it must have more than one element).[2] If the domain has a multiplicative identity (which we may call 1), this nontriviality is equivalent to saying that 1 ≠ 0.[3] A commutative domain with 1 ≠ 0 is called an integral domain.[4]

A finite (nontrivial) domain is automatically a finite field by Wedderburn's little theorem.

Zero divisors have a topological interpretation, at least in the case of commutative rings: a ring R is an integral domain, if and only if it is reduced and its spectrum Spec R is an irreducible topological space. The first property is often considered to encode some infinitesimal information, whereas the second one is more geometric.

An example: the ring k[x, y]/(xy), where k is a field, is not a domain, as the images of x and y in this ring are zero divisors. Geometrically, this corresponds to the fact that the spectrum of this ring, which is the union of the lines x = 0 and y = 0, is not irreducible. Indeed, these two lines are its irreducible components.

Constructions of domains[edit]

One way of proving that a ring is a domain is by exhibiting a filtration with special properties.

Theorem: If R is a filtered ring whose associated graded ring gr(R) is a domain, then R itself is a domain.

This theorem needs to be complemented by the analysis of the graded ring gr(R).


  • The ring nZ is a domain (for each integer n > 1) but not an integral domain since 1 \not \in n\mathbb{Z}.[2]
  • The quaternions form a noncommutative domain. More generally, any division algebra is a domain, since all its non-zero elements are invertible.
  • The set of all integral quaternions is a noncommutative ring which is a subring of quaternions, hence a noncommutative domain.
  • The matrix ring of order greater than one is never a domain, since it has zero divisors, and even nilpotent elements. For example, the square of the matrix unit E12 is zero.
  • The tensor algebra of a vector space, or equivalently, the algebra of polynomials in noncommuting variables over a field,  \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. Indeed, it has two natural filtrations, by the degree of the derivative and by the total degree, and the associated graded ring for either one is isomorphic to the ring of polynomials in two variables. By the theorem above, the Weyl algebra is a 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[edit]

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


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

Given a field K and 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 2007).

For many special classes of groups, the answer is affirmative. Farkas and Snider proved in 1976 that if G is a torsion-free polycyclic-by-finite group and char K = 0 then the group ring K[G] is a domain. Later (1980) Cliff removed the restriction on the characteristic of the field. In 1988, Kropholler, Linnell and Moody generalized these results to the case of torsion-free solvable and solvable-by-finite groups. Earlier (1965) work of Michel Lazard, whose importance was not appreciated by the specialists in the field for about 20 years, had dealt with the case where K is the ring of p-adic integers and G is the pth congruence subgroup of GL(n,Z).

See also[edit]


  1. ^ Polcino M. & Sehgal (2002), p. 65.
  2. ^ a b Lanski (2005), p. 343, Definition 10.18.
  3. ^ Jacobson (2009), p. 90, Section 2.2. Note that if 1=0, then a=1a=0a=0 showing that all elements are 0.
  4. ^ Rowen (1994), p. 99.