Inalgebra, adomain is anonzeroring in whichab = 0 impliesa = 0 orb = 0.[1] (Sometimes such a ring is said to "have thezero-product property".) Equivalently, a domain is a ring in which 0 is the only leftzero divisor (or equivalently, the only right zero divisor). Acommutative domain is called anintegral domain.[1][2] Mathematical literature contains multiple variants of the definition of "domain".[3]
The ring 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, the ring is a domain if and only if is prime.
The set of allLipschitz quaternions, that is, quaternions of the form wherea,b,c,d are integers, is a noncommutative subring of the quaternions, hence a noncommutative domain.
Similarly, the set of allHurwitz quaternions, that is, quaternions of the form wherea,b,c,d are either all integers or allhalf-integers, is a noncommutative domain.
Amatrix ring Mn(R) forn ≥ 2 is never a domain: ifR is nonzero, such a matrix ring has nonzero zero divisors and evennilpotent elements other than 0. For example, the square of thematrix unitE12 is 0.
Thetensor algebra of avector space, or equivalently, the algebra of polynomials in noncommuting variables over a field, is a domain. This may be proved using an ordering on the noncommutative monomials.
IfR is a domain andS is anOre extension ofR thenS is a domain.
Suppose thatG is agroup andK is afield. Is thegroup ringR =K[G] a domain? The identity
shows that an elementg of finiteordern > 1 induces a zero divisor1 −g inR. Thezero divisor problem asks whether this is the only obstruction; in other words,
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 ifG is a torsion-freepolycyclic-by-finite group andcharK = 0 then the group ringK[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-freesolvable and solvable-by-finite groups. Earlier (1965) work ofMichel Lazard, whose importance was not appreciated by the specialists in the field for about 20 years, had dealt with the case whereK is the ring ofp-adic integers andG is thepthcongruence subgroup ofGL(n,Z).
Zero divisors have a topological interpretation, at least in the case of commutative rings: a ringR is an integral domain if and only if it isreduced and itsspectrum SpecR is anirreducible topological space. The first property is often considered to encode some infinitesimal information, whereas the second one is more geometric.
An example: the ringk[x,y]/(xy), wherek is a field, is not a domain, since the images ofx andy in this ring are zero divisors. Geometrically, this corresponds to the fact that the spectrum of this ring, which is the union of the linesx = 0 andy = 0, is not irreducible. Indeed, these two lines are its irreducible components.
^Some authors also consider thezero ring to be a domain: see Polcino M. & Sehgal (2002), p. 65. Some authors apply the term "domain" also torngs with the zero-product property; such authors considernZ to be a domain for each positive integern: see Lanski (2005), p. 343. But integral domains are always required to be nonzero and to have a 1.