In algebra, a domain is a nonzero ring in which implies or .[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.[2] Mathematical literature contains multiple variants of the definition of "domain".[3]
Z/6Z
n
Z/nZ
n
a+bi+cj+dk
a+bi+cj+dk
K\langlex1,\ldots,xn\rangle,
Suppose that G is a group and K is a field. Is the group ring a domain? The identity
(1-g)(1+g+ … +gn-1)=1-gn,
shows that an element g of finite order induces a zero divisor 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 2017).
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 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 .
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, where k is a field, is not a domain, since 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 and, is not irreducible. Indeed, these two lines are its irreducible components.