Concept
In algebra, a domain is a nonzero ring in which ab = 0 implies a = 0 or b = 0. (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. Mathematical literature contains multiple variants of the definition of "domain". 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. A finite domain is automatically a finite field, by Wedderburn's little theorem. The quaternions form a noncommutative domain. More generally, any division algebra is a domain, since all its nonzero elements are invertible. The set of all integral quaternions is a noncommutative ring which is a subring of quaternions, hence a noncommutative domain. A matrix ring Mn(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, 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. 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 > 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 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).