Domain (ring theory)

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In mathematics, and more specifically in algebra, a domain is a ring such that ab = 0 implies that either 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 is equivalent to saying that 1 ≠ 0[3] Thus a domain is a nontrivial ring without left or right zero divisors. A commutative domain with 1 ≠ 0 is called an integral domain.[4]

A finite 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).

Examples[edit]

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

 (1-g)(1+g+\ldots+g^{n-1})=1-g^n,

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]

Notes[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.

References[edit]