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Ore condition

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In mathematics, especially in the area of algebra known as ring theory, the Ore condition is a condition introduced by Øystein Ore, in connection with the question of extending beyond commutative rings the construction of a field of fractions, or more generally localization of a ring. The right Ore condition for a domain R, and any pair a, b of non-zero elements, is the requirement that the sets aR and bR should intersect in more than the element 0. The left Ore condition is defined similarly. A domain that satisfies the right Ore condition is called a right Ore domain.

For every right Ore domain R, there is a unique (up to natural R-isomorphism) division ring D containing R as a subring such that every element of D is of the form rs−1, for r in R and s nonzero in R. Such a division ring D is called a ring of right fractions of R, and R is called a right order in D. In particular, any domain satisfying either Ore condition R can be considered a subring of a division ring. However, the converse does not hold: for example, if F is any field, and G = \langle x,y \rangle\, is the free monoid on two symbols x and y, then the monoid ring F[G]\, does not satisfy any Ore condition, but is a free ideal ring and thus indeed a subring of a division ring, by (Cohn 1995, Cor 4.5.9). In fact, a subring R of a division ring D is a right Ore domain if and only if D is a flat left R-module (Lam 2007, Ex. 10.20).

A different, stronger version of the Ore conditions is usually given for the case where R is not a domain, namely that there should be a common multiple

c = au = bv

with u, v not zero divisors. On this case Ore's theorem guarantees the existence of an over-ring called the (right or left) classical ring of quotients.

[edit] Multiplicative sets

The Ore condition can be generalized to other multiplicative subsets, and is presented in textbook form in (Lam 1999, §10) and (Lam 2007, §10). A subset S of a ring R is called a right denominator set if it satisfies the following for every a,b in R, and s, t in S:

  1. st in S, the set S is multiplicative
  2. aSsR is not empty, the set S is right permutable
  3. If sa = 0, then there is some u in S with au = 0, the set is right reversible

If S is a right denominator set, then one can construct the ring of right fractions RS−1 similarly to the commutative case. If S is taken to be the set of regular elements (those elements a in R such that if b in R is nonzero, then ab and ba are nonzero), then the right Ore condition is simply the requirement that S be a right denominator set.

Many properties of commutative localization hold in this more general setting. If S is a right denominator set for a ring R, then the left R-module RS−1 is flat. Furthermore, if M is a right R-module, then the S-torsion, torS(M) = { m in M : ms = 0 for some s in S }, is an R-submodule isomorphic to Tor1(M,RS−1), and the module MR RS−1 is naturally isomorphic to a module MS−1 consisting of "fractions" as in the commutative case.

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