Ore Condition - Application

Application

Since it is well known that each integral domain is a subring of a field of fractions (via an embedding) in such a way that every element is of the form rs−1 with s nonzero, it is natural to ask if the same construction can take a noncommutative domain and associate a division ring (a noncommutative field) with the same property. It turns out that the answer is sometimes "no", that is, there are domains which do not have an analogous "right division ring of fractions".

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. The notion of a ring of left fractions and left order are defined analogously, with elements of D being of the form s−1r.

It is important to remember that the definition of R being a right order in D includes the condition that D must consist entirely of elements of the form rs−1. Any domain satisfying one of the Ore conditions can be considered a subring of a division ring, however this does not automatically mean R is a left order in D, since it is possible D has an element which is not of the form s−1r. Thus it is possible for R to be a right-not-left Ore domain. Intuitively, the condition that all elements of D be of the form rs−1 says that R is a "big" R-submodule of D. In fact the condition ensures R_R is an essential submodule of D_R. Lastly, there is even an example of a domain in a division ring which satisfies neither Ore condition (see examples below).

Another natural question is: "When is a subring of a division ring right Ore?" One characterization is that 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. In this case, Ore's theorem guarantees the existence of an over-ring called the (right or left) classical ring of quotients.