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field of fractions

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Idea

For a commutative integral domain DD, the set of all nonzero elements D ×D^\times is a multiplicative set, the corresponding commutative localization D(D ×) 1DD\to (D^\times)^{-1} D is an injective homomorphism of rings and Q(D)=(D ×) 1DQ(D) = (D^\times)^{-1} D is a field, called the field of fractions or the quotient field of DD. Its elements are fractions a/ba/b where aDa\in D and bD ×b\in D^\times which are by the definition the equivalence classes of pairs (a,b)D×D (a,b) \in D\times D^\sharp and (a,b)(c,d)(a,b)\sim (c,d) iff ad=bca d = b c. The addition is given by the formula

ab+cd=ad+bcbd \frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd}

and multiplication by (a/b)(c/d)=(ac)/(bd)(a/b)(c/d) = (ac)/(bd).

The field of fractions Q(D)Q(D) is unique minimal field in which the integral domain DD is embedded in the sense that every field KDK\supset D contains the subfield isomorphic to Q(D)Q(D), namely consisting of all the fractions a/da/d with aDa\in D, dD d\in D^\sharp taken in the sense of division in KK.

In the noncommutative case, the notion of a quotient skewfield is not always, and not uniquely defined. One class where this notion works well is the case or (left or right) Ore domains where the set of all nonzero elements form a (left or right) Ore set and the Ore localization at that set defines a skewfield of fractions?.

Not every noncommutative integral domain can be embedded at all into a division ring. Suppose it does, say DFD\hookrightarrow F. Then all the say left fractions d 1bd^{-1} b with d,bD,d0d,b\in D, d\neq 0 do not form a subfield, namely we can not in general add a fractions to a fraction with common denominator, so this candidate for a subfield may have noninvertible elements.

Revised on July 7, 2014 08:18:30 by Urs Schreiber (192.76.8.26)