field of fractions

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

$\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd}$

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

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

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 $D\hookrightarrow F$. Then all the say left fractions $d^{-1} b$ with $d,b\in D, d\neq 0$ 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 October 17, 2012 18:22:20
by Zoran Škoda
(161.53.130.104)