Polynomials and fields · Definition

Field of fractions

modern algebrarings

It is the way to divide in a ring where one cannot always do so: building fractions out of an integral domain, just as the rationals come from the integers.

Definition

The field of fractions \(\operatorname{Frac}(D)\) of an integral domain \(D\) is the field made of the fractions \(a/b\), with \(a,b\in D\) and \(b\neq0\), where \(a/b=c/d\) when \(ad=bc\).

\[\frac ab=\frac cd\iff ad=bc,\qquad \frac ab+\frac cd=\frac{ad+bc}{bd},\qquad \frac ab\cdot\frac cd=\frac{ac}{bd}\]

The rationals are built from the integers by forming pairs \((a,b)\) with \(b\neq0\) and declaring that \((a,b)\) and \((c,d)\) represent the same fraction when \(ad=bc\). The same recipe works in any integral domain \(D\): the classes of pairs form the field \(\operatorname{Frac}(D)\), with the usual addition and multiplication. The absence of zero divisors in \(D\) is what makes the relation transitive and keeps the product of two nonzero denominators nonzero.

The most common examples are \(\operatorname{Frac}(\mathbb{Z})=\mathbb{Q}\), the field of rational functions \(\operatorname{Frac}(F[x])=F(x)\) over a field \(F\), and \(\operatorname{Frac}(\mathbb{Z}[i])=\mathbb{Q}(i)\). The domain \(D\) is identified with the fractions \(a/1\), so \(D\) sits inside its field of fractions, and if \(D\) is already a field nothing is added.

It is the smallest field containing \(D\): any field \(K\) containing \(D\) contains the quotients \(ab^{-1}\), that is, a copy of \(\operatorname{Frac}(D)\). This universal property is what is used when a result about polynomials is proved by passing to rational functions, for example to factor in \(\mathbb{Z}[x]\) using \(\mathbb{Q}[x]\).

Think about it

Why is there no field of fractions of \(\mathbb{Z}_6\)?

Because \(\mathbb{Z}_6\) has zero divisors: from \(2\cdot3=0\) and inverting \(2\) one would get \(3=0\), absurd. No field contains a ring with zero divisors.

What is the field of fractions of a field \(F\)?

\(F\) itself: every nonzero denominator is already invertible, so \(a/b=ab^{-1}\) lies in \(F\).

Exercises 10

References

  • David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004. §7.5
  • Joseph A. Gallian, Contemporary Abstract Algebra, 9th edition, Cengage, 2017. cap. 15
  • I. N. Herstein, Abstract Algebra, 3rd edition, Wiley, 1996. §3.6
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §21
  • Serge Lang, Algebra, Revised 3rd edition, Springer, 2002. cap. II §4