core27 min studyLesson 144 in Mathematics

Integral domains and fraction fields

Construct fractions from a domain and identify the exact role of nonzero denominators and cancellation.

On this page

What you’ll understand

  • Recognize an integral domain and prove its cancellation property.
  • Construct fractions as equivalence classes of numerator-denominator pairs.
  • Distinguish rational functions from polynomial functions and their evaluation domains.

The rational numbers complete integer arithmetic by allowing division by every nonzero integer. The same construction works for any integral domain, including polynomial rings. The resulting fraction field contains the original ring and supplies exactly the inverses needed for its nonzero elements. The construction is not simply a license to write arbitrary fraction symbols: we must decide when two fractions are equal, verify the operations, and explain why zero divisors would break the proposed embedding.

A domain is the right starting point

An integral domain is a nonzero commutative ring with identity and no nonzero zero divisors. Thus a product is zero only if at least one factor is zero. Subtracting equal products proves that every nonzero factor is cancellable. Fields are domains, and the integers are a domain without being a field. A polynomial ring over a domain is also a domain, because the leading coefficients of two nonzero polynomials multiply to a nonzero leading coefficient.

The nonzero condition excludes the zero ring, where the identity equals zero. Commutativity matters for the simple fraction construction used here. Our fractions will move denominators freely through products; more general noncommutative constructions require additional hypotheses. Residue rings modulo a composite positive integer are not domains, since nontrivial factors of the modulus produce nonzero classes with zero product. These distinctions determine which algebraic procedures are available before any calculation begins.

ab=0⟹a=0 or b=0,ac=bc, c≠0⟹a=bab=0\Longrightarrow a=0\ \text{or}\ b=0,\qquad ac=bc,\ c\neq0\Longrightarrow a=b
The zero-product implication and nonzero cancellation hold in an integral domain. They do not assert that c already has an inverse in the domain.↔ Scroll wide equations horizontally.

Fractions are classes, not unique pairs

Start with pairs consisting of a numerator a and a nonzero denominator b. Declare the pair for a over b equivalent to the pair for c over d when ad equals bc. This is the cross-multiplication rule familiar from rational numbers. Reflexivity and symmetry are immediate; transitivity uses cancellation of a nonzero denominator. That is one place where the absence of zero divisors enters the construction rather than merely simplifying a later theorem.

Define addition by a common denominator and define multiplication by multiplying numerators and denominators. Denominator products remain nonzero in a domain, and algebra shows that equivalent input pairs give equivalent results. The class of a over one embeds the original ring injectively. A nonzero fraction has nonzero numerator, so exchanging numerator and denominator gives its inverse. These properties make the collection a field, called the fraction field of the domain.

ab=cd  ⟺  ad=bc,ab+cd=ad+bcbd,(ab)−1=ba(a≠0)\frac ab=\frac cd\iff ad=bc,\qquad \frac ab+\frac cd=\frac{ad+bc}{bd},\qquad \left(\frac ab\right)^{-1}=\frac ba\quad(a\neq0)
All denominators are nonzero elements of a fixed domain. The reciprocal formula applies only to a nonzero fraction.↔ Scroll wide equations horizontally.

The same idea creates rational functions

Taking the fraction field of polynomials over a field gives rational functions. A rational function is an algebraic fraction class, not merely a formula evaluated at a chosen real input. For example, the fractions with numerator x squared minus one and denominator x minus one, and with numerator x plus one and denominator one, represent the same rational function. The polynomial identity after cross multiplication proves the equality without selecting a numerical x.

However, the two displayed expressions have different immediate evaluation domains as real formulas: the first cannot be substituted at one without cancellation. Keep the rational-function identity separate from a claim that an unevaluated denominator can be set to zero. Finally, a ring with a nonzero zero divisor cannot embed in any field while preserving its products and distinct elements. Its nonzero annihilating factors would become nonzero field elements with product zero, contradicting the field's domain property.

Add fractions in a polynomial fraction field

Compute 1x−1+1x+1\frac{1}{x-1}+\frac{1}{x+1} in Q(x)\mathbb{Q}(x), then state where the displayed real-valued formula can be evaluated.

  1. Both denominators are nonzero polynomials in the domain Q[x]\mathbb{Q}[x]. Their product is x2−1x^2-1, also a nonzero polynomial, so it is a valid denominator in the fraction field.

  2. Bring the terms to the common denominator: the numerator becomes (x+1)+(x−1)=2x(x+1)+(x-1)=2x. The fraction-field sum is therefore 2x/(x2−1)2x/(x^2-1).

  3. As a real-valued expression, its denominator vanishes at one and negative one. Both original summands and the resulting displayed formula are defined at every other real input. The algebraic computation did not treat the indeterminate x as a fixed real number.

The result

The sum is 2x/(x2−1)2x/(x^2-1) in the fraction field; its displayed real formula is defined for x≠±1x\neq\pm1.

Explore the model

Interactive lab

Interactive lab · function

Locate a forbidden evaluation

Set coefficient a to one in the lab's y=ax2y=ax^2 plot. Use the position control to locate where x2x^2 becomes zero, and predict why the expression 1/x21/x^2 cannot be evaluated there. The lab plots the denominator polynomial, not its reciprocal or a general rational function.

Locate a forbidden evaluationThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA nonzero polynomial is a valid fraction-field denominator even though it may vanish at particular numerical inputs.

Check your understanding

Why can a ring with nonzero elements a and b satisfying ab = 0 not embed injectively in a field?

Try it yourself

In Q(x)\mathbb{Q}(x), show that (x2−4)/(x−2)(x^2-4)/(x-2) equals x+2x+2. Does the first displayed formula permit direct substitution of x = 2?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Factor the numerator as a polynomial identity before discussing numerical evaluation.

Compare with the solution

The identity x2−4=(x−2)(x+2)x^2-4=(x-2)(x+2) proves equality of the two fraction classes, since cross multiplication gives identical polynomials. The expression with denominator x−2x-2 is not directly evaluable at two, even though the equivalent polynomial x+2x+2 is and has value four there. The field identity and the domain of a chosen formula answer different questions.

Take it with you

  • A domain permits cancellation and embeds in its fraction field.
  • Fractions are equivalence classes; cross multiplication specifies their equality.
  • Algebraic equality of rational functions does not authorize substitution into a zero denominator.

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