Construct division without losing equality

Try

The integer pairs (1,2) and (2,4) should represent the same fraction. Their cross products agree. To generalize this construction, start with a domain R and pairs (a,b) with b nonzero. Declare (a,b) equivalent to (c,d) when ad=bc.

Before adding or multiplying fractions, prove this really is an equivalence relation. Otherwise a “fraction” might depend on which chain of equal representations was used.

Predict

Suppose ad=bc and cf=de. Predict how to derive af=be. Multiplying the equations gives a common factor d that must be canceled. This is where the domain hypothesis enters, and it is why nonzero denominators alone are not sufficient in a ring with zero divisors.

Experiment

Track representative-independent arithmetic in the integer example.

Operation	Pair formula	Example
Addition	(a,b)+(c,d)=(ad+bc,bd)	1/2+1/3=5/6
Multiplication	(a,b)(c,d)=(ac,bd)	(1/2)(2/3)=1/3
Nonzero inverse	(a,b)⁻¹=(b,a)	(3/5)⁻¹=5/3

Replace one half by two fourths and recompute the first row. The raw numerator and denominator change, but cross multiplication verifies equality. The formulas must preserve equivalence for arbitrary representatives, not only these examples.

Understand

Reflexivity and symmetry of the relation are immediate from equality. For transitivity, ad=bc and cf=de imply adf=bcf=bde. Since d is nonzero in a domain, cancellation gives af=be. The product bd is nonzero, so the arithmetic formulas remain within the allowed pairs.

To check multiplication respects representatives, suppose ab′=a′b and cd′=c′d. Multiplying gives acb′d′=a′c′bd, exactly the cross-product equality for the products. Addition follows by distributing the proposed cross products and using the same two equalities. Thus the operations descend to equivalence classes.

The resulting set Frac(R) is a field. Zero is 0/1, one is 1/1, additive inverse is −a/b, and a nonzero a/b has inverse b/a. A fraction is zero exactly when its numerator is zero. Associativity and distributivity follow by putting representatives over common denominators and applying the ring laws. The map R→Frac(R), a↦a/1, is an injective unital homomorphism because equality of such fractions means equality of their numerators.

This field has a useful universal property. If φ:R→K is an injective unital map into a field, it extends uniquely by a/b↦φ(a)φ(b)⁻¹. Injectivity keeps denominators nonzero. Cross multiplication proves well-definedness, and the formulas preserve operations. Any extension must send b⁻¹ to the inverse of φ(b), so no other extension is possible.

Without the domain condition, the naive construction fails. In Z/6Z the nonzero denominators two and three have zero product, so pair multiplication is not closed. More general localizations exist, but require choosing allowed denominators and a suitable relation; that theory is beyond this construction.

Apply

Compare F[x] with its fraction field F(x). A polynomial inverse of x would contradict additivity of degree, while the formal fraction 1/x is available in F(x). Rational functions are formal fraction classes: equality is cross multiplication, not agreement at every numerical input where a denominator might vanish.

Write down one place where each assumption is used: commutativity, no zero divisors, nonzero denominator, and injectivity of the map into K. This makes the universal property a proved statement rather than an informal claim that fractions “behave as expected.”

Continue

Field extensions next study larger fields as vector spaces over smaller ones. Algebraic elements behave differently from the transcendental symbol x used in a rational-function field.