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Abstract Algebra: Theory and Applications

Thomas W. Judson

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Section 18.1 Fields of Fractions

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Every field is also an integral domain; however, there are many integral domains that are not fields. For example, the integers Z form an integral domain but not a field. A question that naturally arises is how we might associate an integral domain with a field. There is a natural way to construct the rationals Q from the integers: the rationals can be represented as formal quotients of two integers. The rational numbers are certainly a field. In fact, it can be shown that the rationals are the smallest field that contains the integers. Given an integral domain D, our question now becomes how to construct a smallest field F containing D. We will do this in the same way as we constructed the rationals from the integers.
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An element p/q∈Q is the quotient of two integers p and q; however, different pairs of integers can represent the same rational number. For instance, 1/2=2/4=3/6. We know that
ab=cd
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if and only if ad=bc. A more formal way of considering this problem is to examine fractions in terms of equivalence relations. We can think of elements in Q as ordered pairs in Z×Z. A quotient p/q can be written as (p,q). For instance, (3,7) would represent the fraction 3/7. However, there are problems if we consider all possible pairs in Z×Z. There is no fraction 5/0 corresponding to the pair (5,0). Also, the pairs (3,6) and (2,4) both represent the fraction 1/2. The first problem is easily solved if we require the second coordinate to be nonzero. The second problem is solved by considering two pairs (a,b) and (c,d) to be equivalent if ad=bc.
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If we use the approach of ordered pairs instead of fractions, then we can study integral domains in general. Let D be any integral domain and let
S={(a,b):a,b∈D and b≠0}.
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Define a relation on S by (a,b)∼(c,d) if ad=bc.

Proof.

Since D is commutative, ab=ba; hence, ∼ is reflexive on D. Now suppose that (a,b)∼(c,d). Then ad=bc or cb=da. Therefore, (c,d)∼(a,b) and the relation is symmetric. Finally, to show that the relation is transitive, let (a,b)∼(c,d) and (c,d)∼(e,f). In this case ad=bc and cf=de. Multiplying both sides of ad=bc by f yields
afd=adf=bcf=bde=bed.
Since D is an integral domain, we can deduce that af=be or (a,b)∼(e,f).
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We will denote the set of equivalence classes on S by FD. We now need to define the operations of addition and multiplication on FD. Recall how fractions are added and multiplied in Q:
ab+cd=ad+bcbd;ab⋅cd=acbd.
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It seems reasonable to define the operations of addition and multiplication on FD in a similar manner. If we denote the equivalence class of (a,b)∈S by [a,b], then we are led to define the operations of addition and multiplication on FD by
[a,b]+[c,d]=[ad+bc,bd]
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and
[a,b]⋅[c,d]=[ac,bd],
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respectively. The next lemma demonstrates that these operations are independent of the choice of representatives from each equivalence class.

Proof.

We will prove that the operation of addition is well-defined. The proof that multiplication is well-defined is left as an exercise. Let [a1,b1]=[a2,b2] and [c1,d1]=[c2,d2]. We must show that
[a1d1+b1c1,b1d1]=[a2d2+b2c2,b2d2]
or, equivalently, that
(a1d1+b1c1)(b2d2)=(b1d1)(a2d2+b2c2).
Since [a1,b1]=[a2,b2] and [c1,d1]=[c2,d2], we know that a1b2=b1a2 and c1d2=d1c2. Therefore,
(a1d1+b1c1)(b2d2)=a1d1b2d2+b1c1b2d2=a1b2d1d2+b1b2c1d2=b1a2d1d2+b1b2d1c2=(b1d1)(a2d2+b2c2).

Proof.

The additive and multiplicative identities are [0,1] and [1,1], respectively. To show that [0,1] is the additive identity, observe that
[a,b]+[0,1]=[a1+b0,b1]=[a,b].
It is easy to show that [1,1] is the multiplicative identity. Let [a,b]∈FD such that a≠0. Then [b,a] is also in FD and [a,b]⋅[b,a]=[1,1]; hence, [b,a] is the multiplicative inverse for [a,b]. Similarly, [−a,b] is the additive inverse of [a,b]. We leave as exercises the verification of the associative and commutative properties of multiplication in FD. We also leave it to the reader to show that FD is an abelian group under addition.
It remains to show that the distributive property holds in FD; however,
[a,b][e,f]+[c,d][e,f]=[ae,bf]+[ce,df]=[aedf+bfce,bdf2]=[aed+bce,bdf]=[ade+bce,bdf]=([a,b]+[c,d])[e,f]
and the lemma is proved.
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The field FD in Lemma 18.3 is called the field of fractions or field of quotients of the integral domain D.

Proof.

We will first demonstrate that D can be embedded in the field FD. Define a map ϕ:D→FD by ϕ(a)=[a,1]. Then for a and b in D,
ϕ(a+b)=[a+b,1]=[a,1]+[b,1]=ϕ(a)+ϕ(b)
and
ϕ(ab)=[ab,1]=[a,1][b,1]=ϕ(a)ϕ(b);
hence, ϕ is a homomorphism. To show that ϕ is one-to-one, suppose that ϕ(a)=ϕ(b). Then [a,1]=[b,1], or a=a1=1b=b. Finally, any element of FD can be expressed as the quotient of two elements in D, since
ϕ(a)[ϕ(b)]−1=[a,1][b,1]−1=[a,1]⋅[1,b]=[a,b].
Now let E be a field containing D and define a map ψ:FD→E by ψ([a,b])=ab−1. To show that ψ is well-defined, let [a1,b1]=[a2,b2]. Then a1b2=b1a2. Therefore, a1b1−1=a2b2−1 and ψ([a1,b1])=ψ([a2,b2]).
If [a,b] and [c,d] are in FD, then
ψ([a,b]+[c,d])=ψ([ad+bc,bd])=(ad+bc)(bd)−1=ab−1+cd−1=ψ([a,b])+ψ([c,d])
and
ψ([a,b]⋅[c,d])=ψ([ac,bd])=(ac)(bd)−1=ab−1cd−1=ψ([a,b])ψ([c,d]).
Therefore, ψ is a homomorphism.
To complete the proof of the theorem, we need to show that ψ is one-to-one. Suppose that ψ([a,b])=ab−1=0. Then a=0b=0 and [a,b]=[0,b]. Therefore, the kernel of ψ is the zero element [0,b] in FD, and ψ is injective.
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Example 18.5.

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Since Q is a field, Q[x] is an integral domain. The field of fractions of Q[x] is the set of all rational expressions p(x)/q(x), where p(x) and q(x) are polynomials over the rationals and q(x) is not the zero polynomial. We will denote this field by Q(x).
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We will leave the proofs of the following corollaries of Theorem 18.4 as exercises.
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