## Section18.1Fields of Fractions

Every field is also an integral domain; however, there are many integral domains that are not fields. For example, the integers $${\mathbb Z}$$ 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 $${\mathbb Q}$$ 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 $$D\text{,}$$ our question now becomes how to construct a smallest field $$F$$ containing $$D\text{.}$$ We will do this in the same way as we constructed the rationals from the integers.

An element $$p/q \in {\mathbb Q}$$ is the quotient of two integers $$p$$ and $$q\text{;}$$ however, different pairs of integers can represent the same rational number. For instance, $$1/2 = 2/4 = 3/6\text{.}$$ We know that

\begin{equation*} \frac{a}{b} = \frac{c}{d} \end{equation*}

if and only if $$ad = bc\text{.}$$ A more formal way of considering this problem is to examine fractions in terms of equivalence relations. We can think of elements in $${\mathbb Q}$$ as ordered pairs in $${\mathbb Z} \times {\mathbb Z}\text{.}$$ A quotient $$p/q$$ can be written as $$(p, q)\text{.}$$ For instance, $$(3, 7)$$ would represent the fraction $$3/7\text{.}$$ However, there are problems if we consider all possible pairs in $${\mathbb Z} \times {\mathbb Z}\text{.}$$ There is no fraction $$5/0$$ corresponding to the pair $$(5,0)\text{.}$$ Also, the pairs $$(3,6)$$ and $$(2,4)$$ both represent the fraction $$1/2\text{.}$$ 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 $$ad = bc\text{.}$$

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

\begin{equation*} S = \{ (a, b) : a, b \in D \text{ and } b \neq 0 \}\text{.} \end{equation*}

Define a relation on $$S$$ by $$(a, b) \sim (c, d)$$ if $$ad = bc\text{.}$$

Since $$D$$ is commutative, $$ab = ba\text{;}$$ hence, $$\sim$$ is reflexive on $$D\text{.}$$ Now suppose that $$(a,b) \sim (c,d)\text{.}$$ Then $$ad=bc$$ or $$cb = da\text{.}$$ Therefore, $$(c,d) \sim (a, b)$$ and the relation is symmetric. Finally, to show that the relation is transitive, let $$(a, b) \sim (c, d)$$ and $$(c, d) \sim (e,f)\text{.}$$ In this case $$ad = bc$$ and $$cf = de\text{.}$$ Multiplying both sides of $$ad = bc$$ by $$f$$ yields

\begin{equation*} a f d = a d f = b c f = b d e = bed\text{.} \end{equation*}

Since $$D$$ is an integral domain, we can deduce that $$af = be$$ or $$(a,b ) \sim (e, f)\text{.}$$

We will denote the set of equivalence classes on $$S$$ by $$F_D\text{.}$$ We now need to define the operations of addition and multiplication on $$F_D\text{.}$$ Recall how fractions are added and multiplied in $${\mathbb Q}\text{:}$$

\begin{align*} \frac{a}{b} + \frac{c}{d} & = \frac{ad + b c}{b d};\\ \frac{a}{b} \cdot \frac{c}{d} & = \frac{ac}{b d}\text{.} \end{align*}

It seems reasonable to define the operations of addition and multiplication on $$F_D$$ in a similar manner. If we denote the equivalence class of $$(a, b) \in S$$ by $$[a, b]\text{,}$$ then we are led to define the operations of addition and multiplication on $$F_D$$ by

\begin{equation*} [a, b] + [c, d] = [ad + b c,b d] \end{equation*}

and

\begin{equation*} [a, b] \cdot [c, d] = [ac, b d]\text{,} \end{equation*}

respectively. The next lemma demonstrates that these operations are independent of the choice of representatives from each equivalence class.

We will prove that the operation of addition is well-defined. The proof that multiplication is well-defined is left as an exercise. Let $$[a_1, b_1] = [a_2, b_2]$$ and $$[c_1, d_1] =[ c_2, d_2]\text{.}$$ We must show that

\begin{equation*} [a_1 d_1 + b_1 c_1,b_1 d_1] = [a_2 d_2 + b_2 c_2,b_2 d_2] \end{equation*}

or, equivalently, that

\begin{equation*} (a_1 d_1 + b_1 c_1)( b_2 d_2) = (b_1 d_1) (a_2 d_2 + b_2 c_2)\text{.} \end{equation*}

Since $$[a_1, b_1] = [a_2, b_2]$$ and $$[c_1, d_1] =[ c_2, d_2]\text{,}$$ we know that $$a_1 b_2 = b_1 a_2$$ and $$c_1 d_2 = d_1 c_2\text{.}$$ Therefore,

\begin{align*} (a_1 d_1 + b_1 c_1)( b_2 d_2) & = a_1 d_1 b_2 d_2 + b_1 c_1 b_2 d_2\\ & = a_1 b_2 d_1 d_2 + b_1 b_2 c_1 d_2\\ & = b_1 a_2 d_1 d_2 + b_1 b_2 d_1 c_2\\ & = (b_1 d_1) (a_2 d_2 + b_2 c_2)\text{.} \end{align*}

The additive and multiplicative identities are $$[0,1]$$ and $$[1,1]\text{,}$$ respectively. To show that $$[0,1]$$ is the additive identity, observe that

\begin{equation*} [a, b] + [0, 1] = [ a 1 + b 0, b 1] = [a,b]\text{.} \end{equation*}

It is easy to show that $$[1, 1]$$ is the multiplicative identity. Let $$[a, b] \in F_D$$ such that $$a \neq 0\text{.}$$ Then $$[b, a]$$ is also in $$F_D$$ and $$[a,b] \cdot [b, a] = [1,1]\text{;}$$ hence, $$[b, a]$$ is the multiplicative inverse for $$[a, b]\text{.}$$ Similarly, $$[-a,b]$$ is the additive inverse of $$[a, b]\text{.}$$ We leave as exercises the verification of the associative and commutative properties of multiplication in $$F_D\text{.}$$ We also leave it to the reader to show that $$F_D$$ is an abelian group under addition.

It remains to show that the distributive property holds in $$F_D\text{;}$$ however,

\begin{align*} [a, b] [e, f] + [c, d][ e, f ] & = [a e, b f ] + [c e, d f]\\ & = [a e d f + b f c e, b d f^2 ]\\ & = [a e d + b c e, b d f ]\\ & = [a d e + b c e, b d f ]\\ & = ( [a, b] + [c, d] ) [ e, f ] \end{align*}

and the lemma is proved.

The field $$F_D$$ in Lemma 18.3 is called the field of fractions or field of quotients of the integral domain $$D\text{.}$$

We will first demonstrate that $$D$$ can be embedded in the field $$F_D\text{.}$$ Define a map $$\phi : D \rightarrow F_D$$ by $$\phi(a) = [a, 1]\text{.}$$ Then for $$a$$ and $$b$$ in $$D\text{,}$$

\begin{equation*} \phi( a + b ) = [a+b, 1] = [a, 1] + [b, 1] = \phi(a ) + \phi(b) \end{equation*}

and

\begin{equation*} \phi( a b ) = [a b, 1] = [a, 1] [b, 1] = \phi(a ) \phi(b); \end{equation*}

hence, $$\phi$$ is a homomorphism. To show that $$\phi$$ is one-to-one, suppose that $$\phi(a) = \phi( b)\text{.}$$ Then $$[a, 1] = [b, 1]\text{,}$$ or $$a = a1 = 1b = b\text{.}$$ Finally, any element of $$F_D$$ can be expressed as the quotient of two elements in $$D\text{,}$$ since

\begin{equation*} \phi(a) [\phi(b)]^{-1} = [a, 1] [b, 1]^{-1} = [a, 1] \cdot [1, b] = [a, b]\text{.} \end{equation*}

Now let $$E$$ be a field containing $$D$$ and define a map $$\psi :F_D \rightarrow E$$ by $$\psi([a, b]) = a b^{-1}\text{.}$$ To show that $$\psi$$ is well-defined, let $$[a_1, b_1] = [a_2, b_2]\text{.}$$ Then $$a_1 b_2 = b_1 a_2\text{.}$$ Therefore, $$a_1 b_1^{-1} = a_2 b_2^{-1}$$ and $$\psi( [a_1, b_1]) = \psi( [a_2, b_2])\text{.}$$

If $$[a, b ]$$ and $$[c, d]$$ are in $$F_D\text{,}$$ then

\begin{align*} \psi( [a, b] + [c, d] ) & = \psi( [ad + b c, b d ] )\\ & = (ad + b c)(b d)^{-1}\\ & = a b^{-1} + c d^{-1}\\ & = \psi( [a, b] ) + \psi( [c, d] ) \end{align*}

and

\begin{align*} \psi( [a, b] \cdot [c, d] ) & = \psi( [ac, b d ] )\\ & = (ac)(b d)^{-1}\\ & = a b^{-1} c d^{-1}\\ & = \psi( [a, b] ) \psi( [c, d] )\text{.} \end{align*}

Therefore, $$\psi$$ is a homomorphism.

To complete the proof of the theorem, we need to show that $$\psi$$ is one-to-one. Suppose that $$\psi( [a, b] ) = ab^{-1} = 0\text{.}$$ Then $$a = 0b = 0$$ and $$[a, b] = [0, b]\text{.}$$ Therefore, the kernel of $$\psi$$ is the zero element $$[ 0, b]$$ in $$F_D\text{,}$$ and $$\psi$$ is injective.

### Example18.5.

Since $${\mathbb Q}$$ is a field, $${\mathbb Q}[x]$$ is an integral domain. The field of fractions of $${\mathbb Q}[x]$$ is the set of all rational expressions $$p(x)/q(x)\text{,}$$ 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 $${\mathbb Q}(x)\text{.}$$

We will leave the proofs of the following corollaries of Theorem 18.4 as exercises.