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## 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 \}. \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. \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}. \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], \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). \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). \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]. \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]. \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] ). \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.