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 $D$, 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.

An element $p/q \in {\mathbb Q}$ 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 \begin{equation*}\frac{a}{b} = \frac{c}{d}\end{equation*} 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 ${\mathbb Q}$ as ordered pairs in ${\mathbb Z} \times {\mathbb Z}$. 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 ${\mathbb Z} \times {\mathbb Z}$. 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 $ad = bc$.

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 \mbox{ and } b \neq 0 \}.\end{equation*} Define a relation on $S$ by $(a, b) \sim (c, d)$ if $ad=bc$.

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

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

##### 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 $p(x)/q(x)$, 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 ${\mathbb Q}(x)$.

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