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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 \(D\), 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 \(D\).

Proof
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.