# 8.1: Field of Fractions

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

In the history of number systems, there is a clear progression: Faced with a void where there could be more numbers, more numbers are invented. First came the natural numbers (the counting numbers), and when people found that certain subtraction problems had no solution, negative numbers were introduced to fill the void. Relevant to our current discussion, the rational numbers come about when one notices that some division problems amongst integers don't have a solution.

Suppose that \(R\) is an integral domain. Then the only impediment to division is a lack of actual quotients: if the quotients were to exist, they would have to be unique.

Consider \(\mathbb{Z}\). This is of course an integral domain, but wouldn't it be nice if \(2\) had a multiplicative inverse? We'll *extend* \(\mathbb{Z}\) by including \(\frac{1}{2}\). But when we include \(\frac{1}{2}\), we also have to include all possible sums and products in order to ensure that we still have a ring; the operations of addition and multiplication need to be closed, of course. So in addition to \(\frac{1}{2}\), we also need to include every number \(\frac{n}{2^m}\), with \(n\in \mathbb{Z}\) and \(m\in \mathbb{N}\) in order to ensure that the set is closed under multiplication and addition. Call this set \(R\). Then \(R\) is a commutative ring with unity. It's also an integral domain, but still not a field, since, for example, \(3\) has no multiplicative inverse.

In that case, we can go ahead and include the multiplicative inverse of *every* positive integer, along with all possible sums and products of those inverses. The resulting ring, of course, is the rational numbers, \(\mathbb{Q}\).

We would like to extend this construction to an arbitrary integral domain: Starting from an integral domain \(D\), we introduce inverses and the appropriate sums and products until every element has an inverse. In fact, this involves copying the whole notion of fractions.

First we construct a ring \(D'\). For an integral domain \(D\), \(D'\) is the set \(D\times (D\setminus \{0\})\). Each pair \((a,b)\) in \(D'\) can be thought of as fractions \(\frac{a}{b}\); note that we disallow \(b=0\). The operation \(+\) defined by \((a,b)+(x,y) = (ay+bx, by)\) and multiplication is defined by \((a,b)\cdot (x,y) = (ax, by)\). Then \(D'\) is a commutative ring; we leave it as an exercise to show this is true.

Exercise 8.1.0

Let \(D\) be an integral domain. Show that \(D'\) is an integral domain. (In particular, check all of the ring axioms, and then show that there are no zero-divisors in \(D'\).)

With rational numbers, it is important to notice that many different fractions are the same: currently our ring \(D'\) is much too large! For example, we haven't introduced any mechanism for cancellation of numerator and denominator: \((a,b)\cdot (b,a)=(ab,ab)\neq 1\) in \(D'\).

We'll construct the actual ring of fractions as a quotient of \(D'\). To construct a quotient, we only need to identify a suitable ideal \(I\); the quotient will then just be \(D'/\mathord I\). The ideal should contain everything that is 'equivalent to \(0\)' in the ring of fractions. Thinking by analogy to the rationals, we see that this is the set \(I=\{(0,x) \mid x\in D\}\).

This set is easily shown to be an ideal: \((0,a)+(0,b)=(0ab, ab)=(0,ab)\in I\). And for any \((x,y)\in D'\), we have \((x,y)\cdot (0,a)=(0,ya)\in I\). Then \(I\) is an ideal.

We can now check that \((a,b)\cdot (b,a)=(1,1)\) in the quotient \(D'/\mathord I\): \((a,b)\cdot (b,a)=(ab, ab)\), and \((ab,ab)-(1,1)=(ab-ab,ab)=(0,ab)=0\), as desired.

We then define the field of fractions as \(Q=D'/\mathord I\). This is in fact a field: For any \(a,b \neq 0\), we have \((a,b)^{-1}=(b,a)\), so every non-zero element in \(Q\) has a multiplicative inverse.

Definition 8.1.1: Field of Fractions

The *field of fractions* of an integral domain \(D\) is \(D'/\mathord I\), with \(D'\) and \(I\) as defined above.

### Contributors

- Tom Denton (Fields Institute/York University in Toronto)