## FANDOM

1,165 Pages

Every integral domain can be embedded in a field (see proof below). That is, using concepts from set theory, given an arbitrary integral domain (such as the integers), one can construct a field that contains a subset isomorphic to the integral domain. Such a field is called the field of fractions of the given integral domain.

## Examples

• The rational numbers $\Q$ is the field of fractions of the integers $\Z$ ;
• The Gaussian rational numbers $\Q[i]$ is the field of fractions of the Gaussian integers $\Z[i]$
Theorem. (Field of Fractions) Every integral domain can be embedded in a field.
Proof. Let $R$ be an integral domain. That is, a commutative ring with unity in which the zero-product rule holds. Now consider the set $\bar F=R\times R^*$ , the set of all ordered pairs of elements in $R$ , excluding those in which the second element is 0.

We will now define a binary relation $\cong$ on $\bar F$ , which we claim to be an equivalence relation by the following criteria: $(p,q)\cong(\hat p,\hat q)$ , if and only if $p\cdot\hat q=\hat p\cdot q$ , for all $(p,q),(\hat p,\hat q)\in\bar F$ .

To show that $\cong$ is an equivalence relation, we must show that the reflexive, symmetric, and transitive properties hold. To that end, let $(p,q),(\hat\cdot{p},\hat{q}),(\tilde{p},\tilde{q})\in\bar F$ .

(Reflexive) Since $p\cdot q=p\cdot q$ , $(p,q)\cong(p,q)$ .

(Symmetric) Suppose that $(p,q)\cong(\hat p,\hat q)$ . Then $p\cdot\hat q=\hat p\cdot q$ , and so by the symmetric property of equality, $\hat p\cdot q=p\cdot\hat q$ . Thus, $(\hat p,\hat q)\cong(p,q)$

(Transitive) Now suppose that $(p,q)\cong(\hat p,\hat q)$ and $(\hat p,\hat q)\cong(\tilde p,\tilde q)$ . Then $p\cdot\hat q=\hat p\cdot q$ and $\hat p\cdot\tilde q=\tilde p\cdot\hat q$ . Multiplying these equations together, we obtain $p\hat q\hat p\tilde q=\hat{p}q\tilde p\hat q$ . Thus, $p\hat q\hat p\tilde q-\hat{p}q\tilde p\hat q=0$ , and so $\hat p\hat q(p\hat p\tilde q-\hat{p}q\tilde p)=0$ . Since $\hat q\ne0$ , we must have $\hat p(p\tilde q-q\tilde p)=0$ , as $R$ has the zero-product rule.

In the case that $\hat p\ne0$ , we will have $p\tilde q-q\tilde p=0$ , which implies $p\tilde q=\tilde pq$ and that $(p,q)\cong(\tilde p,\tilde q)$ .

Otherwise, if $\hat p=0$ , then $p\cdot\hat q=0\cdot q=0$ , which then implies that $p=0$ . Similarly, we will also find that $\tilde p=0$ . This is the special case that $p\tilde q=0=\tilde pq$ , and so $(p,q)\cong(\tilde p,\tilde q)$ .

Thus, $\cong$ is an equivalence relation, and so we will now define $F=\bar F/\cong$ , the set of all equivalence classes, and use the notation $\frac{p}{q}$ to denote the element $[p,q]$ , the equivalence class containing $(p,q)$ .

We now must determine addition and multiplication operations on $F$ and show that $F$ is a field. We claim that addition and multiplication can be given by:

• $\frac{p}{q}+\frac{r}{s}=\frac{ps+qr}{qs}$ , for all $\frac{p}{q},\frac{r}{s}\in F$ ;
• $\frac{p}{q}\cdot\frac{r}{s}=\frac{pr}{qs}$ , for all $\frac{p}{q},\frac{r}{s}\in F$ ;

Since $\frac{p}{q}$ and $\frac{r}{s}$ may not be unique representations of elements in $F$ , it is necessary to show that the above rules for addition and multiplication provide for well-defined operations.

To that end, let $\frac{p}{q},\frac{\hat p}{\hat q},\frac{r}{s},\frac{\hat r}{\hat s}$ be elements of $F$ , and suppose that $\frac{p}{q}=\frac{\hat p}{\hat q}$ and $\frac{r}{s}=\frac{\hat r}{\hat s}$ . We will now show that $\frac{ps+qr}{qs}=\frac{\hat p\hat s+\hat q\hat r}{\hat q\hat s}$ and $\frac{pr}{qs}=\frac{\hat p\hat r}{\hat q\hat s}$ .

Since $\frac{p}{q}=\frac{\hat p}{\hat q}$ and $\frac{r}{s}=\frac{\hat r}{\hat s}$ , we have $p\hat q=\hat pq$ and $r\hat s=\hat rs$ . Then: