Math Wiki

Field of fractions

1,010pages on
this wiki
Add New Page
Talk0 Share

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.


  • 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:


See also

Ad blocker interference detected!

Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.