FANDOM


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:


Tombstone

See also

Community content is available under CC-BY-SA unless otherwise noted.