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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 is the field of fractions of the integers ;
• The Gaussian rational numbers is the field of fractions of the Gaussian integers Theorem. (Field of Fractions) Every integral domain can be embedded in a field.
Proof. Let be an integral domain. That is, a commutative ring with unity in which the zero-product rule holds. Now consider the set , the set of all ordered pairs of elements in , excluding those in which the second element is 0.

We will now define a binary relation on , which we claim to be an equivalence relation by the following criteria: , if and only if , for all .

To show that is an equivalence relation, we must show that the reflexive, symmetric, and transitive properties hold. To that end, let .

(Reflexive) Since , .

(Symmetric) Suppose that . Then , and so by the symmetric property of equality, . Thus, (Transitive) Now suppose that and . Then and . Multiplying these equations together, we obtain . Thus, , and so . Since , we must have , as has the zero-product rule.

In the case that , we will have , which implies and that .

Otherwise, if , then , which then implies that . Similarly, we will also find that . This is the special case that , and so .

Thus, is an equivalence relation, and so we will now define , the set of all equivalence classes, and use the notation to denote the element , the equivalence class containing .

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

• , for all ;
• , for all ;

Since and may not be unique representations of elements in , it is necessary to show that the above rules for addition and multiplication provide for well-defined operations.

To that end, let be elements of , and suppose that and . We will now show that and .

Since and , we have and . Then: