# Abstract Algebra/Fraction Fields

We know from experience that we arrive at the idea of fractions by merely considering the idea of the quotient of two integers. The motivation behind this is simply to arrive at a multiplicative inverse for every non-zero element. Thus, we can consider an integral domain R and construct its field of fractions. However, we can also try to make this work for any commutative ring, even if it has zero divisors other than 0. There is a slight alteration required because we cannot define $\frac{a}{b}\frac{c}{d}$ when bd=0. Thus, we must place restrictions in case b and d are zero divisors in the case of multiplication. In this case, it is called the localization of a ring.

## Definitions

A multiplicative subset of a commutative ring R is a subset that does not contain 0, does contain 1, and is closed under multiplication. Some examples of multiplicative sets are the set of nonzero elements of an integral domain, the set of elements of a commutative ring that are not zero divisors, and R\P where P is a prime ideal of the commutative ring R.

Let S be a multiplicative subset. We will consider the Cartesian product R×S. Define the equivalence relation on this product: (a,b)~(c,d) whenever there exists an s such that s(ad-bc)=0.

If it is an integral domain, then (a,b) could be regarded as a/b. Now to check that this is an equivalence relation, it is obvious that it is reflexive and symmetric. To prove that it is transitive, let (a,b)~(c,d) and let (c,d)~(e,f). Then there are elements s and t within S such that s(ad-bc)=0 and such that t(cf-de)=0. This implies that stfad-stfbc=0 and that sbtcf-sbtde=0. Adding the two, we get stfad-sbtde=0, or std(af-be)=0, implying that (a,b)~(e,f).

We can thus use these equivalence classes to define the fraction: $\frac{a}{b}$ is the equivalence class containing (a,b).

Now we set this to be a ring. First, we define addition to be $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$ and multiplication to be $\frac{a}{b}\frac{c}{d}=\frac{ac}{bd}$. The additive identity is $\frac{0}{1}$ and the additive inverse is $\frac{-a}{b}$. The multiplicative identity is simply $\frac{1}{1}$.

Now we prove below that it is indeed a ring.

## Theorem

The set of fractions with addition and multiplication as defined is a commutative ring, and if R is an integral domain, then the fractions are also. If R is an integral domain and S is R\{0}, then the set of fractions is a field.