# 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.

## DefinitionsEdit

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 $\frac{a}{b} \cdot \frac{c}{d}=\frac{ac}{bd}$. 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:

## TheoremEdit

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. And if additionally S is R\{0}, then the set of fractions is a field.

### ProofEdit

First, we note that

1. $(\frac{a}{b} + \frac{c}{d}) + \frac{e}{f} = \frac{ad + bc}{bd} + \frac{e}{f} = \frac{(ad+bc)f + bde}{bdf} = \frac{adf + b(cf+de)}{bdf} = \frac{a}{b} + \frac{cf+de}{df} = \frac{a}{b} + (\frac{c}{d} + \frac{e}{f})$
2. $\frac{a}{b} + \frac{0}{1} = \frac{a \cdot 1 + b \cdot 0}{b \cdot 1} = \frac{a}{b}$
3. $\forall c \in S : 1_R (0\cdot 1 - 0\cdot c) = 0 \Rightarrow \frac{0}{1} = \frac{0}{c} (*)$ and therefore $\frac{a}{b} + \frac{-a}{b} = \frac{ab + b(-a)}{bb} = \frac{0}{bb} = \frac{0}{1}$

, from which follows that $(R \times S, +)$ is a group.

It is abelian because of the definition of the sum in S and R is commutative.

Furthermore, $(R \times S, \cdot)$ is a monoid because

1. $\frac{1}{1} \cdot \frac{a}{b} = \frac{1a}{1b} = \frac{a}{b}$
2. and $(\frac{a}{b} \cdot \frac{c}{d}) \cdot \frac{e}{f} = \frac{ace}{bdf} = \frac{a}{b} \cdot (\frac{c}{d} \cdot \frac{e}{f})$, where two (not difficult) intermediate steps are left to the reader.

And, also the distributive laws hold, because

$\frac{a}{b} \left( \frac{c}{d} + \frac{e}{f} \right) = \left( \frac{c}{d} + \frac{e}{f} \right) \frac{a}{b} = \frac{acf +ade}{bdf}$ and
$\frac{acf +ade}{bdf} = \frac{acbf + adbe}{bdfb} = \frac{ac}{bd} + \frac{ae}{bf}$

, which shows that we have indeed found a ring.

The ring is commutative because of the definition of the product in S R is commutative.

Let now R be an integral domain, and let $\frac{a}{b}, \frac{c}{d} \in S$. Then, because of $(*)$ and since $bd \in S$, $\frac{a}{b} \cdot \frac{c}{d} = 0 \Leftrightarrow \frac{ac}{bd} = 0 \Leftrightarrow \frac{ac}{bd} = \frac{0}{bd} \Leftrightarrow \exists s \in S: s(ac\cdot bd - 0 \cdot bd) = sacbd = 0$. But since R was assumed to be an integral domain, and $s, b, d \neq 0$ since $s, b, d \in S$, the last statement is exactly equivalent to $\exists s \in S: a = 0 \vee c = 0$, which is in turn equivalent to $\exists s \in S: sab = s(ab - 0c) = 0 \vee \exists s \in S: scd = s(cd - 0d) = 0$, and this is equivalent to $\frac{a}{b} = 0 \vee \frac{c}{d} = 0$, which shows that the fraction set is an integral domain if R is one.

Let's assume now that S = R \ {0}, and let $\frac{a}{b} \neq 0 \Leftrightarrow a \neq 0$, where the last equivalence is due to (*) and $\frac{a}{b} = 0 \Leftrightarrow \exists s \in S: s(a1 - b0) = as1 = as = 0 \Leftrightarrow a=0$, where the last equivalence is in turn due to the fact that R is an integral domain and S does not contain zero. Then $ab \neq 0$ due to the fact that R is an integral domain, and thereofore $ab \in S$ since S = R \ {0}, and $\frac{a}{b} \cdot \frac{b}{a} = \frac{ab}{ab}$. But since $\forall s \in S: s(ab1 - ba1) = 0$, we have $\frac{ab}{ab} = 1$ and $\frac{a}{b} \cdot \frac{b}{a} = 1$ and therefore, by noting that we have assumed R to be commutative, we have that every element of R\{0} is invertible.

From this follows that the set of fractions is indeed a field, because we have already checked all field axioms, QED.