Last updates: 08 January 2012

Let $A$ be a commutative ring.
A **zero divisor** is an element $a\in A$
such that there exists $b\ne 0$ with
$ab=0$.

An **integral domain** is a commutative ring with no zero
divisors except 0.

Let $A$ be an integral domain. The **field of fractions**
of $A$ is the set
$$\mathbb{F}=\left\{\frac{a}{b}\phantom{\rule{0.5em}{0ex}}\right|\phantom{\rule{0.5em}{0ex}}a,b\in A\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}b\ne 0\},$$
with
$$\frac{a}{b}=\frac{c}{d}\phantom{\rule{2em}{0ex}}\text{if}\phantom{\rule{2em}{0ex}}ad=bc,$$
and operations given by
$$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}.$$

Let $A$ be an integral domain. Let $\mathbb{F}$
be the field of fractions over $A$.

- The relation $=$ is an equivalence relation, the operations on $\mathbb{F}$ are well defined and $\mathbb{F}$ is a field.
- The map $$\begin{array}{cccc}\iota :& A& \u27f6& \mathbb{F}\\ & a& \u27fc& \frac{a}{1}\end{array}$$ is an injective homomorphism.
- If $\mathbb{K}$ is a field with an injective ring homomorphism
$\zeta :A\to \mathbb{K}$ then there is a
unique ring homomorphism $\phi :\mathbb{F}\to \mathbb{K}$ such that $\zeta =\phi \circ \iota $.

Many curricula introduce the field of fractions in primary school, when calculations with fractions are introduced. The rational numbers $\mathbb{Q}$ are the field of fraction of the integers $\mathbb{Z}$.

Part (c) of Theorem 1.1 is the universal property for fields of fractions.

[Bou]
N. Bourbaki, *Théorie des Ensembles*, Chapter III,
Masson, Springer-Verlag, 1970
MR??????