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Fraction Rings and Fraction Fields 📂Abstract Algebra

Fraction Rings and Fraction Fields

Definition 1

For a ring $\left( A , + , \cdot \right)$, let’s say $S := A^{\ast} = A \setminus \left\{ 0 \right\}$ is a subset $S \subset A$ that excludes the identity element $0$ for addition $+$ in $A$.

Field of Fractions

If $A$ is an integral domain, $$ (a,s) \equiv (b,t) \iff at = bs $$ When we define the equivalence relation $\equiv$ on the Cartesian product $A \times S$ of $A$ and $S$ as above, let’s denote the equivalence class of $(a,s)$ as $a/s$, and represent the set of those equivalence classes as $S^{-1} A := A \times S / \equiv$. With the new operations $\oplus$ and $\odot$ defined as $$ \begin{align*} {{ a } \over { s }} \oplus {{ b } \over { t }} :=& {{ at + bs } \over { st }} \\ {{ a } \over { s }} \odot {{ b } \over { t }} :=& {{ ab } \over { st }} \end{align*} $$ we define the field $\left( S^{-1} A , \oplus , \odot \right)$ as the Field of Fractions of $A$.

Ring of Fractions

When $A$ has a unity $1$ and $1 \in S$ and $\left( S , \cdot \right)$ is a magma, $$ (a,s) \equiv (b,t) \iff \left( at - bs \right) u = 0 $$ for some $u \in S$, when we define the equivalence relation $\equiv$ on the Cartesian product $A \times S$ of $A$ and $S$ as above, let’s denote the equivalence class of $(a,s)$ as $a/s$, and represent the set of those equivalence classes as $S^{-1} A := A \times S / \equiv$. With the new operations $\oplus$ and $\odot$ defined as $$ \begin{align*} {{ a } \over { s }} \oplus {{ b } \over { t }} :=& {{ at + bs } \over { st }} \\ {{ a } \over { s }} \odot {{ b } \over { t }} :=& {{ ab } \over { st }} \end{align*} $$ we define the ring $\left( S^{-1} A , \oplus , \odot \right)$ as the Ring of Fractions of $A$.

Explanation

The motive for the field of fractions is naturally the rational number field $\mathbb{Q}$, obtained in the same way from the integer ring $\mathbb{Z}$, by abstracting what one usually sees first when studying equivalence relations in set theory.

The premise of $A$ being an integral domain for the field of fractions is to ensure through the Cancellation Law that $\equiv$ is an equivalence relation, namely, it’s necessary when demonstrating transitivity among the properties of binary relations $\equiv$ being reflexive, symmetric, and transitive. The ring of fractions is a generalization of the field of fractions, where $A$ is relaxed to a ring having unity and the equivalence relation is defined slightly differently.

It is trivial from the definition that if $A$ is an integral domain, the ring of fractions of $A$ is a field of fractions. Consequently, if $A$ is a differential ring, it naturally becomes a differential field.


  1. Atiyah. (1994). Introduction to Commutative Algebra: p36~37. ↩︎