Definition 2

Let $\mathbf{A}$ be a ring, $S$ a subset of $\mathbf{A}$ and $\overline{S}$ the multiplicative subset generated by $S$. The ring of fractions of $\mathbf{A}$ defined by $S$ and denoted by $\mathbf{A}[S^{-1}]$ is the quotient set of $\mathbf{A} \times \overline{S}$ under the equivalence relation (1) with the ring structure defined by

for $a, b$ in $\mathbf{A}$, $s, t$ in $\overline{S}$. The canonical mapping of $\mathbf{A}$ to $\mathbf{A}[S^{-1}]$ is the homomorphism $a \mapsto a/1$, which makes $\mathbf{A}[S^{-1}]$ into an $A$-algebra.

In this chapter we usually denote this canonical mapping by $i_A^S$; the proof of Proposition 1 shows that the ordered pair $(\mathbf{A}[S^{-1}], i_A^S)$ satisfies the conditions of the statement of this proposition.

Remarks

(1) Clearly $\mathbf{A}[\overline{S}^{-1}] = \mathbf{A}[S^{-1}]$.

(2) Two elements of $\mathbf{A}[S^{-1}]$ can always be written in the form $a/s$ and $a'/s$ ($a, a'$ in $\mathbf{A}$, $s \in \overline{S}$) with the same "denominator" $s$, for if $b/t$ and $b'/t'$ are two elements of $\mathbf{A}[S^{-1}]$, then $b/t = bt'/tt'$ and $b'/t' = b't/tt'$.

(3) The kernel of $i_A^S$ is the set of $a \in A$ such that there exists $s \in \overline{S}$ satisfying $sa = 0$; for $i_A^S$ to be injective, it is necessary and sufficient that $S$ contain no divisor of zero in $A$.

(4) If $S$ contains a nilpotent element, then $0 \in \overline{S}$ and the ring $A[S^{-1}]$ is reduced to 0; this follows easily from Definition 2.

(5) For $i_A^S$ to be a bijection, it is necessary and sufficient that every element $s \in S$ be invertible in $A$: the condition is obviously necessary, since $s/1$ is invertible in $A[S^{-1}]$; it is sufficient, since for all $t \in \overline{S}$, $t$ is therefore invertible in $A$ and $a/t = at^{-1}/1$ in $A[S^{-1}]$; hence $i_A^S$ is surjective and it has already been seen in Remark 3 that it is injective. Then $A$ and $A[S^{-1}]$ are identified by means of $i_A^S$.

Example (7). If $R$ is the set of elements in $A$ which are not divisors of 0, the ring $A[R^{-1}]$ is precisely what we have called the ring of fractions of $A$ (Algebra, Chapter I, §9, no. 4); to avoid any confusion we shall often call it the total ring of fractions of $A$. In particular, if $A$ is an integral domain, $A[R^{-1}]$ is the field of fractions of $A$ (loc. cit.).