Total ring of fractions

Remarks

The canonical map $R \to Q(R)$, $a \mapsto a/1$, is injective, because no element of $S$ kills a nonzero element. If $S'$ is a multiplicative set and $R \to S'^{-1}R$ is injective, then $S' \subseteq S$.

Example: if $R$ is an integral domain, then $S = R \setminus \set{0}$ and $Q(R)$ is the field of fractions of $R$. For $R = \Z \times \Z$, the non-zero-divisors are the pairs $(a, b)$ with $a \ne 0$ and $b \ne 0$, and $Q(R) \cong \Q \times \Q$. For $R = \Z/6\Z$, the non-zero-divisors are $1$ and $5$, which are already units, so $Q(R) = R$.

Non-example: in $R = \Z \times \Z$, the set $T = \set{(1,0)^n : n \ge 0} = \set{(1,1), (1,0)}$ is multiplicative, but $(1, 0)$ is a zero divisor. The localization $T^{-1}R \cong \Z$ is not the total ring of fractions, and $R \to T^{-1}R$ kills $(0, 1)$.