Given a commutative ring with unity, $\mathcal R$, and a multiplicative set $S,$ to define the fraction ring $S^{-1}\mathcal R$ one defines an equivalence relation $(a,s)\sim (a^{\prime},s^{\prime}),$ where $ a,a^{\prime} \in \mathcal R$, and $s,s^{\prime} \in S$ if the following equation is satisfied: $$ k(as^{\prime}-sa^{\prime})=0, \text {for some} \,k \in S.$$
In the book that I am learning from, there is the statement: "If $0\in S,$ then the only element (which means equivalence class) of $S^{-1} \mathcal R$ is $0/1.$" It is this statement that I don't understand. If $0 \in S,$ then the above equation is satisfied for all pairs $(a,s), (a^{\prime},s^{\prime}).$ Two other things are also unclear to me:
- Is $k \in S$ in the above equation supposed to be unique for each such two pairs or equivalence class ?
- Must one assume that $\mathcal R$ is an integral domain ?
Thanks for your comment.
Note that two different variables have the same name $s$ in your definition of the fraction ring.
Yes, and that is why there is only one equivalence class.
$\mathfrak{s}$ is not supposed to be unique (if $S$ contains some product $\mathfrak{ts}$, then $\mathfrak{ts}(as^{\prime}-sa^{\prime})=0$ as well.)
There is a priori no reason to ask for $\mathcal R$ to be an integral domain, in fact otherwise the sentence "$ \mathfrak{s}(as^{\prime}-sa^{\prime})=0, \text {for some} \,\mathfrak{s} \in S$" would seem a bit silly.