Fraction equivalence is transitive in localizations (general non-domain case)

Question

Let $A$ be a ring (commutative with $1$), let $S$ be a multiplicatively closed subset of $A$, i.e $S$ is contained in $A$ , $1\in S$ and $a,b\in S$ implies $ab\in S$, for every $a,b\in A$. Consider the following relation defined in $A\times S$: $$(a,s)\equiv(b,t) \iff \exists u\in S:u(at-bs)=0$$ I proved this relation to be reflexive and symmetric, but i have difficulties in proving its transitivity. Can someone suggest me a trick?

Answer 1

Let $(a,s),(b,t),(c,u)$ and $x, y$ be such that $(a,s)\equiv (b,t)\pmod S$, and $(b,t)\equiv(c,u)\pmod S$, that is to say,
\begin{align} x(at-bs)=0,\\ y(bu-ct)=0 \end{align} Now multiply the first equation by $uy$ and the second by $sx$, to obtain:
$$xy(aut-bus)=0,$$ $$xy(bus-sct)=0.$$
Add the two equations together and you will find what you want.

The following is what you will have when adding two equations together: $xyt(au-cs)=0$, which means that $(a,s)\equiv(c,u)\pmod S$, hence the transitivity follows at once.