Let $R$ be a commutative ring with identity $1$ and $T$ a non-empty multiplicatively closed subset of $R$, and let $M$ be an $R$-module. Define a relation $\sim$ on $M\times T$ as follows: $(m, t)\sim (m', t') \iff \exists s \in T \text{ such that } stm' = st'm.$
Prove the relation $\sim$ on $M\times T$ is an equivalence relation.
(1) Reflexive
$(m, t)\sim (m, t) \iff \exists s \in T \text{ such that } stm = stm.$ (it is clear)
(2) Symmetric
$(m, t)\sim (m', t') \iff \exists s \in T \text{ such that } stm' = st'm.$ We can see that $(m', t')\sim (m, t)$, i.e. $\exists s \in T \text{ such that } st'm = stm'.$ (it is clear)
(3) Transitive
$(m, t)\sim (m', t')$ and $(m', t')\sim (m'', t'')$ mean $$ \exists s_1 \in T \text{ such that } s_1tm' = s_1t'm $$ and $$ \exists s_2 \in T \text{ such that } s_2t'm'' = s_2t''m'. $$ We must prove there exist $s_3\in T$ such that $s_3tm'' = s_3t''m$. How to obtain that?
Recalling that we're in a commutative setting, let's equalize the $s_1tm^\prime$ and $s_2t^{\prime\prime}m^\prime$ term by multiplying the first equation by $s_2t^{\prime\prime}$ and the second by $s_1t$. This gives us \begin{align} s_2t^{\prime\prime} \cdot s_1tm^\prime &= s_2t^{\prime\prime} \cdot s_1t^\prime m \\ s_1t \cdot s_2t^\prime m^{\prime\prime} &= s_1t \cdot s_2t^{\prime\prime}m^\prime \end{align} Rearranging, we get $s_1s_2t^\prime t^{\prime\prime}m = s_1s_2tt^{\prime\prime} m^\prime = s_1s_2t^\prime t m^{\prime\prime}$. Because $T$ is multiplicatively closed, $s_1s_2t^\prime$ is still in $T$, giving us the desired mediator between $t^{\prime\prime}m$ and $tm^{\prime\prime}$.