Let $R$ be a commutative ring (that does not necessarily include $1$), and let $S$ be a multiplicative subset of $R$.
Show that the relation defined on the set $R \times S$ by $$(a, x) \sim (b, y) \iff w(ay - bx) = 0, \text{ for some } w \in S $$ is an equivalence relation on $R \times S$.
Obviously if $R$ is a field (or even an integral domain), this is somewhat trivial; but the possibility of zero divisors is making this kind of frustrating.
Reflexivity and symmetry are easy enough. Transitivity is giving me some trouble, though...
I'll lay it out: Suppose that $(a,x) \sim (b,y)$ and $(b,y) \sim (c, z)$. Then there exist $u,v \in S$ such that $u(ay - bx) = 0 = v(bz - cy)$. We seek a $w \in S$ such that $w(az - cx) = 0$.
Now, if $R$ is a field, we can simply choose $w = y$ since (after solving for $az$ and $cx$), we have $$ az - cx = bxzy^{-1} - bzxy^{-1} \implies y(az - cx) = bxz - bxz = 0. $$ I tried to reverse engineer this observation to treat the more general case, but I just ended up pushing around symbols for an eternity...
(With some help from @JonathanZ supports MonicaC)
Take $w = uvy \in S$. Then \begin{align*} uvy(az - cx) &= uvyaz - uvycx \\ &= (uay)vz - uvcyx \\ &= ubxvz - uvcyx. \end{align*} Since $$ v(bz - cy) = 0 \implies uvx(bz - cy) = 0 \implies uvbxz - uvxcy = 0, $$ the proof is complete.