Construction on the equivalent class for localization

Question

I have a question about the following example about the localization :

If $x$ is an element of a commutative ring $R$ and $ S = \left\{1, x,x^{2},...\right\}$ then $S^{-1}R$ can be identified (is canonically isomorphic to) $ R[x^{-1}]=R[s]/(xs-1)$.

My main question is it is possible to set the equivalence relation (on $R\times S$) well, the construction is naturally giving the equivalence relation like this,

$$(r,x^{s}) \sim (r',x^{t}) \overset{Def}\iff u(rx^t-r'x^s)=0~ for~ some~ u\in S .......(*)$$

If $R$ is an integral domain, just put $u=1$(Here is just quotient ring...). But, otherwise, I cannot sure that the element in $u \in S$ always exist whenever I pick a pair, $(r,x^{s}).$ Of course, since clearly $R$ is just commutative ring, and need not to be integral domain, I (vaguely) expect an (nonzero) element $x^{p} \in S$ (for some $p\in \mathbb{Z}^{+}$) such that $x^{p}(rx^t-r'x^s)=0$. However, I am not sure that the positive integer $p$ really exist for any (not specific) commutative ring $R$.

Or, is it no problem such positive integer $p$ does not exist (even though such situation is strange...)? Suppose that, an above definition $(*)$ , $(r,x^s) \in R \times S$ does not match any pair $(r',x^{t})$ but, at least, there exists $u=1 (\in S)$ such that $(r,x^s) \sim (r,x^s)$ (which holds reflexive, symmetric, and transitive). Anyway, I can construct equivalent class $[(r,x^s)]$ by following these three conditions.

Answer 1

What you seem to be asking is whether sometimes the equivalence class $[(r,x^s)]$ can have only one element.

Suppose $x^{s+1}\neq x^s$ and $x^{s+1}\neq 0$. Then $(r,x^s)\equiv (rx,x^{s+1})$.

Let's now check the remaining cases.

Case 1

Suppose $x^{s+1}=x^s$, which is the same as saying that $x^s (x-1)=0$.

If $x-1=0$, then $x=1$, which is a trivial case. Here every equivalence class $[(r,x^s)]$ can have only one element.

If $x-1\neq 0$, then $(r,x^s)\equiv (r+x-1,x^s)$.

Case 2

Suppose $x^{s+1}=0$ but $x^s\neq 0$.

Then clearly $x\neq 0$. Furthermore, $(r,x^s)\equiv(r+x,x^s)$.