Example for a ring $R[\frac{1}{a}],a\in R$ such that $\frac{u}{a^m} = \frac{v}{a^n}$ but $ua^n \ne va^m$

Question

For a ring $R$, adjoint a inverse of an element $a\in R$ by taking $R[\frac{1}{a}]\cong R[x]/<ax-1>$.

I am stucking on constructing an explicit example such that $\frac{u}{a^m} = \frac{v}{a^n}$ but $ua^n \ne va^m$. I know that to do this I require $a$ is not a nilpotent (I have deduced that if it is nilpotent then we would always have a zero ring) and $R$ not an integral domain.

So could someone tell me how to construct it, please?

Thanks for any help.

Answer 1

Let $R$ be any ring - not an integral domain - and $a$ a zero-divisor, say $ab=0$. Consider $R[\frac{1}{a}]$.

Then $\frac{b}{1}=\frac{0}{1}$ in $R[\frac{1}{a}]$, but $b \cdot 1 \neq 0 \cdot 1$ in $R$.