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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.

Xam
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Y.X.
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    What makes you think this is possible? – Mr. Chip Sep 26 '17 at 13:20
  • @Mr.Chip Because I have just find the sufficient condition of two fractions are equal in such a ring, see https://math.stackexchange.com/questions/2441481/two-fractions-in-r-frac1x-are-equal-iff-exists-n-ge-0xn-cdot-axi – Y.X. Sep 26 '17 at 13:35
  • You want $ua^n \neq va^m$ to hold in $R$ or in $R[\frac{1}{a}]$? – MooS Sep 26 '17 at 14:22
  • @MooS I need it in $R$. – Y.X. Sep 26 '17 at 14:24
  • Then it is very easy. In fact, let $a$ be a nilpotent. Then $\frac{1}{1}=\frac{0}{1}$, but $1 \cdot 1 \neq 0 \cdot 1$. – MooS Sep 26 '17 at 14:26

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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 $R$.

MooS
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