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From this problem we have that $A$ is an ideal of $R$ consisting of nilpotent elements. Then multiplication is well-defined in the factor ring and $(u+A)(s+A) = 1+A$ for some $s \in R$. So $us+A = 1+A$ and $us-1 \in A$. By hypothesis, $(us-1)^n = 0$ for some $n \in \mathbb{N}$. I am stuck because I know that $R$ is not necessarily a division ring so I can't just use inverses to deduce that $us-1=0$ and $us=1$ which implies that $u$ is a unit in $R$. Help?

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  • $\begingroup$ Hint: Do you know that 1 plus a nilpotent is a unit? If so, then the fact that $us$ and $su$ are both units (being 1 more than nilpotents) is sufficient to conclude that $u$ is also a unit. The product must be checked in both orders, because the ring $R$ might not be Dedekind-finite. $\endgroup$ Dec 1, 2019 at 2:11

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You already arrived at $us = 1- t$ where $t$ is an element of $A$, hence nilpotent.

It is then clear that $1 - t$ is a unit in $R$: $$(1-t)(1+t+t^2+\dotsc)=1.$$ Note that the "infinite sum" is in fact a finite sum, due to the fact that $t$ is nilpotent.

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