# If $u+A$ is a unit in $R/A$, show that $u$ is a unit in $R$.

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?

• 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. Dec 1, 2019 at 2:11

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.