I would like to ask about the correctness of the proof of 2.4. Proposition. I think it is not true. This proposition is stated as follows:
Let $R$ be a local ring. Then, every element in $R$ is either invertible or nilpotent.
Proof. Let $a$ be a non-invertible element in $R$. Then, $1-a$ is invertible since $R$ is a local ring, so there exists $u\in R$ such that $(1-a)u=1$ and that can be held when $u=1+a+a^2+\cdots+a^{n-1}\in R$. Then, $a^n=0$, so $a$ is nilpotent.
I think the bold line is wrong. Any counterexample or reference or technique is very much appreciated. Thank you in advance.
You're right, this proof is nonsense and the proposition is false. An easy counterexample is the ring $k[[x]]$ of formal power series over a field $k$, with unique maximal ideal $(x)$; we can take $a = x$ and then $1 - x$ is invertible with inverse $1 + x + x^2 + \dots $ but $x$ is not nilpotent.