Nilpotent Element a in a Ring and Unit Element $1-ab$
Problem 171
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
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Proof 1.
Since $a$ is nilpotent, we have $a^n=0$ for some positive integer $n$.
Then for any $b \in R$, we have $(ab)^n=a^nb^n=0$ since $R$ is commutative.
Then we have the following equality:
\[(1-ab)(1+(ab)+(ab)^2+\cdots+(ab)^{n-1})=1.\]
Therefore $1-ab$ is a unit in $R$.
Proof 2.
There exists $n \in \N$ such that $a^n=0$ since $a$ is nilpotent.
Assume that $1-ab$ is not a unit for some $b \in R$.
Then there exists a prime ideal $\frakp$ of $R$ such that $\frakp \ni 1-ab$.
Since $a^n=0\in \frakp$, we have $a\in \frakp$ since $\frakp$ is an prime ideal.
Then $ab \in \frakp$ and we have
\[1=(1-ab)+ab \in \frakp.\]
However, this implies that $\frakp=R$ and this is a contradiction. Thus $1-ab$ is a unit of the ring $R$ for all $b \in R$.
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