Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.14
Let $R$ be a commutative ring and let $x \in R$ be nilpotent – that is, $x^n = 0$ for some positive integer $n$. Prove the following.
(1) $x$ is either zero or a zero divisor.
(2) $rx$ is nilpotent for all $r \in R$.
(3) $1+x$ is a unit in $R$.
(4) The sum of a unit and a nilpotent element is a unit.
Solution:
(1) Say $m$ is minimal such that $x^m = 0$. If $m = 1$, then $x = 0$. If $m > 1$, then $x \neq 0$, $x^{m-1} \neq 0$ and $x \cdot x^{m-1} = 0$, so that $x$ is a zero divisor.
(2) Since R is commutative, we have $$(rx)^m = r^mx^m = 0.$$ (3) Note that \begin{align*} (1 – (-x))(\sum_{i=0}^{m-1} (-x)^i) =&\ (\sum_{i=0}^{m-1} (-x)^i) – (\sum_{i=0}^{m-1} (-x)^{i+1}) \\=&\ (\sum_{i=0}^{m-1} (-x)^i) – (\sum_{i=1}^{m} (-x)^i \\=&\ 1 + (\sum_{i=1}^{m-1} (-x)^i) – (\sum_{i=1}^{m-1} (-x)^i) – (-x)^m \\=&\ 1 – (-1)^mx^m = 1.\end{align*} Thus $1+x$ is a unit.
(4) Let $u$ be a unit and $x$ nilpotent. Then $u^{-1}x$ is nilpotent, so $1+u^{-1}x$ is a unit, and thus $$u(1+u^{-1}x) = u+x$$ is a unit.
