## Basic properties of nilpotent ring elements

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.

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 $(1 - (-x))(\sum_{i=0}^{m-1} (-x)^i)$ $= (\sum_{i=0}^{m-1} (-x)^i) - (\sum_{i=0}^{m-1} (-x)^{i+1})$ $= (\sum_{i=0}^{m-1} (-x)^i) - (\sum_{i=1}^{m} (-x)^i$ $= 1 + (\sum_{i=1}^{m-1} (-x)^i) - (\sum_{i=1}^{m-1} (-x)^i) - (-x)^m$ $= 1 - (-1)^mx^m$ $= 1$. 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.