$1 + x$ is a unit if $x$ is a nilpotent element
by Hidenori
Proposition
Let $x$ be a nilpotent element of a ring $A$. Show that $1 + x$ is a unit of $A$. Deduce that the sum of a nilpotent element and a unit is a unit.
Solution
Let $x$ be a nilpotent element and $x^n = 0$ for some $n \in \mathbb{N}$. Then $(1 + x)(1 - x + x^2 - x^3 + \cdots + (-x)^n) = 1 + (-x)^n = 1 + (-1)^n \cdot 0 = 1$. Therefore, $1 + x$ is a unit.
Let $a$ be a unit.
\[\begin{align*} (a + x)(\sum_{i=0}^{n-1} a^i(-x)^{n-1-i}) &= \sum_{i=0}^{n-1} a^{i+1}(-x)^{n-1-i} + \sum_{i=0}^{n-1} a^{i+1}x(-x)^{n-1-i} \\ &= \sum_{i=1}^{n} a^{i}(-x)^{n-i} - \sum_{i=0}^{n-1} a^{i+1}(-x)^{n-i} \\ &= -(-x)^n + (\sum_{i=1}^{n-1} a^{i}(-x)^{n-i}) + a^n \\ &= a^n \end{align*}\]Therefore, $(a + x)(\sum_{i=0}^{n-1} a^i(-x)^{n-1-i})/a^n = 1$, so $a + x$ is a unit.
(Alternative solution for the second part) Let $a$ be a unit and $x$ be a nilpotent element with $x^n = 0$ for some $n \in \mathbb{N}$. Then $x / a$ is nilpotent, so $1 + x / a$ is a unit. Thus $a(1 + x / a) = a + x$ is a unit.
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