Unity plus Negative of Nilpotent Ring Element is Unit

Theorem

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $x \in R$ be nilpotent.

Then $1_R - x$ is a unit of $R$.

By definition of nilpotent element:

$x^n = 0_R$

for some $n \in \Z_{>0}$.

$\ds a^n - b^n$

$=$

$\ds \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j$

$\ds $

$=$

$\ds \paren {a - b} \circ \paren {a^{n - 1} + a^{n - 2} \circ b + a^{n - 3} \circ b^2 + \dotsb + a \circ b^{n - 2} + b^{n - 1} }$

for $a, b \in R$.

Putting $a = 1_R$ and $b = x$, we have:

$\ds \paren {1_R}^n - x^n$

$=$

$\ds \paren {1_R - x} \circ \paren {1_R^{n - 1} + 1_R^{n - 2} \circ x + 1_R^{n - 3} \circ x^2 + \dotsb + 1_R \circ x^{n - 2} + x^{n - 1} }$

$\ds \leadsto \ \ $

$\ds 1_R - 0_R$

$=$

$\ds \paren {1_R - x} \circ \paren {1_R + 1_R \circ x + 1_R \circ x^2 + \dotsb + 1_R \circ x^{n - 2} + x^{n - 1} }$

as $x^n = 0_R$, $1_R^k = 1_R$

$\ds \leadsto \ \ $

$\ds 1_R$

$=$

$\ds \paren {1_R - x} \circ \paren {1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1} }$

Thus by definition $1_R - x$ has a product inverse $1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1}$.

Hence by definition $1_R - x$ is a unit of $R$.

$\blacksquare$