Question

1. 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. 

Answer 1

Since $x$ is a nilpotent element, $x^n = 0$ for some positive integer $n$. Compute

$(1 + x)(1 - x + x^2 - x^3 + \dots + (-x)^{n - 1}) = 1 + x - x - x^2 + x^2 + x^3 - \dots + (-1)^{n - 1}x^n = 1 + (-1)^{n - 1}x^n = 1 + (-1)^{n - 1}0 = 1 + 0 = 1,$i. e. there exists the right multiplicative inverse of $1 + x$. Obviously, we have the same left multiplicative inverse of $1 + x$ and $(1 + x)^{-1} = 1 - x + x^2 - x^3 + \dots + (-x)^{n - 1}$. Hence we have proved that $1 + x$ is a unit.

Let $u$ be a unit of ring $A$. Then similarly we have

$(u + x)u^{-1}(1 - u^{-1} + u^{-2}x^2 - u^{-3}x^3 + \dots + (-ux)^{n-1}) = uu^{-1} + u^{-1}x - u^{-1}x + u^{-2}x + \dots + (-u)^{n - 1}x^n = 1 + (-u)^{n - 1}x^n + 1 + (-u)^{n - 1}0 = 1 + 0 = 1,$i. e. there exists the multiplicative inverse $(u + x)^{-1} = u^{-1}(1 - u^{-1} + u^{-2}x^2 - u^{-3}x^3 + \dots + (-ux)^{n-1})$. Hence, the sum of a nilpotent element and a unit is a unit.