Let x be a nilpotent element of the commutative ring R i

Proof:
a) If x is not 0, since x^m=0, then x*(x^(m-1))=0. let y=x^(m-1), then xy=0. So x is a zero divisor. Thus either x is 0, or x is a zero divisor.
b) Since R is commutative, then (rx)^m=r^m * x^m =r^m * 0 =0. Thus rx is nilpotent for all r in R.
c) Let x'=-x. Since x is nilpotent, then x' is nilpotent and we have x^m=(-x)^m=(x')^m=0.
Let y=1+x'+x'^2+...+x'^(m-1), then we have
(1-x')y=1+x'+x'^2+...+x'^(m-1) - (x'+x'^2+x'^3+...+x'^m)
=1-x'^m=1-0=1.
Thus 1-x'=1-(-x)=1+x is a unit in R. Its inverse is y.
d) If a is a unit, x is a nilpotent element, then from (c), we can similarly induce that a+x is a unit.
Since a is a unit, then we can find some b, such that ab=1. Then we have
a+x=b^(-1)(ab+bx)=b^(-1)(1+bx)
From (b), x is nilpotent, then bx is also nilpotent.
From (c), bx is nilpotent, then 1+bx is a unit. So we can find some y, such that (1+bx)y=1.
Thus b^(-1)(1+bx)*yb=b^(-1)*1*b=1
So b^(-1)(1+bx)=a+x is a unit and its inverse is yb