Question

If x is a nilpotent element of a commutative ring R, prove that 1+x is a unit in R.

Answer 1

So basically we have \(x^m=0\) and we want to show that there exists a \(v\) such that \((1+x)v=1\).

Answer 2

I believe if you multiply (1+x) by:\[\left(1-x+x^2-x^3+\cdots +(-1)^{m-1}x^{m-1}\right)\]you should end up with:\[(1+x)(1-x+x^2-x^3+\cdots +(-1)x^{m-1})=1+(-1)^{m-1}x^m\]

Answer 3

That seems to work. Followup question: Prove that the sum of any nilpotent element with any unit is a unit.

Answer 4

Let u be a unit, and x be the nilpotent element. Since u is a unit, there exists a v such that uv = 1. So rewrite (u+x) as:\[(u+x)=(u+1\cdot x)=(u+uvx)=u(1+vx)\]Since x^m = 0, and the ring is commutative, it follows that vx is also nilpotent of order m. Now you can use the previous problem to get the inverse of 1+vx, and tack on a v to that to get rid of the u.

Answer 5

Perfect, thanks.