1. ## Rings theory....

I've got this question that I don't seem to be able to solve on my own :-\
R is a ring. x is an nilpotent (?) element in it - meaning that there exists some n in N such that x^n = 0 (and x is not 0 of course).

There were some questions posed, and I got stuck on the last one:
prove that the element (x+u), where u is comutative with x (meaning xu=ux), is reversible (?) (meaning, that there exists an element r in R such that r*(x+u) = (x+u)*r = 1).
I try all the tricks up my sleeve and still am stuck.

a question before I've proved (1+x) is reversible, noticing that:
(1+x)(1-x+x^2-....+x^n-1) = 1 (since x^n is 0)... but this question seems much harder.

Thank you very much for reading/responding!

2. Originally Posted by aurora
I've got this question that I don't seem to be able to solve on my own :-\
R is a ring. x is an nilpotent (?) element in it - meaning that there exists some n in N such that x^n = 0 (and x is not 0 of course).
If $x^n = 0$ then $x^n + 1 = (x+1)(x^{n-1}-x^{n-2}+...\pm x \mp 1)$ if $n$ is odd. However, it is safe to assume that $n$ is odd because if $x^n = 0 \implies x^{n+1}=0$. This shows that $1+x$ is an invertible element in the ring. The problem asks to show $u+x$ is invertible if $u$ is invertible. Notice that $u+x$ is invertible if and only if $u^{-1}(u+x) = 1 + u^{-1}x$ is invertible. But $u^{-1}x$ is nilpotent because $(u^{-1}x)^n = u^{-n}x^n =0$. By the above result it means $1+u^{-1}x$ is invertible and consequently $u+x$ is invertible.

3. Thank you for the response.
However, I was not given the fact that u is invertible - only that it commutes with x (ux=xu)....
:-\

I'm sorry - I just reread the question and realized I am being given the fact that u is invertible :-)
Sorry again, thanks!

4. Then it's wrong! $0x=x0 ,$ but $0+x=x$ is nilpotent so not invertible...

5. Indeed, I've written above - I missed the fact that u is invertible .

Thank you.