Let R be a ring and suppose r ∈R such that r^2 = 0. Show that (1+r) has a multiplicative inverse in R.
A multiplicative inverse if (1+r)*x = 1 where x is some element in R.
The Attempt at a Solution
We know we have to use two facts.
1. Multiplicative inverse (1+r)*x=1
2. the fact that r^2 = 0.
So I first tried, well, why don't we use (1+r)(1+r) = 0
However, this only got me to 1+r^2+2r = 0 with r^2 =0 we have 1+2r = 0
But that doesn't appear to be what we wanted to show.
Althought, we could subtract -1 from both sides, and subtract r from both sides to get
r = -1-r
r = -(1+r)
But this still isn't what we wanted to show.
If I divide both sides by r, then I get 1 = -1/r (1+r) and this would work, however, I can't assume element r can be divided to get 1. (at least I dont think I can based off the few multiplicative axioms I have for rings/fields).
I have to think that I am on the right path, but any hints to get me closer?