# Nilpotent elements of a non-commutative ring can form an ideal

Let R be a non-commutative ring with identity such that the identity map is the only ring automorphism of R. Prove that the set N of all nilpotent elements of R is an ideal of R.

-
Welcome to MathSE. I see that you are relatively new here. So I wanted to let you know a few things about MathSE. We like to know where the problem is from what you've tried on a problem; this prevents people from wasting their time telling you thinks you already know, and helps make sure the answers are at an appropriate level. If this is homework, please consider adding the [homework] tag; people will still help, so don't worry. Also, posting questions in the imperative ("Compute", "Prove", "Show") is considered rude by some of the members, so please consider editing the post. Thank you. – Arturo Magidin Feb 4 '12 at 3:44
Can you give an example of a non-commutative ring that satisfies the hypothesis? I can't think of one. At a minimum, all invertible elements have to be central (else you have nontrivial inner automorphisms), which is kind of a weird condition. Matrix rings obviously don't work, neither do (integral) group rings since they're generated by the group elements which are invertible, and "free non-commutative" rings on > 1 generator have automorphisms that permute the generators. I'm running out of ideas for non-commutative rings. – Ted Feb 4 '12 at 3:59

2. If $x$ is nilpotent, then $1-x$ is invertible.