# Nilpotent elements of a non-commutative ring with trivial automorphism group 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 $R$.

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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

## 1 Answer

Hints:

1. In such a ring, every invertible element is central, else there is a nontrivial inner automorphism.

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

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