# In a non-commutative ring (possibly without identity) with no nontrivial automorphisms, do nilpotent elements form an ideal?

In an old exam appeared this statement:

True/False: "Let $$R$$ be a ring with the property that the unique ring automorphism is the identity. Then the set of all nilpotent elements form an ideal".

I've seen in Nilpotent elements of a non-commutative ring with trivial automorphism group form an ideal that the result is valid when the ring has multiplicative identity. I want to know if the result is true even omitting the hypothesis of the ring having 1. Some intuition tells me that, as the examples of non-commutative rings with trivial automorphism group are hard to construct (see for example Is there a non-commutative ring with a trivial automorphism group?), the result should be true because it is an exam question.

A way of thinking to solve the exercise was to embed $$R$$ via $$r\mapsto (r,0)$$ in the ring $$R\times \mathbb{Z}$$ as in Hungerford Theorem III 1.10. The ring $$R\times \mathbb{Z}$$ has the usual sum but the product is $$(r_1,n_1)(r_2,n_2)=(r_1r_2+n_2r_1+n_1r_2, n_1 n_2).$$

Now I would like to show that $$R\times\mathbb{Z}$$ has only the trivial automorphism, because the nilpotent elements of $$R\times\mathbb{Z}$$ are on the copy of $$R\times\{0\}$$ so I may use the result in $$R\times\mathbb{Z}$$ which has 1.

Can you help me to finish or provide me a counterexample if this is false? Thank you

Note that you don't need to show that $$R\times \mathbb{Z}$$ has no nontrivial automorphisms; you only need to show it has no nontrivial inner automorphisms, since only inner automorphisms are used in the proof in the case that $$R$$ has unit. Now note that any inner automorphism of $$R\times \mathbb{Z}$$ maps $$R\times\{0\}$$ to itself (since $$R\times\{0\}$$ is a two-sided ideal), and so restricts to an automorphism of the rng $$R\times\{0\}\cong R$$. By hypothesis, this automorphism is the identity. But any inner automorphism of $$R\times\mathbb{Z}$$ must also fix the unit $$(0,1)$$, and so since $$(0,1)$$ and $$R\times\{0\}$$ generate the whole ring, it must be the identity.