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

Question

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

Answer 1

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.