If R is a ring with idempotent then eR is clean ring if and only if eRe is clean ring

Question

I am reading this paper on clean rings: Clean general rings, by W.K. Nicholson, Y. Zhou, and I am puzzled by the proof of Corollary 11:

Corollary 11. If $R$ is a ring and $e^2=e \in R$, then $eR$ is a clean general ring if and only if $eRe$ is a clean ring.
Proof. Observe that the map $\theta: eR \to eRe$ given by $\theta(x)=xe$ is an onto ring morphism with kernel $K=eR(1-e)$. Hence $eR/K \cong eRe$ and $K^2=0$. If $eRe$ is clean this implies that $eR$ is clean by Theorem 10; conversely if $eR$ is clean so is its image $eRe$. $\square$

I would appreciate any clarification on the following:

• What is a ring morphism, did the writer mean a ring homomorphism? If so, how did they prove that the mapping preserved the multiplication? (i.e. $\phi(xy)=\phi(x) \cdot \phi(y)$)

• Additionally, they mentioned that every element in $K$ (the kernel) is a nilpotent element, in what way did this benefit the proof?

• Lastly, they conclude that if $eRe$ is clean, this implies that $eR$ is clean by Theorem 10:

Theorem 10. Let $I$ be a general ring and let $A \lhd I$.
(1) If $I$ is clean, then $A$ and $I/A$ are both clean and idempotents lift modulo $A$. (2) The converse is true if $A \subseteq J$ or if all primitive factors of $I$ are artinian.

In what manner did the theorem aid the proof?

I look forward to any elucidation you can provide on these questions.

Answer 1

what is a ring morphism, did the writter mean a ring homomorphism?

Yes, morphism and homomorphism usually mean the same thing.

if so how did they prove that the mapping preserved the multiplication?

(i.e.) ϕ(xy)=ϕ(x)⋅ϕ(y)

That's actually quite easy once you introduce $\xi,\eta\in R$ with $x=e\xi$ and $y=e\eta$.

additionally, they mentiond that every element in K (the kerenl) is a nilpotent element, in what way did this benefit the proof?

lastly the conclude that if eRe is clean this implies that eR is clean by [theorem 10][3] in what manner did the theorem aid the proof.

It seems that they want to use Theorem 10(2), and $K^2=0$ implies $K\subset J$, because nilpotent ideals are contained in the Jacobson radical $J$. The last fact I found here, I'm not an expert.