A question on clean rings

Question

Recall that a ring $R$ is called clean if every element of $R$ is a sum of a unit of $R$ and an idempotent of $R$. We call a module $M$ clean if its endomorphism ring $End(M)$ is a clean ring. Nicholson has a great paper on clean rings titled 'Extensions of clean rings' (Communications in algebra, 29(6), 2589-2595 (2001)). Nicholson proved that if $R$ is a ring and $e^2=e\in R$ is such that the corner rings $eRe$ and $(1-e)R(1-e)$ are both clean rings, then $R$ is a clean ring. Nicholson said that this result can be generalized using the principle of the mathematical induction and the aforementioned result as follows:

If $1=e_1+e_2+\ldots+e_n$ is a ring $R$ where $e_i$ are orthogonal (i.e., $e_ie_j=e_je_i=0 \,\, \forall i\neq j$) idempotents and each $e_iRe_i$ is clean, then $R$ is clean.

I'm stuck with this statement. I can't see how to prove the latter statement by induction and the Nicholson's aforementioned result. I appreciate any help.

Answer 1

The case where $n = 1$ is trivial and the case where $n = 2$ is the initial statement. If we assume our theorem is true for $n - 1$, then observe $$1 - e_n = e_1 + e_2 + \ldots + e_{n-1}$$ and apply the theorem to show that if $$e_k R e_k$$ is clean for all $k < n$, then so is $$(1 - e_n)R(1 - e_n)$$ and thus we're back to the case $n = 2$.

The only thing I've swept under the rug here is assuming that $(1 - e_n)$ is unity in $(1 - e_n)R(1 - e_n)$, but this is an easy verification by the theory of corner rings.