Let $R$ be a ring with identity which for every $xin R$ there exist two idempotent elements $e_1,e_2$ such that $x=e_1+e_2$ and $e_1e_2=e_2e_1$. Prove that: $x^3=x$ for every $xin R$.

1. Answering the question

You don't need $R$ to have an identity. More generally, let me show:

Problem. Let $R$ be any nonunital ring. Assume that for every $x in R$, there exist two idempotent elements $e_1$ and $e_2$ of $R$ such that $x = e_1 + e_2$ and $e_1 e_2 = e_2 e_1$. Prove that every $x in R$ satisfies $6x = 0$ and $x^3 = x$.

Solution. Let $x in R$ be arbitrary. Then, (by assumption) we know that there exist two idempotent elements $e_1$ and $e_2$ of $R$ such that $x = e_1 + e_2$ and $e_1 e_2 = e_2 e_1$. Consider these $e_1$ and $e_2$. Since $e_1$ and $e_2$ are idempotent, we have $e_1^2 = e_1$ and $e_2^2 = e_2$. Squaring both sides of the equality $x = e_1 + e_2$, we obtain $x^2 = left(e_1 + e_2right)^2 = underbrace{e_1^2}_{=e_1} + underbrace{e_2^2}_{=e_2} + e_1 e_2 + underbrace{e_2 e_1}_{=e_1 e_2} = underbrace{e_1 + e_2}_{=x} + underbrace{e_1 e_2 + e_1 e_2}_{= 2 e_1 e_2} = x + 2 e_1 e_2$. Hence, $x^2 - x = 2 e_1 e_2$. Now,

$x^3 - x = underbrace{x}_{=e_1 + e_2}underbrace{left(x^2 - xright)}_{= 2 e_1 e_2} + underbrace{left(x^2 - xright)}_{= 2 e_1 e_2} = underbrace{left(e_1 + e_2right) 2 e_1 e_2}_{= 2 e_1 e_1 e_2 + 2 e_2 e_1 e_2} + 2 e_1 e_2$

$= 2 underbrace{e_1 e_1}_{= e_1^2 = e_1} e_2 + 2 underbrace{e_2 e_1}_{= e_2^2} e_2 + 2 e_1 e_2 = 2 e_1 e_2 + 2 e_2 underbrace{e_1 e_1}_{= e_1^2 = e_1} + 2 e_1 e_2 = 2 e_1 e_2 + 2 underbrace{e_2 e_1}_{= e_1 e_2} + 2 e_1 e_2$

$= 2 e_1 e_2 + 2 e_1 e_2 + 2 e_1 e_2 = 3 cdot underbrace{2 e_1 e_2}_{= x^2 - x} = 3 cdot left(x^2-xright)$.

We now forget that we fixed $x$. We thus have shown that
begin{equation}
x^3 - x = 3 cdot left(x^2-xright) qquad text{ for every $x in R$.}
label{darij.sol.1}
tag{1}
end{equation}
Now, let $x in R$ again. Applying eqref{darij.sol.1} to $-x$ instead of $x$, we obtain
begin{equation}
left(-xright)^3 - left(-xright) = 3 cdot left(left(-xright)^2-left(-xright)right) .
end{equation}
Adding this equality to eqref{darij.sol.1}, we obtain
begin{equation}
x^3 - x + left(-xright)^3 - left(-xright) = 3 cdot left(x^2 - xright) + 3 cdot left(left(-xright)^2-left(-xright)right) .
end{equation}
After cancelling terms, this simplifies to $0 = 6x^2$. Thus, $6x^2 = 0$.

We now forget that we fixed $x$. We thus have shown that
begin{equation}
6x^2 = 0 qquad text{ for every $x in R$.}
label{darij.sol.2}
tag{2}
end{equation}
Now, let $x in R$ again. Then, (by assumption) we know that there exist two idempotent elements $e_1$ and $e_2$ of $R$ such that $x = e_1 + e_2$ and $e_1 e_2 = e_2 e_1$. Consider these $e_1$ and $e_2$. Since $e_1$ is idempotent, we have $e_1^2 = e_1$. But eqref{darij.sol.2} (applied to $e_1$ instead of $x$) gives $6 e_1^2 = 0$. Thus, $6 underbrace{e_1}_{=e_1^2} = 6 e_1^2 = 0$. Similarly, $6 e_2 = 0$. Now, $6 underbrace{x}_{= e_1 + e_2} = 6left(e_1 + e_2right) = underbrace{6 e_1}_{=0} + underbrace{6 e_2}_{=0} = 0$. Finally, recall that $x^2 - x = 2 e_1 e_2$ (we have already shown this above), and eqref{darij.sol.1} yields
begin{equation}
x^3 - x = 3 cdot underbrace{left(x^2 - xright)}_{= 2 e_1 e_2} = 3 cdot 2 e_1 e_2 = underbrace{6 e_1}_{=0} e_2 = 0 ,
end{equation}
so that $x^3 = x$. Thus the problem is solved. $blacksquare$

Addendum. Let $R$ be as in the Problem above. Then, the ring $R$ is commutative.

Proof. The Problem shows that $x^3 = x$ for all $x in R$. Hence, a classical fact yields that $R$ is commutative. This proves the Addendum. $blacksquare$

2. Generalizing to $n$ idempotents

Here is a generalization of the "$6x=0$" part of the above problem:

Theorem 1. Let $n$ be a nonnegative integer. Let $R$ be a nonunital ring
such that every element of $R$ is a sum of $n$ pairwise commuting idempotents.
Then, $left( n+1right) !x=0$ for all $xin R$.

The proof of this theorem will require several auxiliary results, which in my
opinion are interesting on their own.

We let $mathbb{N}$ be the set $left{ 0,1,2,ldotsright} $ of all
nonnegative integers. If $ninmathbb{N}$, then $left[ nright] $ shall
denote the $n$-element set $left{ 1,2,ldots,nright} $. We recall the
product rule:

Proposition 2. Let $minmathbb{N}$. Let $R$ be a unital ring. Let $I$ be
a finite set. For each $uinleft[ mright] $ and $vin I$, let $P_{u,v}$ be
an element of $R$. Then,
begin{equation}
left( sum_{iin I}P_{1,i}right) left( sum_{iin I}P_{2,i}right)
cdotsleft( sum_{iin I}P_{m,i}right) =sum_{left( i_{1},i_{2}
,ldots,i_{m}right) in I^{m}}P_{1,i_{1}}P_{2,i_{2}}cdots P_{m,i_{m}}.
end{equation}

(This is a known fact, and is easily proven by induction on $m$; intuitively
it is obvious anyway.)

Let us first prove a basic property of sums of idempotents in unital
commutative rings:

Proposition 3. Let $R$ be a unital commutative ring. Let $ninmathbb{N}$.
Let $e_1 ,e_2 ,ldots,e_n $ be $n$ idempotents in $R$. Let $x=e_1
+e_2 +cdots+e_n $. Then,
begin{equation}
xleft( x-1right) left( x-2right) cdotsleft( x-nright) =0.
end{equation}
(Here, of course, $1,2,ldots,n$ denote the corresponding elements of $R$.)

Proof of Proposition 3. For each $uinleft[ nright] $ and $vinleft{
0,1right} $, we define an element $P_{u,v}$ of $R$ by
begin{equation}
P_{u,v}=
begin{cases}
1-e_{u}, & text{if }v=0;\
e_{u}, & text{if }v=1.
end{cases}
end{equation}
Then, each $uinleft[ nright] $ satisfies
begin{equation}
sum_{iinleft{ 0,1right} }P_{u,i}=underbrace{P_{u,0}}
_{substack{=1-e_{u}\text{(by the definition of }P_{u,2}text{)}
}}+underbrace{P_{u,1}}_{substack{=e_{u}\text{(by the definition of
}P_{u,1}text{)}}}=left( 1-e_{u}right) +e_{u}=1.
end{equation}
Multiplying these equalities for all $uinleft[ nright] $, we obtain
begin{equation}
left( sum_{iinleft{ 0,1right} }P_{1,i}right) left( sum
_{iinleft{ 0,1right} }P_{2,i}right) cdotsleft( sum_{iinleft{
0,1right} }P_{n,i}right) =underbrace{1cdot1cdotcdotscdot1}_{ntext{
times}}=1.
end{equation}
Hence,
begin{align}
1 & =left( sum_{iinleft{ 0,1right} }P_{1,i}right) left(
sum_{iinleft{ 0,1right} }P_{2,i}right) cdotsleft( sum
_{iinleft{ 0,1right} }P_{n,i}right) nonumber\
& =sum_{left( i_{1},i_{2},ldots,i_{n}right) inleft{ 0,1right}
^{n}}P_{1,i_1}P_{2,i_2}cdots P_{n,i_n}
label{darij.pf.prop.3.1}
tag{3}
end{align}
(by Proposition 2, applied to $m=n$ and $I=left{ 0,1right} $).

Next, I claim that
begin{equation}
left( e_{u}-i_{u}right) left( P_{1,i_1}P_{2,i_2}cdots P_{n,i_n}
right) =0
label{darij.pf.prop.3.2a}
tag{4}
end{equation}
for each $left( i_1 ,i_2 ,ldots,i_n right) inleft{ 0,1right}
^{n}$ and each $uinleft[ nright] $.

[Proof of eqref{darij.pf.prop.3.2a}: Let $left( i_1 ,i_2 ,ldots
,i_n right) inleft{ 0,1right} ^{n}$ and $uinleft[ nright] $. We
shall show that $left( e_{u}-i_{u}right) P_{u,i_{u}}=0$.

Note that $e_{u}$ is idempotent (since $e_1 ,e_2 ,ldots,e_n $ are $n$
idempotents), so that $e_{u}^{2}=e_{u}$.

We have $i_{u}inleft{ 0,1right} $ (since $left( i_1 ,i_2
,ldots,i_n right) inleft{ 0,1right} ^{n}$), so that we have either
$i_{u}=1$ or $i_{u}=0$. Thus, we are in one of the following two cases:

Case 1: We have $i_{u}=1$.

Case 2: We have $i_{u}=0$.

Let us first consider Case 1. In this case, we have $i_{u}=1$. Thus,
$P_{u,i_{u}}=P_{u,1}=e_{u}$ (by the definition of $P_{u,1}$). Thus,
begin{equation}
left( e_{u}-underbrace{i_{u}}_{=1}right) underbrace{P_{u,i_{u}}}
_{=e_{u}}=left( e_{u}-1right) e_{u}=e_{u}^{2}-e_{u}=0
end{equation}
(since $e_{u}^{2}=e_{u}$). Thus, $left( e_{u}-i_{u}right) P_{u,i_{u}}=0$
is proven in Case 1.

Next, let us consider Case 2. In this case, we have $i_{u}=0$. Thus,
$P_{u,i_{u}}=P_{u,0}=1-e_{u}$ (by the definition of $P_{u,0}$). Thus,
begin{equation}
left( e_{u}-underbrace{i_{u}}_{=0}right) underbrace{P_{u,i_{u}}
}_{=1-e_{u}}=left( e_{u}-0right) left( 1-e_{u}right) =e_{u}-e_{u}
^{2}=0
end{equation}
(since $e_{u}^{2}=e_{u}$). Thus, $left( e_{u}-i_{u}right) P_{u,i_{u}}=0$
is proven in Case 2.

We have now proven $left( e_{u}-i_{u}right) P_{u,i_{u}}=0$ in both Cases 1
and 2. Thus, $left( e_{u}-i_{u}right) P_{u,i_{u}}=0$ always holds.

But
begin{equation}
P_{1,i_1 }P_{2,i_2 }cdots P_{n,i_n }=prod_{vinleft[ nright]
}P_{v,i_{v}}=P_{u,i_{u}}prod_{substack{vinleft[ nright] ;\vneq
u}}P_{v,i_{v}}
end{equation}
(here, we have split off the factor for $v=u$ from the product, since $R$ is
commutative). Hence,
begin{equation}
left( e_{u}-i_{u}right) underbrace{left( P_{1,i_1 }P_{2,i_2 }cdots
P_{n,i_n }right) }_{=P_{u,i_{u}}prod_{substack{vinleft[ nright]
;\vneq u}}P_{v,i_{v}}}=underbrace{left( e_{u}-i_{u}right) P_{u,i_{u}}
}_{=0}prod_{substack{vinleft[ nright] ;\vneq u}}P_{v,i_{v}}=0.
end{equation}
This proves eqref{darij.pf.prop.3.2a}.]

Next, I claim that
begin{equation}
left( x-sum_{jinleft[ nright] }i_{j}right) left( P_{1,i_1
}P_{2,i_2 }cdots P_{n,i_n }right) =0
label{darij.pf.prop.3.2}
tag{5}
end{equation}
for each $left( i_1 ,i_2 ,ldots,i_n right) inleft{ 0,1right}
^{n}$.

[Proof of eqref{darij.pf.prop.3.2}: Let $left( i_1 ,i_2 ,ldots
,i_n right) inleft{ 0,1right} ^{n}$. Recall that $x=e_1
+e_2 +cdots+e_n =sum_{jinleft[ nright] }e_{j}$. Thus,
begin{equation}
x-sum_{jinleft[ nright] }i_{j}=sum_{jinleft[ nright] }e_{j}
-sum_{jinleft[ nright] }i_{j}=sum_{jinleft[ nright] }left(
e_{j}-i_{j}right) =sum_{uinleft[ nright] }left( e_{u}-i_{u}right)
.
end{equation}
Multiplying both sides of this equality by $P_{1,i_1 }P_{2,i_2 }cdots
P_{n,i_n }$, we find
begin{align*}
& left( x-sum_{jinleft[ nright] }i_{j}right) left( P_{1,i_1
}P_{2,i_2 }cdots P_{n,i_n }right) \
& =left( sum_{uinleft[ nright] }left( e_{u}-i_{u}right) right)
left( P_{1,i_1 }P_{2,i_2 }cdots P_{n,i_n }right) \
& =sum_{uinleft[ nright] }underbrace{left( e_{u}-i_{u}right)
left( P_{1,i_1 }P_{2,i_2 }cdots P_{n,i_n }right) }
_{substack{=0\text{(by eqref{darij.pf.prop.3.2a})}}}=0.
end{align*}
This proves eqref{darij.pf.prop.3.2}.]

Now, let $y=xleft( x-1right) left( x-2right) cdotsleft( x-nright)
$. Then, I claim that
begin{equation}
yleft( P_{1,i_1 }P_{2,i_2 }cdots P_{n,i_n }right) =0
label{darij.pf.prop.3.3}
tag{6}
end{equation}
for each $left( i_1 ,i_2 ,ldots,i_n right) inleft{ 0,1right}
^{n}$.

[Proof of eqref{darij.pf.prop.3.3}: Let $left( i_1 ,i_2 ,ldots
,i_n right) inleft{ 0,1right} ^{n}$. Let $m=sum_{jinleft[
nright] }i_{j}$. Then, $m$ is a sum of $n$ elements of the set $left{
0,1right} $ (since $i_1 ,i_2 ,ldots,i_n $ are $n$ elements of the set
$left{ 0,1right} $), and thus is an integer between $0$ and $n$. In
other words, $minleft{ 0,1,ldots,nright} $. Hence, $x-m$ is a factor
of the product $xleft( x-1right) left( x-2right) cdotsleft(
x-nright) $. Thus, the product $xleft( x-1right) left( x-2right)
cdotsleft( x-nright) $ is a multiple of $x-m$. In other words, $y$ is a
multiple of $x-m$ (since $y=xleft( x-1right) left( x-2right)
cdotsleft( x-nright) $). In other words, there exists a $zin R$ such
that $y=zleft( x-mright) $. Consider this $z$.

From eqref{darij.pf.prop.3.2}, we obtain $left( x-sum_{jinleft[
nright] }i_{j}right) left( P_{1,i_1 }P_{2,i_2 }cdots P_{n,i_n
}right) =0$. In view of $m=sum_{jinleft[ nright] }i_{j}$, this
rewrites as $left( x-mright) left( P_{1,i_1 }P_{2,i_2 }cdots
P_{n,i_n }right) =0$. Now,
begin{equation}
underbrace{y}_{=zleft( x-mright) }left( P_{1,i_1 }P_{2,i_2 }cdots
P_{n,i_n }right) =zunderbrace{left( x-mright) left( P_{1,i_1
}P_{2,i_2 }cdots P_{n,i_n }right) }_{=0}=0.
end{equation}
This proves eqref{darij.pf.prop.3.3}.]

Now,
begin{align*}
& xleft( x-1right) left( x-2right) cdotsleft( x-nright) \
& =y=ycdot1\
& =ycdotsum_{left( i_1 ,i_2 ,ldots,i_n right) inleft{
0,1right} ^{n}}P_{1,i_1 }P_{2,i_2 }cdots P_{n,i_n }\
& qquadleft(
begin{array}
[c]{c}
text{here, we have multiplied both sides of}\
text{the equality eqref{darij.pf.prop.3.1} by }y
end{array}
right) \
& =sum_{left( i_1 ,i_2 ,ldots,i_n right) inleft{ 0,1right}
^{n}}underbrace{yleft( P_{1,i_1 }P_{2,i_2 }cdots P_{n,i_n }right)
}_{substack{=0\text{(by eqref{darij.pf.prop.3.3})}}}=0.
end{align*}
This proves Proposition 3. $blacksquare$

Now let us generalize Proposition 3 by replacing the "global" commutativity of
$R$ by the "local" commutativity of our $n$ idempotents; this is a cheap generalization:

Proposition 4. Let $R$ be a unital ring. Let $ninmathbb{N}$. Let
$e_1 ,e_2 ,ldots,e_n $ be $n$ pairwise commuting idempotents in $R$. Let
$x=e_1 +e_2 +cdots+e_n $. Then,
begin{equation}
xleft( x-1right) left( x-2right) cdotsleft( x-nright) =0.
end{equation}
(Here, of course, $1,2,ldots,n$ denote the corresponding elements of $R$.)

Proof of Proposition 4. We know that $R$ is a unital ring, thus a
$mathbb{Z}$-algebra. Let $S$ be the $mathbb{Z}$-subalgebra of $R$ generated
by $e_1 ,e_2 ,ldots,e_n $. Then, $S$ is a $mathbb{Z}$-algebra generated
by $n$ pairwise commuting elements (since its $n$ generators $e_1
,e_2 ,ldots,e_n $ pairwise commute), and thus is commutative itself
(because any $mathbb{Z}$-algebra generated by pairwise commuting elements is
commutative). Thus, $S$ is a unital commutative ring. Moreover, the $n$
elements $e_1 ,e_2 ,ldots,e_n $ belong to $S$ (since they together
generate $S$ as a $mathbb{Z}$-algebra), and thus their sum $x=e_1
+e_2 +cdots+e_n $ belongs to $S$ as well. Hence, Proposition 3 (applied to
$S$ instead of $R$) shows that $xleft( x-1right) left( x-2right)
cdotsleft( x-nright) =0$. This proves Proposition 4. $blacksquare$

Proposition 5. Let $R$ be a unital ring. Let $ein R$ be idempotent. Let
$pinmathbb{Z}$. Let $x=pein R$. Then, each $ninmathbb{N}$ satisfies
begin{equation}
xleft( x-1right) left( x-2right) cdotsleft( x-nright) =left(
pleft( p-1right) left( p-2right) cdotsleft( p-nright) right) e.
label{darij.eq.prop.5.1}
tag{7}
end{equation}

Proof of Proposition 5. We shall prove eqref{darij.eq.prop.5.1} by
induction on $n$:

Induction base: We have $x=pe$. In other words, eqref{darij.eq.prop.5.1}
holds for $n=0$. This concludes the induction base.

Induction step: Let $N$ be a positive integer. Assume that
eqref{darij.eq.prop.5.1} holds for $n=N-1$. We must prove that
eqref{darij.eq.prop.5.1} holds for $n=N$ as well.

We have assumed that eqref{darij.eq.prop.5.1} holds for $n=N-1$. In other
words,
begin{equation}
xleft( x-1right) left( x-2right) cdotsleft( x-left( N-1right)
right) =left( pleft( p-1right) left( p-2right) cdotsleft(
p-left( N-1right) right) right) e.
end{equation}
But $e$ is idempotent, so that $e^{2}=e$. We have
begin{equation}
ecdotleft( pe-Nright) =ecdot pe-eN=punderbrace{e^{2}}_{=e}
-Ne=pe-Ne=left( p-Nright) e.
end{equation}
Now,
begin{align*}
& xleft( x-1right) left( x-2right) cdotsleft( x-Nright) \
& =underbrace{left( xleft( x-1right) left( x-2right) cdotsleft(
x-left( N-1right) right) right) }_{=left( pleft( p-1right)
left( p-2right) cdotsleft( p-left( N-1right) right) right)
e}cdotleft( underbrace{x}_{=pe}-Nright) \
& =left( pleft( p-1right) left( p-2right) cdotsleft( p-left(
N-1right) right) right) underbrace{ecdotleft( pe-Nright)
}_{=left( p-Nright) e}\
& =underbrace{left( pleft( p-1right) left( p-2right) cdotsleft(
p-left( N-1right) right) right) cdotleft( p-Nright) }_{=pleft(
p-1right) left( p-2right) cdotsleft( p-Nright) }e\
& =left( pleft( p-1right) left( p-2right) cdotsleft( p-Nright)
right) e.
end{align*}
In other words, eqref{darij.eq.prop.5.1} holds for $n=N$ as well. This
completes the induction step. Thus, eqref{darij.eq.prop.5.1} is proven by
induction; i.e., Proposition 5 is proven. $blacksquare$

Corollary 6. Let $R$ be a unital ring. Let $ein R$ be idempotent. Let
$ninmathbb{N}$. Assume that $left( n+1right) e$ is a sum of $n$ pairwise
commuting idempotents in $R$. Then, $left( n+1right) !e=0$.

Proof of Corollary 6. We have assumed that $left( n+1right) e$ is a sum
of $n$ pairwise commuting idempotents in $R$. In other words, there exist $n$
pairwise commuting idempotents $e_1 ,e_2 ,ldots,e_n $ such that $left(
n+1right) e=e_1 +e_2 +cdots+e_n $. Consider these $e_1 ,e_2
,ldots,e_n $.

Let $x=left( n+1right) e$. Thus, $x=left( n+1right) e=e_1
+e_2 +cdots+e_n $. Hence, Proposition 4 yields
begin{equation}
xleft( x-1right) left( x-2right) cdotsleft( x-nright) =0.
end{equation}
But Proposition 5 (applied to $p=n+1$) yields
begin{align*}
xleft( x-1right) left( x-2right) cdotsleft( x-nright) &
=underbrace{left( left( n+1right) left( left( n+1right)
-1right) left( left( n+1right) -2right) cdotsleft( left(
n+1right) -nright) right) }_{substack{=left( n+1right) nleft(
n-1right) cdots1\=left( n+1right) !}}e\
& =left( n+1right) !e.
end{align*}
Comparing these two equalities, we obtain $left( n+1right) !e=0$. This
proves Corollary 6. $blacksquare$

Our next goal is to extend Corollary 6 to nonunital rings. There are several
ways to do so. The simplest one is to embed a nonunital ring $R$ into a unital
ring, e.g., via the Dorroh
extension. Here is a
slightly different one, in which we don't exactly embed $R$ into a unital
ring, but construct a nonunital ring homomorphism from $R$ into a unital ring
that is "injective enough" for idempotents (despite not generally being injective).

Definition. Let $R$ be a nonunital ring.

(a) We let $operatorname{End} R$ be the unital ring of all endomorphisms
of the $mathbb{Z}$-module $R$ (that is, of all $mathbb{Z}$-linear maps
$Rrightarrow R$).

(b) If $rin R$, then $L_R $ shall denote the map $Rrightarrow
R, xmapsto rx$. This map $L_R $ is an endomorphism of the $mathbb{Z}
$-module $R$, and thus belongs to $operatorname{End} R$. (For evident
reasons, $L_R $ is known as the "left multiplication by $r$".)

(c) We let $L_R $ denote the map $Rrightarrowoperatorname{End}
R, rmapsto L_R $. (This map $L_R $ is known as the "left regular action"
of $R$.)

Proposition 7. Let $R$ be a nonunital ring. Then, the map $L_R
:Rrightarrowoperatorname{End} R$ is a nonunital ring homomorphism.

Proof of Proposition 7. This is well-known and completely straightforward
(just check that $L_R$ is $mathbb{Z}$-linear and preserves products).
$blacksquare$

Note that if $R$ is a unital ring, then the map $L_R :Rrightarrow
operatorname{End} R$ is injective.

Now, we can generalize Corollary 6 to nonunital rings:

Corollary 8. Let $R$ be a nonunital ring. Let $ein R$ be idempotent. Let
$ninmathbb{N}$. Assume that $left( n+1right) e$ is a sum of $n$ pairwise
commuting idempotents in $R$. Then, $left( n+1right) !e=0$.

Proof of Corollary 8. We have assumed that $left( n+1right) e$ is a sum
of $n$ pairwise commuting idempotents in $R$. In other words, there exist $n$
pairwise commuting idempotents $e_1 ,e_2 ,ldots,e_n $ such that $left(
n+1right) e=e_1 +e_2 +cdots+e_n $. Consider these $e_1 ,e_2
,ldots,e_n $.

Consider the map $L_R :Rrightarrowoperatorname{End} R$. This map $L_R $ is
a nonunital ring homomorphism (by Proposition 7). Hence, the images
$L_R left( e_1 right) ,L_R left( e_2 right) ,ldots,L_R left(
e_n right) $ of the $n$ pairwise commuting idempotents $e_1 ,e_2
,ldots,e_n $ under $L_R$ must again be $n$ pairwise commuting idempotents.
For the same reason, the image $L_R left( eright) $ of the idempotent $e$
must again be an idempotent. Also, applying the map $L_R$ to both sides of
the equality $left( n+1right) e=e_1 +e_2 +cdots+e_n $, we obtain
begin{equation}
L_R left( left( n+1right) eright) =L_R left( e_1 +e_2
+cdots+e_n right) =L_R left( e_1 right) +L_R left( e_2 right)
+cdots+L_R left( e_n right)
end{equation}
(since $L_R$ is a nonunital ring homomorphism). In view of $L_R left(
left( n+1right) eright) =left( n+1right) L_R left( eright) $
(which holds since $L_R$ is a nonunital ring homomorphism), this rewrites as
begin{equation}
left( n+1right) L_R left( eright) =L_R left( e_1 right)
+L_R left( e_2 right) +cdots+L_R left( e_n right) .
end{equation}
Thus, $left( n+1right) L_R left( eright) $ is a sum of $n$ pairwise
commuting idempotents in $operatorname{End} R$ (since $L_R left(
e_1 right) ,L_R left( e_2 right) ,ldots,L_R left( e_n right) $
are $n$ pairwise commuting idempotents in $operatorname{End} R$). Hence,
Corollary 6 (applied to $operatorname{End} R$ and $L_R left( eright) $
instead of $R$ and $e$) yields $left( n+1right) !L_R left( eright)
=0$.

But $e$ is idempotent; thus, $e^{2}=e$. The definition of $L_R$ yields
$left( L_R left( eright) right) left( eright) =ee=e^{2}=e$. Now,
applying the map $left( n+1right) !L_R left( eright) in
operatorname{End} R$ to the element $ein R$, we find
begin{equation}
left( left( n+1right) !L_R left( eright) right) left( eright)
=left( n+1right) !underbrace{left( L_R left( eright) right)
left( eright) }_{=e}=left( n+1right) !e.
end{equation}
Hence, $left( n+1right) !e=underbrace{left( left( n+1right)
!L_R left( eright) right) }_{=0}left( eright) =0left( eright)
=0$. This proves Corollary 8. $blacksquare$

We can now prove Theorem 1 at last:

Proof of Theorem 1. Let $xin R$. We must prove that $left( n+1right)
!x=0$.

We know that $x$ is a sum of $n$ pairwise commuting idempotents (since every
element of $R$ is a sum of $n$ pairwise commuting idempotents). In other
words, there exist $n$ pairwise commuting idempotents $e_1 ,e_2
,ldots,e_n $ such that $x=sum_{i=1}^{n}e_i $. Consider these $e_1
,e_2 ,ldots,e_n $.

Let $iinleft{ 1,2,ldots,nright} $. Then, $e_i in R$ is an
idempotent. Moreover, $left( n+1right) e_i $ is a sum of $n$ pairwise
commuting idempotents in $R$ (since every element of $R$ is a sum of $n$
pairwise commuting idempotents). Hence, Corollary 8 (applied to $e=e_i $)
yields $left( n+1right) !e_i =0$.

Now, forget that we fixed $i$. We thus have shown that $left( n+1right)
!e_i =0$ for each $iinleft{ 1,2,ldots,nright} $. Summing up these
equalities over all $iinleft{ 1,2,ldots,nright} $, we obtain
$sum_{i=1}^{n}left( n+1right) !e_i =0$. Now,
begin{equation}
left( n+1right) !underbrace{x}_{=sum_{i=1}^{n}e_i }=left( n+1right)
!sum_{i=1}^{n}e_i =sum_{i=1}^{n}left( n+1right) !e_i =0.
end{equation}
This proves Theorem 1. $blacksquare$

See Rings in which each element is a sum of $n$ commuting idempotents for further developments about rings $R$ satisfying the conditions of Theorem 1.