# Ring and sum of idempotent elements

Let $R$ be a ring with identity which for every $x\in R$ 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 $x\in R$.

• Nice question! I can so far only show that $6x^2 = 0$. Indeed, add your equality $x^3-3x^2+2x=0$ to the analogous equality $\left(-x\right)^3-3\left(-x\right)^2+2\left(-x\right)=0$ for $-x$ in lieu of $x$. – darij grinberg Apr 30 '15 at 7:36
• Ah! Applying $6x^2 = 0$ to $e_1$ instead of $x$, we obtain $6e_1^2 = 0$, which simplifies to $6e_1 = 0$ since $e_1$ is idempotent. Now you're done. – darij grinberg Apr 30 '15 at 7:37
• You've got the solution. – hamid kamali Apr 30 '15 at 7:47
• I've posted a cleaned-up version of this argument as an answer. – darij grinberg Apr 30 '15 at 7:49
• your solution was complete. So I edit the question. – hamid kamali Apr 30 '15 at 8:00

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 $$x^2 = \left(e_1 + e_2\right)^2 = \underbrace{e_1^2}_{=e_1} + \underbrace{e_2^2}_{=e_2} + e_1 e_2 + \underbrace{e_2 e_1}_{=e_1 e_2} = \underbrace{e_1 + e_2}_{=x} + \underbrace{e_1 e_2 + e_1 e_2}_{= 2 e_1 e_2} = x + 2 e_1 e_2$$. Hence, $$x^2 - x = 2 e_1 e_2$$. Now,

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

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

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

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

We now forget that we fixed $$x$$. We thus have shown that $$$$6x^2 = 0 \qquad \text{ for every x \in R.} \label{darij.sol.2} \tag{2}$$$$ Now, let $$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} = 6\left(e_1 + e_2\right) = \underbrace{6 e_1}_{=0} + \underbrace{6 e_2}_{=0} = 0$$. Finally, recall that $$x^2 - x = 2 e_1 e_2$$ (we have already shown this above), and \eqref{darij.sol.1} yields $$$$x^3 - x = 3 \cdot \underbrace{\left(x^2 - x\right)}_{= 2 e_1 e_2} = 3 \cdot 2 e_1 e_2 = \underbrace{6 e_1}_{=0} e_2 = 0 ,$$$$ so that $$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, $$\left( n+1\right) !x=0$$ for all $$x\in R$$.

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

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

(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 $$n\in\mathbb{N}$$. Let $$e_1 ,e_2 ,\ldots,e_n$$ be $$n$$ idempotents in $$R$$. Let $$x=e_1 +e_2 +\cdots+e_n$$. Then, $$$$x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) =0.$$$$ (Here, of course, $$1,2,\ldots,n$$ denote the corresponding elements of $$R$$.)

Proof of Proposition 3. For each $$u\in\left[ n\right]$$ and $$v\in\left\{ 0,1\right\}$$, we define an element $$P_{u,v}$$ of $$R$$ by $$$$P_{u,v}= \begin{cases} 1-e_{u}, & \text{if }v=0;\\ e_{u}, & \text{if }v=1. \end{cases}$$$$ Then, each $$u\in\left[ n\right]$$ satisfies $$$$\sum_{i\in\left\{ 0,1\right\} }P_{u,i}=\underbrace{P_{u,0}} _{\substack{=1-e_{u}\\\text{(by the definition of }P_{u,2}\text{)} }}+\underbrace{P_{u,1}}_{\substack{=e_{u}\\\text{(by the definition of }P_{u,1}\text{)}}}=\left( 1-e_{u}\right) +e_{u}=1.$$$$ Multiplying these equalities for all $$u\in\left[ n\right]$$, we obtain $$$$\left( \sum_{i\in\left\{ 0,1\right\} }P_{1,i}\right) \left( \sum _{i\in\left\{ 0,1\right\} }P_{2,i}\right) \cdots\left( \sum_{i\in\left\{ 0,1\right\} }P_{n,i}\right) =\underbrace{1\cdot1\cdot\cdots\cdot1}_{n\text{ times}}=1.$$$$ Hence, \begin{align} 1 & =\left( \sum_{i\in\left\{ 0,1\right\} }P_{1,i}\right) \left( \sum_{i\in\left\{ 0,1\right\} }P_{2,i}\right) \cdots\left( \sum _{i\in\left\{ 0,1\right\} }P_{n,i}\right) \nonumber\\ & =\sum_{\left( i_{1},i_{2},\ldots,i_{n}\right) \in\left\{ 0,1\right\} ^{n}}P_{1,i_1}P_{2,i_2}\cdots P_{n,i_n} \label{darij.pf.prop.3.1} \tag{3} \end{align} (by Proposition 2, applied to $$m=n$$ and $$I=\left\{ 0,1\right\}$$).

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

[Proof of \eqref{darij.pf.prop.3.2a}: Let $$\left( i_1 ,i_2 ,\ldots ,i_n \right) \in\left\{ 0,1\right\} ^{n}$$ and $$u\in\left[ n\right]$$. 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 $$i_{u}\in\left\{ 0,1\right\}$$ (since $$\left( i_1 ,i_2 ,\ldots,i_n \right) \in\left\{ 0,1\right\} ^{n}$$), 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, $$$$\left( e_{u}-\underbrace{i_{u}}_{=1}\right) \underbrace{P_{u,i_{u}}} _{=e_{u}}=\left( e_{u}-1\right) e_{u}=e_{u}^{2}-e_{u}=0$$$$ (since $$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, $$$$\left( e_{u}-\underbrace{i_{u}}_{=0}\right) \underbrace{P_{u,i_{u}} }_{=1-e_{u}}=\left( e_{u}-0\right) \left( 1-e_{u}\right) =e_{u}-e_{u} ^{2}=0$$$$ (since $$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 $$$$P_{1,i_1 }P_{2,i_2 }\cdots P_{n,i_n }=\prod_{v\in\left[ n\right] }P_{v,i_{v}}=P_{u,i_{u}}\prod_{\substack{v\in\left[ n\right] ;\\v\neq u}}P_{v,i_{v}}$$$$ (here, we have split off the factor for $$v=u$$ from the product, since $$R$$ is commutative). Hence, $$$$\left( e_{u}-i_{u}\right) \underbrace{\left( P_{1,i_1 }P_{2,i_2 }\cdots P_{n,i_n }\right) }_{=P_{u,i_{u}}\prod_{\substack{v\in\left[ n\right] ;\\v\neq u}}P_{v,i_{v}}}=\underbrace{\left( e_{u}-i_{u}\right) P_{u,i_{u}} }_{=0}\prod_{\substack{v\in\left[ n\right] ;\\v\neq u}}P_{v,i_{v}}=0.$$$$ This proves \eqref{darij.pf.prop.3.2a}.]

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

[Proof of \eqref{darij.pf.prop.3.2}: Let $$\left( i_1 ,i_2 ,\ldots ,i_n \right) \in\left\{ 0,1\right\} ^{n}$$. Recall that $$x=e_1 +e_2 +\cdots+e_n =\sum_{j\in\left[ n\right] }e_{j}$$. Thus, $$$$x-\sum_{j\in\left[ n\right] }i_{j}=\sum_{j\in\left[ n\right] }e_{j} -\sum_{j\in\left[ n\right] }i_{j}=\sum_{j\in\left[ n\right] }\left( e_{j}-i_{j}\right) =\sum_{u\in\left[ n\right] }\left( e_{u}-i_{u}\right) .$$$$ Multiplying both sides of this equality by $$P_{1,i_1 }P_{2,i_2 }\cdots P_{n,i_n }$$, we find \begin{align*} & \left( x-\sum_{j\in\left[ n\right] }i_{j}\right) \left( P_{1,i_1 }P_{2,i_2 }\cdots P_{n,i_n }\right) \\ & =\left( \sum_{u\in\left[ n\right] }\left( e_{u}-i_{u}\right) \right) \left( P_{1,i_1 }P_{2,i_2 }\cdots P_{n,i_n }\right) \\ & =\sum_{u\in\left[ n\right] }\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 $$y=x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right)$$. Then, I claim that $$$$y\left( P_{1,i_1 }P_{2,i_2 }\cdots P_{n,i_n }\right) =0 \label{darij.pf.prop.3.3} \tag{6}$$$$ for each $$\left( i_1 ,i_2 ,\ldots,i_n \right) \in\left\{ 0,1\right\} ^{n}$$.

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

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

Now, \begin{align*} & x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) \\ & =y=y\cdot1\\ & =y\cdot\sum_{\left( i_1 ,i_2 ,\ldots,i_n \right) \in\left\{ 0,1\right\} ^{n}}P_{1,i_1 }P_{2,i_2 }\cdots P_{n,i_n }\\ & \qquad\left( \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) \in\left\{ 0,1\right\} ^{n}}\underbrace{y\left( 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 $$n\in\mathbb{N}$$. 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, $$$$x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) =0.$$$$ (Here, of course, $$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 $$x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) =0$$. This proves Proposition 4. $$\blacksquare$$

Proposition 5. Let $$R$$ be a unital ring. Let $$e\in R$$ be idempotent. Let $$p\in\mathbb{Z}$$. Let $$x=pe\in R$$. Then, each $$n\in\mathbb{N}$$ satisfies $$$$x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) =\left( p\left( p-1\right) \left( p-2\right) \cdots\left( p-n\right) \right) e. \label{darij.eq.prop.5.1} \tag{7}$$$$

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, $$$$x\left( x-1\right) \left( x-2\right) \cdots\left( x-\left( N-1\right) \right) =\left( p\left( p-1\right) \left( p-2\right) \cdots\left( p-\left( N-1\right) \right) \right) e.$$$$ But $$e$$ is idempotent, so that $$e^{2}=e$$. We have $$$$e\cdot\left( pe-N\right) =e\cdot pe-eN=p\underbrace{e^{2}}_{=e} -Ne=pe-Ne=\left( p-N\right) e.$$$$ Now, \begin{align*} & x\left( x-1\right) \left( x-2\right) \cdots\left( x-N\right) \\ & =\underbrace{\left( x\left( x-1\right) \left( x-2\right) \cdots\left( x-\left( N-1\right) \right) \right) }_{=\left( p\left( p-1\right) \left( p-2\right) \cdots\left( p-\left( N-1\right) \right) \right) e}\cdot\left( \underbrace{x}_{=pe}-N\right) \\ & =\left( p\left( p-1\right) \left( p-2\right) \cdots\left( p-\left( N-1\right) \right) \right) \underbrace{e\cdot\left( pe-N\right) }_{=\left( p-N\right) e}\\ & =\underbrace{\left( p\left( p-1\right) \left( p-2\right) \cdots\left( p-\left( N-1\right) \right) \right) \cdot\left( p-N\right) }_{=p\left( p-1\right) \left( p-2\right) \cdots\left( p-N\right) }e\\ & =\left( p\left( p-1\right) \left( p-2\right) \cdots\left( p-N\right) \right) e. \end{align*} In other words, \eqref{darij.eq.prop.5.1} holds for $$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 $$e\in R$$ be idempotent. Let $$n\in\mathbb{N}$$. Assume that $$\left( n+1\right) e$$ is a sum of $$n$$ pairwise commuting idempotents in $$R$$. Then, $$\left( n+1\right) !e=0$$.

Proof of Corollary 6. We have assumed that $$\left( n+1\right) e$$ 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 $$\left( n+1\right) e=e_1 +e_2 +\cdots+e_n$$. Consider these $$e_1 ,e_2 ,\ldots,e_n$$.

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

(b) If $$r\in R$$, then $$L_R$$ shall denote the map $$R\rightarrow R,\ x\mapsto rx$$. 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 $$R\rightarrow\operatorname{End} R,\ r\mapsto L_R$$. (This map $$L_R$$ is known as the "left regular action" of $$R$$.)

Proposition 7. Let $$R$$ be a nonunital ring. Then, the map $$L_R :R\rightarrow\operatorname{End} R$$ 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 $$L_R :R\rightarrow \operatorname{End} R$$ is injective.

Now, we can generalize Corollary 6 to nonunital rings:

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

Proof of Corollary 8. We have assumed that $$\left( n+1\right) e$$ 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 $$\left( n+1\right) e=e_1 +e_2 +\cdots+e_n$$. Consider these $$e_1 ,e_2 ,\ldots,e_n$$.

Consider the map $$L_R :R\rightarrow\operatorname{End} R$$. 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 $$L_R \left( e\right)$$ of the idempotent $$e$$ must again be an idempotent. Also, applying the map $$L_R$$ to both sides of the equality $$\left( n+1\right) e=e_1 +e_2 +\cdots+e_n$$, we obtain $$$$L_R \left( \left( n+1\right) e\right) =L_R \left( e_1 +e_2 +\cdots+e_n \right) =L_R \left( e_1 \right) +L_R \left( e_2 \right) +\cdots+L_R \left( e_n \right)$$$$ (since $$L_R$$ is a nonunital ring homomorphism). In view of $$L_R \left( \left( n+1\right) e\right) =\left( n+1\right) L_R \left( e\right)$$ (which holds since $$L_R$$ is a nonunital ring homomorphism), this rewrites as $$$$\left( n+1\right) L_R \left( e\right) =L_R \left( e_1 \right) +L_R \left( e_2 \right) +\cdots+L_R \left( e_n \right) .$$$$ Thus, $$\left( n+1\right) L_R \left( e\right)$$ is a sum of $$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( e\right)$$ instead of $$R$$ and $$e$$) yields $$\left( n+1\right) !L_R \left( e\right) =0$$.

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

We can now prove Theorem 1 at last:

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

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 $$i\in\left\{ 1,2,\ldots,n\right\}$$. Then, $$e_i \in R$$ is an idempotent. Moreover, $$\left( n+1\right) e_i$$ 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 $$\left( n+1\right) !e_i =0$$.

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

• I think you could prove Propositions 3 and 4 more simply as follows: prove by induction on $m$ that $x(x-1)\ldots(x-m)\in\mathrm{span}_{\mathbb Z}\{e_{i_0}\ldots e_{i_m}\mid 1\leq i_0<i_1<\ldots<i_m\leq n\}$. – stewbasic Nov 25 '18 at 22:20
• @stewbasic: Nice observation! But the induction step uses computations fairly similar to mine, at least the way I would do it. – darij grinberg Nov 25 '18 at 22:35