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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 $x\in R$.

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    $\begingroup$ 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 $x$. $\endgroup$ – darij grinberg Apr 30 '15 at 7:36
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    $\begingroup$ 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. $\endgroup$ – darij grinberg Apr 30 '15 at 7:37
  • $\begingroup$ You've got the solution. $\endgroup$ – hamid kamali Apr 30 '15 at 7:47
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    $\begingroup$ I've posted a cleaned-up version of this argument as an answer. $\endgroup$ – darij grinberg Apr 30 '15 at 7:49
  • $\begingroup$ your solution was complete. So I edit the question. $\endgroup$ – hamid kamali Apr 30 '15 at 8:00
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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 $x^2 = \left(e_1 + e_2\right)^2 = \underbrace{e_1^2}_{=e_1} + \underbrace{e_2^2}_{=e_2} + e_1 e_2 + \underbrace{e_2 e_1}_{=e_1 e_2} = \underbrace{e_1 + e_2}_{=x} + \underbrace{e_1 e_2 + e_1 e_2}_{= 2 e_1 e_2} = x + 2 e_1 e_2$. 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 \begin{equation} x^3 - x = 3 \cdot \left(x^2-x\right) \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(-x\right)^3 - \left(-x\right) = 3 \cdot \left(\left(-x\right)^2-\left(-x\right)\right) . \end{equation} Adding this equality to \eqref{darij.sol.1}, we obtain \begin{equation} 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) . \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, $6 \underbrace{x}_{= e_1 + e_2} = 6\left(e_1 + e_2\right) = \underbrace{6 e_1}_{=0} + \underbrace{6 e_2}_{=0} = 0$. Finally, recall that $x^2 - x = 2 e_1 e_2$ (we have already shown this above), and \eqref{darij.sol.1} yields \begin{equation} 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 , \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, $\left( n+1\right) !x=0$ 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, \begin{equation} \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}}. \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 $n\in\mathbb{N}$. Let $e_1 ,e_2 ,\ldots,e_n $ be $n$ idempotents in $R$. Let $x=e_1 +e_2 +\cdots+e_n $. Then, \begin{equation} x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) =0. \end{equation} (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 \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 $u\in\left[ n\right] $ satisfies \begin{equation} \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. \end{equation} Multiplying these equalities for all $u\in\left[ n\right] $, we obtain \begin{equation} \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. \end{equation} 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 \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 $\left( i_1 ,i_2 ,\ldots,i_n \right) \in\left\{ 0,1\right\} ^{n}$ 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, \begin{equation} \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 \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}-0\right) \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_{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}} \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{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. \end{equation} This proves \eqref{darij.pf.prop.3.2a}.]

Next, I claim that \begin{equation} \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} \end{equation} 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, \begin{equation} 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) . \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_{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 \begin{equation} 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} \end{equation} 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, \begin{equation} \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. \end{equation} 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, \begin{equation} x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) =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 $x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) =0$. 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 \begin{equation} 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} \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} 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. \end{equation} But $e$ is idempotent, so that $e^{2}=e$. We have \begin{equation} e\cdot\left( pe-N\right) =e\cdot pe-eN=p\underbrace{e^{2}}_{=e} -Ne=pe-Ne=\left( p-N\right) e. \end{equation} 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 \begin{equation} x\left( x-1\right) \left( x-2\right) \cdots\left( x-n\right) =0. \end{equation} 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 \begin{equation} 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) \end{equation} (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 \begin{equation} \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) . \end{equation} 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 \begin{equation} \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. \end{equation} 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, \begin{equation} \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. \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.

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

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