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Let $R$ be a ring and for $x,y\!\in\!R$, define the commutator as $[x,y]:=xy-yx$. An $r\!\in\!R$ is idempotent iff $r^2=r$.

How can one prove, that if every commutator is idempotent, then the whole ring is commutative, i.e. all commutators are zero?

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You can find a proof of this in "Two elementary generalisations of boolean rings" AMM Feb.1986 –  Theta33 Mar 7 '11 at 8:19
    
Yesss, thank you, that's what I needed, great reference. –  Leon Mar 7 '11 at 15:53

2 Answers 2

This does not answer the general question, but a simple special case, which I find cute:

Suppose $R$ is a finite dimensional algebra over a field of characteristic zero. Pick $x$, $y\in R$. Since $[x,y]$ is a commutator, its action on every finite dimensional $R$-module is of trace zero; on the other hand, since $[x,y]$ is by hypothesis idempotent, its trace on a module is the dimension of its image. It follows that $[x,y]$ acts by zero on every module so, because faithful modules do exist, it must be zero.

Later. Consider the non-commutative polynomial $f(x,y)=[x,y]^2-[x,y]$, and define new polynomial $h$ by $$h(x,y)=f(x,y)-f(y,x)$$ A simple computation shows that $h(x,y)=2[x,y]$.

If we plug elements of $R$ into $f$ we get zero, so the same is true of $h$. The above observation means then that $2[x,y]=0$ for all $x$, $y\in R$. If now $2$ is invertible in $R$, or at least not a divisor of zero, then we see that $R$ is commutative.

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Dear Mariano, This is very nice! Best wishes, –  Matt E Mar 7 '11 at 6:23
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This is a nice answer. But boring computation? $[x,y] = -[y,x]$ hence $[x,y]^2 + [y,x]^2 = 2[x,y]^2$ and $[x,y]+[y,x] = 0$, right? –  Myself Mar 7 '11 at 7:56
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Lovely edit :-) Actually the review-history is as interesting as your answer, it's interesting to see how you arrived at this. –  Myself Mar 7 '11 at 10:40
    
I think you have lost your sign: $h(x,y)=-2[x,y]$, which matters not, though. –  awllower Jul 3 '13 at 9:27
up vote 10 down vote accepted

Proof from "Two elementary generalisations of boolean rings" AMM Feb.1986: Let $Z(R):=\{x\!\in\!R;\: \forall y\!\in\!R\!: xy=yx\}$ denote the center of the ring. We notice that \begin{equation*} \label{eq:7.1} xy-yx=(xy-yx)^2=(yx-xy)^2=yx-xy. \tag*{(1)} \end{equation*} First we show a general lemma, that if $xy\!=\!0$ implies $yx\!=\!0$ $(\ast)$, then every idempotent $e$ is central. For arbitrary $r\!\in\!R$ we have $(e^2\!-\!e)r=0=e(er\!-\!r)$ and by $(\ast)$ we have $0=(er\!-\!r)e=ere\!-\!re$. Similarly, from $r(e^2\!-\!e)=0=(re\!-\!r)e$ by $(\ast)$ we get $0=e(re\!-\!r)=ere\!-\!er$. Therefore $re=ere=er$ which proves $e\!\in\!Z(R)$.

Next we prove, that all commutators are central. This follows from the previous paragraph, because the condition $(\ast)$ holds: if $xy=0$, then $yx$ $=yx-xy$ $=(yx\!-\!xy)^2$ $=(yx)^2$ $=y(xy)x$ $=0$. We have proved, that \begin{equation*} \label{eq:7.2} \forall x,y\in R:\; xy-yx\in Z(R). \tag*{(2)} \end{equation*}

Next we prove that all squares are central. This we get from the equality $x(xy-yx)$ $\overset{(1)}{=}x(yx-xy)$ $\overset{(2)}{=}(yx-xy)x$, which tells us that $x^2y=yx^2$ , i.e. \begin{equation*} \label{eq:7.3} \forall x\in R:\; x^2\in Z(R). \tag*{(3)} \end{equation*}

Next we prove, that $(xy)^2=(yx)^2$. This we achieve via $(3)$ by writing $yxy$ as a sum of squares: $$(xy)^2 =x(yxy)=x\big((yx)^2+y^2-(yx\!-\!y)^2-y^2x\big)$$ \begin{equation*} \label{eq:7.4} \overset{(3)} {=}\big((yx)^2+y^2-(yx-y)^2-y^2x\big)x=(yxy)x=(yx)^2 \tag*{(4)} \end{equation*}

Finally, we show that all elements are central:

$$xy\!-\!yx=(xy\!-\!yx)^2=xy(xy\!-\!yx)\!-\!yx(xy\!-\!yx) \overset{(1)}{=} xy(xy\!-\!yx)\!-\!yx(yx\!-\!xy)$$ $$= (xy)^2\!-\!xy^2x\!-\!(yx)^2\!+\!yx^2y\overset{(4)}{=}-\!xy^2x\!+\!yx^2y \overset{(3)}{=} -\!x^2y^2\!+\!x^2y^2=0$$ This completes the proof.

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Nice, thanks for posting this! –  Myself Mar 7 '11 at 22:04

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