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A ring $R$ is a Boolean ring provided that $a^2=a$ for every $a \in R$. How can we show that every Boolean ring is commutative?

Thanks in advance.

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Every Boolean ring is of characteristic 2, since $a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a\implies a+a=0$.

Now, for any $x,y$ in the ring

$x+y=(x+y)^2=x^2+xy+yx+y^2=x+y+xy+yx$, so $xy+yx=0$ and hence $xy+(xy+yx)=xy$. But since the ring has characteristic 2, $yx=xy$.

I always like to know where these problems come from, their history. This was first proved in a paper by Stone in 1936. Here’s a link to that paper for anyone who is interested:

http://dx.doi.org/10.1090/S0002-9947-1936-1501865-8

His proof is in the first full paragraph on p. 40.

Of course, this is an old chestnut: if you are interested in typical generalizations of this commutativity theorem in a wider, more structural context (to associative, unitary rings) I suggest reading T.Y. Lam’s beautiful Springer GTM 131 “A First Course in Noncommutative Rings”, Chapter 4, §12, in particular the Jacobson-Herstein Theorem (12.9), p. 209: A (unitary, associative) ring $R$ is commutative iff for any $a,b\in R$ one always has $(ab-ba)^{n+1}=ab-ba$ for some $n\in\mathbb N$ ($n$ generally depending on $a,b$). (Further, using Artin’s theorem concerning diassociativity of alternative algebrae, associativity of $R$ may be weakened to alternativity.) Cp. also the exercises given, in particular Ex. 9. Note that the Boolean case is special, as that the ring considered needn’t be unitary a-priori. Kind regards – Stephan F. Kroneck.

Plug $a = x + y$.

**HINT** $\rm\quad\ \ A = X+Y\ \ \Rightarrow\ \ X\ Y = – Y\ X\:.\ $ But $\rm -1 = 1\ $ via $\rm\ A = -1$

As Yuval points out $(x+y)^{2} = x+y$ which implies $x^{2} + y^{2} + x \cdot y + y \cdot x = x+y$. Now from this you have $x \cdot y + y \cdot x =0$.

We want to show that $xy = yx$ for all $x,y \in R$. We know that $(x+y)^2 = x+y$. So $(x+y)^2 = (x+y)(x+y) = xx+xy+yx+yy = x+xy+yx+y = x+x^2y^2+y^2x^2+y$. This equals $x+(xy)+(yx) + y = x+y$ so that $xy = yx$.

When you get to the part where ab=-ba

ab =-ba (1)

Pre-multiply a to both sides

a(ab)=a(-ba)

a^2b=a-ba

ab=a-ba (2)

Post-multiply a to (1)

(ab)a=(-ba)a

aba=-ba^2

aba=-ba (3)

.

.

.

From (2) & (3), you can deduce that ab=ba

If $a,b\in R$,

\begin{align}

2ba

&=4ba-2ba\\

&=4(ba)^2-2ba\\

&=(2ba)^2-2ba\\

&=2ba-2ba\\

&=0,

\end{align}

so

\begin{align}

ab

&=ab+0\\

&=ab+2ba\\

&=[ab+ba]+ba\\

&=[(a+b)^2-a^2-b^2]+ba\\

&=[(a+b)-a-b]+ba\\

&=0+ba\\

&=ba.

\end{align}

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