How to show that every Boolean ring is commutative? 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?
 A: 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}
A: 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, $yx=xy$.
A: 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 $xy = yx$.
A: Plug $a = x + y$.
A: HINT $\rm\quad\ \ A = X+Y\ \ \Rightarrow\ \ X\ Y = - Y\ X\:.\ $ But $\rm -1 = 1\ $ via $\rm\ A = -1$
A: 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.
A: If $x,y\in R$ then $xy=-yx$. Hence,
$xy=(-yx)^2=(-yx)(-yx)=(yx)(yx)=(yx)$.
A: 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.
A: 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 $x \cdot y + y \cdot x =0$.
A: 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
A: I have solved this question myself and wonder what are the solutions on the Internet, and yet since I didn't see any solution like mine (most probably because my solution didn't notice that the ring is of characteristic $2$). Anyway, I will put it here. 
For any $x,y$ in the ring let $a=xy-yx$, then $$xax=x^2yx-xyx^2=xyx-xyx=0$$
and $$xay=x^2y^2-xyxy=xy-(xy)^2=0$$
hence $$0=(xax)y-(xay)x=xa(xy-yx)=xa^2=xa.$$
Seeking for symmetry we find that $y(-a)=0$, and so $$ya=ya+y(-a)=y(a+(-a))=y0=0$$
hence $$0=x(ya)-y(xa)=(xy-yx)a=a^2=a.$$
This completes the proof. 
A: For any $a, b \in R$, we have
$$(a+b)^2 = a+b, \text{ since R is Boolean}$$
But,
$$(a+b)^2 = a^2 + b^2 + ab + ba = a + b + ab+ ba$$
Hence,
$$a+b = a+b+ab+ba $$
which means,
$$ab = - ba  \qquad \qquad \qquad  (1) $$
For any $c, d \in R$, we have
$$(c-d)^2 = c-d, \text{ since R is Boolean}$$
But,
$$(c-d)^2 = c^2 + d^2 - cd - dc = c + d - cd + cd = c+d$$
( The last equality follows from eq. (1) )
Hence,
$$c+d = c-d$$
Which means,
$$d = -d$$
Thus the eq. (1), $\;$ $ab=-ba$ is same as $ab=ba$.
Hence $R$ is commutative.
A: I want to provide a one-liner:
$$
\begin{aligned}
ab - ba &= (ab - ba)(ab - ba)(ab - ba) \\
&= (abab - abba - baab + baba)(ab - ba) \\
&= (ab - aba - bab + ba)(ab - ba) 
\\&= abab - abba - abaab + ababa - babab + babba + baab - baba
\\&= ab - aba - ab + aba - bab + ba + bab - ba
\\&= 0.
\end{aligned}$$
