How to show that every Boolean ring is commutative?

Question

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?

Answer 1

Plug $a = x + y$.

Answer 2

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, $yx=xy$.

Answer 3

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 $x \cdot y + y \cdot x =0$.

Answer 4

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 $xy = yx$.

Answer 5

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

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

I want to provide a one-liner: $$ \begin{aligned} ab - ba &= (ab - ba)(ab - ba)(ab - ba) \\ &= (abab - abba - baab + baba)(ab - ba) \\ &= (ab - aba - bab + ba)(ab - ba) \\&= abab - abba - abaab + ababa - babab + babba + baab - baba \\&= ab - aba - ab + aba - bab + ba + bab - ba \\&= 0. \end{aligned}$$