Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
1  
There's a proof of this in the first chapter of Halmos' Lectures on Boolean Algebras. – Michael Hardy Sep 8 '11 at 12:44
    
This is exercise 15 from chapter 7 Introduction to Rings section 1 Definitions and Examples in Dummit and Foote, 3rd edition. – nilo de roock Sep 19 '14 at 10:25

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, $yx=xy$.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Plug $a = x + y$.

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer
    
I don't work out the rest because, Timothy has already done it before me. – anonymous Nov 14 '10 at 18:46

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 $xy = yx$.

share|cite|improve this answer

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

share|cite|improve this answer

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}

share|cite|improve this answer

protected by user26857 Dec 5 '15 at 10:38

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?