# 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?

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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. –  ndroock1 Sep 19 '14 at 10:25

Every Boolean ring is of characteristic 2, since $a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a$ Now, for any $x,y$ in the ring $x+y=(x+y)^2=x^2+xy+yx+y^2=x+y+xy+yx$ So that $xy+yx=0$ and hence $xy=-yx$. But since the ring has characteristic 2, $yx=-yx=xy$

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I know this is a 3 year old answer but the proof is incomplete. The last part, -yx = xy, should read -yx = yx (because 1 = -1). –  Wug Mar 21 '13 at 23:01
@Wug I don't understand this critique –  Cocopuffs Jul 4 '13 at 19:01
@ Wug $xy = - yx,$ but $-1 = 1.$ So $xy = yx$ –  Rising Star Mar 31 at 2:46
@Wug It shouldn't, it has already been proved that $xy = -yx$ in the very same line... –  LMartin May 19 at 22:55
I think what I meant is that it stops short of the real solution because it isn't true for all rings that xy = yx (it's true in this case, but proof does not make any mention of this). It's also possible that I just misread the last line, such things happen if proofs don't have any line breaks. Who knows, it was 3 years ago. –  Wug May 20 at 10:53

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:

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

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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.

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Plug $a = x + y$.

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

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

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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 $xy = yx$.

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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

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