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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?

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    $\begingroup$ There's a proof of this in the first chapter of Halmos' Lectures on Boolean Algebras. $\endgroup$ Commented Sep 8, 2011 at 12:44
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    $\begingroup$ This is exercise 15 from chapter 7 Introduction to Rings section 1 Definitions and Examples in Dummit and Foote, 3rd edition. $\endgroup$ Commented Sep 19, 2014 at 10:25
  • $\begingroup$ It's also exercise 12.50 in Gallian's Contemporary Abstract Algebra. $\endgroup$
    – Alex D
    Commented Feb 15, 2022 at 6:30

14 Answers 14

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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, $yx=xy$.

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  • $\begingroup$ How do you go from xx+xy+yx+yy to x+xy+yx+y? $\endgroup$
    – NUG
    Commented Feb 22, 2017 at 18:58
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    $\begingroup$ @NUG Since every element in $R$ is idempotent. $\endgroup$
    – Bach
    Commented Jun 10, 2019 at 10:46
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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.

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  • $\begingroup$ Can someone provide a valid link or just copy the main idea? I failed to load the paper but I am eager to take a look at it. Thanks. $\endgroup$
    – Bach
    Commented Jun 10, 2019 at 10:41
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    $\begingroup$ @Bach The document loads for me without problem, but anyways: The idea is very similar to the other answer. First show $ba+ab=0$, use this to show $a=-a$ (with $a=b$ in the first equation), then conclude $ba=-(ab)=ab$. $\endgroup$
    – Lilalas
    Commented Feb 19, 2020 at 8:03
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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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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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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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  • $\begingroup$ How do you go from xx+xy+yx+yy to x+xy+yx+y? $\endgroup$
    – NUG
    Commented Feb 22, 2017 at 18:56
  • $\begingroup$ $xx+xy+yx+yy = x^2 +xy+yx+y^2 = x+xy+yx+y$ because it was given that $a^2 =a$ for all $a \in R$. $\endgroup$ Commented Feb 28, 2018 at 18:56
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    $\begingroup$ Hello, how does $xy+yx$ imply $xy=yx$? Doesn't $xy=yx$ imply $xy-yx=0$, and so how is $xy+yx=0$? $\endgroup$
    – user482939
    Commented Apr 3, 2018 at 22:49
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    $\begingroup$ @numericalorange because the Boolean ring is of characteristic 2, and so 1+1=0 implies that 1 = -1 :) $\endgroup$ Commented Apr 20, 2019 at 3:42
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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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If $x,y\in R$ then $xy=-yx$. Hence,

$xy=(-yx)^2=(-yx)(-yx)=(yx)(yx)=(yx)$.

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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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  • $\begingroup$ I don't work out the rest because, Timothy has already done it before me. $\endgroup$
    – anonymous
    Commented Nov 14, 2010 at 18:46
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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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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.

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First a small claim: If $R$ is Boolean, then $-a=a$ for all $a \in R$.

$\textbf{Proof of claim:}$ We know

$$-a=(-a)^2=(-1)^2a^2=a \Rightarrow -a=a.$$

And we are done with the proof of the claim. Next let $a,b \in R$, we seek to show $ab=ba$. But

\begin{align} (a+b)&=(a+b)^2\\ &=a^2+ab+ba+b^2\\ &=a+ab+ba+b \end{align} This implies $ab+ba=0$, replace $ab$ with $-ab$ and we conclude that

$$ba=ab$$

and we are done.

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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.

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

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