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Boolean ring $R$ is a ring where $x^2 =x$ $\forall$ $x \in R$. I want to show that a Boolean ring is commutative. I have done the following.

$$x + x = ( x + x )^2 = x^2 + x^2 + x^2 + x^2 = x + x + x + x \Rightarrow 2x =0 $$

So any element of $R$ is of order $2$. Thus for any two elements $x$ and $y$ in $R$ we shall get

$$x + y = ( x + y )^2 = x^2 + xy + yx + y^2 = x + xy + yx + y \Rightarrow xy + yx = 0$$

Thus

$$xy = xy + 0 = xy + 2yx = xy + yx + yx = yx$$

I have no problem in the above proof. I want to prove it without using the result that any element of $R$ is of order $2$. I have tried the following.

$$(xy)^2 = xy = x^2 y^2 \Rightarrow x ( xy - yx ) y = 0$$

Can we say from above that $xy - yx = 0$ i.e. $xy = yx$ ? Can we say that $x$ and $y$ have inverse element in the ring $R$?

Supriyo
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  • Why do you not want to use that $2x=0$? – Alex Youcis Oct 04 '13 at 05:25
  • I am trying to follow the book Abstract Algebra by Dummit & Foote. It is asked to prove that Boolean Ring is commutative. In a latter exercise they have asked to show $2x = 0$. So I was thinking that there is a proof of $xy = yx$ without using $2x = 0$ – Supriyo Oct 04 '13 at 05:31

1 Answers1

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From $xy+yx=0$ :

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

Ragib Zaman
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