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**

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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