## Boolean Rings

Posted: April 13, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Problem.  Let $n \geq 1$ be an integer and $R$ a ring such that $2x=0, \ x^{2^n + 1}=x,$ for all $x \in R.$ Prove that $x^2=x,$ for all $x \in R.$

Solution. First note that $2 \mid \binom{2^n}{k},$ for all $0 < k < 2^n,$ and therefore, since $2x=0$ for all $x \in R,$ we will have $(a+b)^{2^n}=a^{2^n}+b^{2^n},$ for all $a,b \in R.$ Also $x^{2^{n+1}+1}=x^{2^n}x^{2^n+1}=x^{2^n}x=x^{2^n+1}=x$ and as a result $x^{2^{n+1}+2}=x^2.$ Therefore for all $x \in R$ we have

$x^2-x=x^2+x=(x^2+x)^{2^n+1}=(x^2+x)^{2^n}(x^2+x)$

$=(x^{2^{n+1}} + x^{2^n})(x^2+x) = x^{2^{n+1}+2} + x^{2^{n+1}+1}+x^{2^n+2} + x^{2^n+1}$

$=x^2+x+x^2+x=2(x^2+x)=0. \ \Box$