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

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