I came across the following Theorem on a paper I'm reading:
"If $R$ is a ring and $x^n=x$ for all $x\in R$ then $R$ is commutative"
However I have no idea of how to prove it. May I get some hints?
6
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10$\begingroup$ For $n > 1$ this is an old theorem of Jacobson. See mathoverflow.net/questions/207757/…. $\endgroup$– KConradCommented Jun 18, 2021 at 2:06
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5$\begingroup$ And for $n=1$ it's false. $\endgroup$– Gerry MyersonCommented Jun 18, 2021 at 3:42
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2$\begingroup$ And for $n=0$ it’s trivial. $\endgroup$– Emil JeřábekCommented Jun 18, 2021 at 7:05
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6$\begingroup$ To be pedantic, the symbol $n$ in this post is undefined. In fact the correct statement of Jacobson's Theorem is "if $R$ is a ring, and for all $x \in R$ there exists an integer $n>1$ with $x^n=x$, then $R$ is commutative". It is difficult to prove and I don't advise attempting it. Try proving that $\forall x \in X\, x^2=x \Rightarrow R$ commutative, and if you find that easy, try proving $\forall x \in X\, x^3=x \Rightarrow R$ commutative, which is a starred exercise in Herstein's Topics in Algebra. $\endgroup$– Derek HoltCommented Jun 18, 2021 at 7:33
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3$\begingroup$ Notice this contains as a special case Wedderburn's theorem that a finite division ring is a field so you shouldn't expect it to be trivial $\endgroup$– Benjamin SteinbergCommented Jun 18, 2021 at 12:12
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