1

Suppose $R$ is a ring with identity and $x^n = x, \,\, \forall x\in R$ for some fixed integer $n \geq 2$. Prove that $x^{n-1}$ commute with any elements of $R$.

In the case $n = 2$, I can show that $x+x = 0$ and $x+y = (x+y)^2 = x + xy +yx + y,$ hence $xy + yx = 0,$ hence $xy = yx.$

Krish
  • 7,102
Rikka
  • 890
  • 1
    Note: the question is not the same, but one of the answers is comprehensive and includes the question asked here. – vadim123 Nov 05 '17 at 01:59
  • Would you please give a more elementary answer, it seems hard for me to follow the answer there. – Rikka Nov 05 '17 at 02:08
  • 1
    @Rikka The proofs there are actually rather elementary. I assume what's causing you difficulty is largely the terminology of "nilradical" and possibly "center". Lemma 1 is basically "if $x^n=x$ and $x^m = 0$ then $x = 0$". Lemma 2 is then "assuming for all $x\in R$, $x^m=0$ implies $x=0$, then $a^2=a$ implies $a$ commutes with everything". Finally, Lemma 3 is exactly your question. The actual proofs given of these lemmas are largely algebraic manipulation. If you are (still) having trouble following them, you should be more specific about what you find confusing. – Derek Elkins left SE Nov 05 '17 at 04:40

0 Answers0