Suppose that there is an integer $n >1$, such that $a^n = a$ for all elements of some ring. If $m$ is a positive integer and $a^m = 0$ for some $a$ , then I have to show that $a=0$. Please suggest.
Suppose for each $a$ there is a positive integer $n$ such that $a^n=a$. Show there are no nonzero nilpotent elements.
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ring-theory
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0@mixedmat$h$: Done. – 2011-08-29
2 Answers
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This is almost trivial if you prove the contrapositive, i.e. if $a\neq 0$ then $a^k \neq 0$ for any $k$.
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Hint. What is $a^{n^2} = (a^n)^n$? What is $a^{n^3}=(a^{n^2})^n$? What is $a^{n^k}$ for any positive integer $k$? And what is $a^{\ell}$ for any $\ell\geq m$?