# Suppose for each $a$ there is a positive integer $n$ such that $a^n=a$. Show there are no nonzero nilpotent elements.

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.

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Please don't post questions in the imperative: you aren't assigning homework, after all. You have a question, how about asking a question instead of telling people what to do? What have you tried, or where are you stuck? –  Arturo Magidin Aug 16 '11 at 17:05
HINT: What is $a^{n^2} = (a^n)^n$? What is $(a^{n^2})^n = a^{n^3}$? What is $a^{n^k}$ for any positive integer $k$? And what is $a^{\ell}$ for any $\ell\geq m$? –  Arturo Magidin Aug 16 '11 at 17:07
Suppose m = n. Then a^n = a^m = 0. So a = 0 Now suppose m < n. Then n = m + k , k>0. Then a ^ n = a ^ (m+k) = a^m*a^k =0*a^k = 0. So a = 0. I am stuck in the case that when m >n. –  Tav Aug 16 '11 at 19:09
Read my hint again. There's no need to divide into different cases. Hint${}^2$: since $n\gt 1$, $\lim_{k\to\infty}n^k = \infty$. And please edit your question to get rid of the imperative order. –  Arturo Magidin Aug 16 '11 at 19:24
The case $m>n$ can be handled with Arturo’s hint and what you’ve already done. Use his hint to find a $k>m$ such that $a^k=a$. –  Brian M. Scott Aug 16 '11 at 19:37

This is almost trivial if you prove the contrapositive, i.e. if $a\neq 0$ then $a^k \neq 0$ for any $k$.
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 $\ell\geq m$?