Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer

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 $\ell\geq m$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.