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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?2011-08-16
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    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 $\ell\geq m$?2011-08-16
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    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.2011-08-16
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    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.2011-08-16
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    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 $a^k=a$.2011-08-16
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    Well, it is actually possible to proceed a different way and get a valid solution. Take $m$ minimal subject to $a^{m} = 0$ (and being a positive integer). Then $m \leq n$ forces $m = 1$ more or less as Tavleen says. But if $m > n$, write $m = n + (m-n)$.......2011-08-16
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    @Arturo: I was looking through unanswereds and saw this question. Perhaps you could change your comment to an answer?2011-08-29
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    @mixedmath: Done.2011-08-29

2 Answers 2

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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 $\ell\geq m$?