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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10Please 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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12HINT: 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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0Suppose 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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0Read 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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0The 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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0Well, 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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0@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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0@mixedmath: 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 $\ell\geq m$?