I need to show that if $n$ is not a prime power,then $\mathbb{Z}/\mathbb{nZ}$ admits idempotents $\neq 0,1$
I noticed this thing for $\mathbb{Z_6}$ and $\mathbb{Z_{12}}$ and few more but how do we show that there will always exists such $a$?
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Mathronaut
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If $n$ is not a prime power, there are at least two distinct primes $p,q$ dividing $n$. How are the idempotents in $\mathbb{Z}6$ and $\mathbb{Z}{12}$ related to $p$ and $q$? – Daniel Fischer Sep 17 '13 at 19:51
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they're power of prime! – Mathronaut Sep 17 '13 at 19:53
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Hmm, too bad, though not entirely unrelated. Try $18$. – Daniel Fischer Sep 17 '13 at 19:54
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2Your question has been discussed in Idempotents in $Z_n$. Basically Chinese Remainder Theorem allows you to reduce to the case, where $n$ is a prime power. The total number of idempotents is $2^k$, where $k$ is the number of distinct prime factors of $n$. – Jyrki Lahtonen Sep 17 '13 at 19:54
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$6^2=6$ in $\mathbb{Z_{18}}$ and $16^{16}=16$ in $\mathbb{Z_{24}}$ – Mathronaut Sep 17 '13 at 19:57
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1No, $6^2 = 36 \equiv 0 \pmod{18}$. But $10^2 \equiv 10 \pmod{18}$. So it's not prime powers. – Daniel Fischer Sep 17 '13 at 19:58
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oops! yeah got it – Mathronaut Sep 17 '13 at 19:59
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1Neeraj, what you could notice from Daniel's example ($10$ an idempotent modulo $18$) is that $10\equiv1\pmod9$ and $10\pmod 0\pmod2$, so $10$ is an (trivial) idempotent modulo both the prime power factors of $18$. That's the way it always pans out. – Jyrki Lahtonen Sep 17 '13 at 20:03
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In the post which is already answered most of the proofs rely on CRT , can we do it without CRT? – Mathronaut Sep 18 '13 at 12:02
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@NeerajBhauryal If you can give a real reason you should "do it without" a very useful staple of ring theory, then I will attempt to do so. – rschwieb Sep 18 '13 at 17:37