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Find the number of idempotents in $\Bbb Z_{p^n}$.

Let $x$ be an idempotent in $\Bbb Z_{p^n}$.

Then $x^2=x\mod p^n\implies p^n \mid x^2-x=x(x-1)$

Since $p\mid p^n\implies p\mid x(x-1)\implies p\mid x$ or $p\mid x-1\implies x\equiv 0\mod p$ or $x\equiv 1\mod p$

But I need the relations in $\Bbb Z_{p^n}$.

How to do it.I need some help.Will you please?

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    We can complete your elementary proof very easily - see my answer. – 2017-01-17

4 Answers 4

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$p^n\mid (1\!-\!x)x\,\Rightarrow\,p^n\mid 1\!-\!x\,$ or $\,p^n\mid x,\ $ by $\,1\!-\!x,\,x\,$ $\rm\color{#c00}{coprime}$ $ $ (by $\ 1\!-\!x\, +\,x\, =\, 1)$

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    Question: If $p$ is prime and $p\mid ab\implies p\mid a or p\mid b$ here $p^n$ is not prime then how do you get this? – 2017-01-18
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    @Ben It's obvious by considering the (unique) prime factorization of $\,ab,\,$ which must contain at least $n$ factors of $p$, which *all* must occur in only one of $\,a\,$ or $\,b,\,$ since they $\rm\color{#c00}{\text{have no common factors}}$ – 2017-01-18
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    okay thank you very much – 2017-01-18
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Since $p$ must divide $x$ or $x-1$ but not both, then so does $p^n$.

EDIT: If $p^n|x(x-1)$ then $p\mid x$ or $p\mid x-1$. Let $r$ be the exponent of $p$ in $x$; that is, $r$ is the only natural such that $p^r\mid x$ but $p^{r+1}\not\mid x$ ($r$ can be $0$). Then $p^{n-r}\mid x-1$. Since $p$ can not divide both $x$ and $x-1$, then $r$ or $n-r$ is $0$. Thus $r$ or $n-r$ is $n$.

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    Sorry ,I don't get that ; It's clear that $p$ does not divide both .But if we suppose that $p|x$ then how do you claim that $p^n\mid x$ – 2017-01-14
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    $2\mid 6$ but $2^m\nmid 6$ for any $m$ – 2017-01-14
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There are no nontrivial idempotents in $\mathbb{Z}/p^n\mathbb{Z}$, because the ring has a unique maximal ideal, namely $p\mathbb{Z}/p^n\mathbb{Z}$.

If $I$ is the unique maximal ideal of a commutative ring $R$ and $e$ is an idempotent, then $1=e+(1-e)$ implies that either $e$ or $1-e$ is invertible, because the sum of two noninvertible elements belongs to $I$ (prove it). The only invertible idempotent is $1$ (prove it).

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    Let $e\neq 0,1$ then $Re,R(1-e)$ are both proper non-trivial ideals of $R$ and hence both are contained in a maximal ideal which is $I$ here ;Then $e+(1-e)=1\in I\implies I=R$ false – 2017-01-15
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    Really a good approach – 2017-01-15
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    @BenStokes Yes, if they're both noninvertible, the ideals they generate are proper, so contained in the unique maximal ideal. – 2017-01-15
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A local ring cannot have non-trivial idempotents.

Indeed, if $\mathfrak m$ is maximal ideal of $R$ and $e$ is a idempotent in $R$, either $e$ is a unit, and $e^2=e$ implies $e=1$, or $e$ is not a unit, so it belongs to the unique maximal ideal and $1-e$ doesn't, hence is a unit, so that $1-e=1$ by the previous case, i.e. $e=0$.

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    Please expand so the OP is benefited by this. – 2017-01-14
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    This is basically egreg's answer – 2017-01-14
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    @Xam I saw this one after posting mine, but I decided to keep it because I feel that the OP is not able to understand this generality. – 2017-01-14
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    @egreg: You (and Xam) are probably right… I've added a few details, slightly different from yours, as I think it's not bad to see different ways to explain the same results. – 2017-01-14