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Let $R$ be a ring. An element $x$ in $R$ is said to be idempotent if $x^2=x$. For a specific $n\in{\bf Z}_+$ which is not very large, say, $n=20$, one can calculate one by one to find that there are four idempotent elements: $x=0,1,5,16$. So here is my question:

Is there a general result which tells the number of the idempotent elements of ${\bf Z}_n$?

4 Answers 4

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If $n=p_1^{m_1}\cdots p_k^{m_k}$ is the factorization of $n$ as a product of powers of distinct primes, then the ring $\mathbb Z/n\mathbb Z$ is isomorphic to the product $\mathbb Z/p_1^{m_1}\mathbb Z\times\cdots\times \mathbb Z/p_k^{m_k}\mathbb Z$. It is easy to reduce the problem of counting idempotent elements in this direct product to counting them in each factor.

Can you do that?

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    Fair enough. Now I see. Thanks!2011-06-16
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In $\Bbb{Z}_n$ the relation $x^2=x$ is equivalent to $(x-1)x\equiv 0 ( mod \ n)$, that is $n | x(x-1)$. This is an easy way to calculate all idempotent elements for small $n$. In general, you need to consider the factorization of $n$ in prime factors and note that $x,x-1$ are coprime, and if one prime number divides one of them, it can't divide the other.

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HINT $\ $ Idempotents in $\rm\:\mathbb Z/n\:$ are closely related to factorizations of $\rm\:n\:$ into coprime factors, i.e. $\rm\: n = a\:b\:,\:$ where $\rm\:gcd(a,b) = 1\:.\ $ Indeed, notice that $\rm\:p^k\ |\ e\:(e-1)\ \Rightarrow\ p^k\ |\ e\:$ or $\rm\:p^k\ |\ e-1\:.$

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    So are you saying the number of idempotent elements in Z_n is 2^(omega(n)) where omega(n) is the number of distinct prime factors of n.? Is there a way to determine which elements of Z_n are idempotent. For example, (by a brute force Mathematica code I see that ) for n=200 the idempotents are {0,1,25,176}. Is there any "formula" for finding the idempotents in Z_n?2015-07-06
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$n\geq 1$ be an integer and consider the ring $R=\mathbb{Z}/\mathbb{Z}_{n}$. First of all, we can see that the number of all the idempotents elements of $R$ is $2^m$, where $m$ is the number of distinct prime divisor of $n$.

$\textbf{Proof:} $Let $n=\prod\limits_{i=1}^{n} p_{i}n^{i}$ be the prime factorization of $n$ and $R_{i}=\mathbb{Z}/p_{i}^{n_{i}}$. By the Chinese Remainder theorem we have $R\cong R_{1}$ x $\cdots$ x $R_{m}$, say $(1)$.

$\textbf{Claim:} $ If $p$ is a prime and $l>0$ is an integer, then the only idempotent element of $\mathbb{Z}/p^{l}\mathbb{Z}$ are $0$ and $1$.

$\textbf{Proof:} $So we wnat to show that modulo $p^{l}$, the equation $x^{2}\equiv x$ (mod $p^{l})$ has only two trivial solutions $x=0,1$. Suppose that $x\neq 0$ is a solution of $x^{2}\equiv x$ (mod $p^{l})$. We will show that $x\equiv 1$ (mod $p^{l})$.

Let $x=p^{r}s$, where $0\leq r<1$, and gcd($s, p)=1$. Then $s(p^{r}s-1)\equiv 0$ (mod $p^{l-r})$ which give us $p^{r}s\equiv 1$ (mod $p^{l-r})$. Thus $r=0$ and hence $x\equiv s\equiv 1$ (mod $p^{l})$. It is clear now from (1) and claim that the number of idempotents of ring $R=\mathbb{Z}/\mathbb{Z}_{n}$ is $2^{m}$.

Now we can solve any question for $\mathbb{Z}/\mathbb{Z}_{n}$.

By the above $R=\mathbb{Z}/\mathbb{Z}_{20}$, we know that $R$ has $4$ idempotents, it is clear two of them being $0, 1$ (mod $20$). Let $R_{1}=\mathbb{Z}/\mathbb{Z}_{4}$, $R_{2}=\mathbb{Z}/\mathbb{Z}_{5}$. Then $R\cong R_{1}$ x $R_{2}$. All idempotents of $R_{1}$ x $R_{2}$ are $(0, 0)$, $(1, 0)$, $(0, 1)$, $(1, 1)$.

So we just need to find preimage of each idempotent in $R$. Obviously, the preimages of $(0, 0)$ and $(1, 1)$ are $0$ and $1$ (mod $20$) respectively. Now let let's find the preimage of, say $b=(0, 1)$. Let $a=m+20\mathbb{Z}$ be the preimage of $b$. Then the image of $a$ is $(m+4\mathbb{Z}, m+5\mathbb{Z})=b=(4\mathbb{Z}, 1+5\mathbb{Z})$. So $m$ divisible by $4$ and it is equiavlent to $1$ (mod $5$). It follows that $a=16+20\mathbb{Z}$. Similarly, we can find another idempotent 5 which is preimage of $(1,0)$.