|
Let $R$ be a ring. An element $x$ in $R$ is said to be idempotent if $x^2=x$. For a specif $n\in{\bf Z}_+$, 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:
|
|||
|
|
|
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? |
|||||||||||||||
|
|
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. |
|||
|
|
|
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 $\rm\:p^k\ |\ e-1\:.$ |
|||
|
|
