# Prove Z_m*n has at least 2 idempotent elements besides 0 and 1. [duplicate]

Here m, n are relatively prime and greater than 1. Z_mn is the ring of nonnegative integers less than mn under modulo m*n addition and multiplication. An idempotent element a is a ring element with the property that a^2 = a. I made a list of idempotent elements for some small integers. I found A215202 on Sloane's OEIS. It gives an even stronger statement about the number of idempotent elements (including 0 and 1 there are 2^omega(n) where omega(n) is the number of distinct primes of n). I read some of the Wikipedia article on idempotent elements. I also read some of the stack exchange proof that Z_p has exactly 2 idempotent elements (this seems easy to understand since a^2 = a implies a^n = a for all positive n).

By the Chinese remainder theorem (since $m$ and $n$ are coprime), $\Bbb Z_{mn} \cong \Bbb Z_m\times \Bbb Z_n$. In the latter, $0$ is represented as $(0,0)$ and $1$ by $(1, 1)$. However, the elements $(1, 0)$ and $(0, 1)$ are also idempotents.
For instance, in $\Bbb Z_2\times \Bbb Z_3 \cong \Bbb Z_6$, the element $(1, 0)$ corresponds to $3$ and the element $(0, 1)$ becomes $4$. In $\Bbb Z_{21}$ we have $15$ and $7$. And so on.
Note that if $\Bbb Z_m$ or $\Bbb Z_n$ themselves have other idempotents, say $a \in \Bbb Z_m$ is idempotent, then $(a,0)$ and $(a,1)$ are also idempotents in $\Bbb Z_m\times \Bbb Z_n$. Specifically, we can immediately conclude that the number of idempotents in $\Bbb Z_{mn}$ is at least $2^j$ where $j$ is the number of different primes in the prime factorization of $mn$ (and this is in fact an equality).