Idempotent elements of $Z_4 \times Z_8$

Question

I am trying to find idempotent elements of the direct product $Z_4 \times Z_8$. By the definition of direct product we have: $$Z_4 \times Z_8=\left\{(0,0),(0,1),.....(2,7),(3,7)\right\}$$ Now by the definition of idempotent element we have: $$(a,b)*(a,b)=(a,b)$$ $\implies$ $$(a^2,b^2)=(a,b)$$ $\implies$ $a=0,1$ and $b=0,1$ So the idempotent elements are $$(0,0),(0,1),(1,0),(1,1)$$ Is this correct? Can we say every $Z_m \times Z_n$, these will be the only idempotent elements?

Answer 1

As you showed the idempotent elements of $\mathbb Z_4\times \mathbb Z_8$ are those of the form $(e_1,e_2)$ where $e_1$ is an idempotent element of $\mathbb Z_4$ and $e_2$ is an idempotent element of $\mathbb Z_8$. (this part works for any product of rings $R\times S$).

I add a proof that the idempotent elements of $\mathbb Z_{p^k}$ are just $0$ and $1$:

Suppose $p^k|(a^2-a) = a(a-1)$. Because $a$ and $a-1$ are coprime $p$ can only divide $a$ or $a-1$ which means $p^k$ divides $a$ or $p^k$ divides $a-1$.

However if $n$ is not a prime power there are more idempotent elements of $\mathbb Z_n$. Let $n=ab$ with $a,b>1$ and $a$ and $b$ coprime. Take $x$ such that $x\equiv 0\bmod a$ and $x\equiv 1\bmod b$ to get an idempotent element that is not $0$ or $1$. It is not hard to show $\mathbb Z_n$ has $2^s$ idempotent elements, where $s$ is the number of distinct prime factors of $n$.