# Nilpotent and Idempotent Elements of Ring $\mathbb{Z}_6 \times \mathbb{Z}_2$

I'd just like to make sure I understand nilpotent and idempotent elements of rings properly.

A nilpotent element is an element $b \in R$ s.t. for some $n$, $b^n = 0$. An idempotent element is an element $b \in R$ s.t. $b^2 = b$.

So, for the ring $\mathbb{Z}_6 \times \mathbb{Z}_2$, its nilpotent and idempotent elements would be the pairs of nilpotent elements of $\mathbb{Z}_6$ and $\mathbb{Z}_2$, and similarly for idempotent elements, right?

From what I see, the nilpotent elements of $\mathbb{Z}_6$ are only $0$, and the nilpotent elements of $\mathbb{Z}_2$ are $0$ and $2$. So, does this means the nilpotent elements of $\mathbb{Z}_6$ $\times$ $\mathbb{Z}_2$ are $(0, 0)$ and $(0, 2)$?

Also, from my calculations, the idempotent elements of $\mathbb{Z}_6$ are $0, 1, 3, 4$, while the idempotent elements of $\mathbb{Z}_2$ are $0$ and $1$. Similar to nilpotent elements, does this mean the idempotent elements of $\mathbb{Z}_6 \times \mathbb{Z}_2$ are the combinations of those elements?

$$(a, b)^n=(a^n, b^n)$$
Thus, $(a, b)^n=(0, 0)$ if and only if $a^n=0$ and $b^n=0$, so only pairs of nilpotent elements are nilpotent. Also, if we have two nilpotent elements $a$ and $b$ such that $a^m=0$ and $b^n=0$, then $(a, b)^{\max(m, n)}=0$ and thus all pairs of nilpotent elements are nilpotent.
Also, $(a, b)^2=(a, b)$ if and only if $a^2=a$ and $b^2=b$, so only and all pairs of idempotent elements are idempotent.