1
$\begingroup$

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?

Thanks for your help.

EDIT: I forgot to see (0, 0) = (0, 2) for the nilpotent elements.

| cite | improve this question | | | | |
$\endgroup$
2
$\begingroup$

Yes, you are right and this is quite easy to prove since we have:

$$(a, b)^n=(a^n, b^n)$$

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.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.