## Examples of nilpotent elements

An element , a ring, is called *nilpotent* if for some positive integer .

- Show that if for some integers , then is nilpotent in .
- If is an integer, show that the element is nilpotent if and only if every prime divisor of also divides . Find the nilpotent elements of explicitly.
- Let be a nonempty set and a field, and let be the ring of functions . Prove that contains no nonzero nilpotent elements.

- Suppose , where . Now mod .
- Suppose is nilpotent. Then for some and . Now if is a prime dividing , then divides , so that it divides . Thus every prime dividing divides . Let and , where for all and is some integer. Let . Then , where for each . Thus for some integer , and we have mod .
The nilpotent elements of , where , are as follows: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66.

- Suppose is nilpotent. If , then there exists such that . Let be minimal such that ; note that . Then , where are not zero. Thus contains zero divisors, a contradiction. Thus no nonzero element of is nilpotent.

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## Comments

Hello, don’t you mean $1 \leq e_{i} \leq d_{i}$ so that every prime divisor of $n$ divides $a$ ?

Yes. Thanks!

I think it’s not necessarily true that $1 \leq e_i \leq d_i$. For example, 6 is nilpotent in $Z/12Z$. However, I think the rest of the argument works without this assumption.

Also a minor typo, I think you meant $a^t = (p_1^{d_1} \cdots p_k^{d_k}m)^t

Thanks!