Nonzero nilpotents of $\mathbb{Z}_{210}$?

Question

An exercise asks me to write two nilpotents in $\mathbb{Z}_{210}$. I honestly doubt that there is any nonzero nilpotent in $\mathbb{Z}_{210}$, since $210=2\cdot 3 \cdot 5\cdot 7$ and by CRT there should be no nonzero nilpotent element. Am I missing anything?

Answer 1

You're quite right.

In general, the nilpotent elements of $\mathbb{Z}_{n}$ correspond to the multiples of $rad(n)$. In particular, the only nilpotent element is $0$ when $n$ is square- free.

Answer 2

If $a$ is nilpotent that means $a^k$ is a multiple of $3\cdot2\cdot5\cdot7$ for some $k$ which means each of those primes must divide $a$ which means $a$ is zero.