Question

Find the nil radical of `ZZ_35`?

Answer 1

`{ 0 }`

Explanation:

The nilradical of a ring is the set of its nilpotent elements.

A nilpotent element of a ring is an element such `x` that `x^n = 0` for some integer `n`.

In the case of `ZZ_35`, the nilpotent elements are essentially numbers in the range `0` to `34` which have a power divisible by `35`.

Note that `35 = 5 * 7` has no repeated factor. So the only way that `x^n -= 0` modulo `35` is for `x` to be divisible by both `5` and `7` itself and therefore `x -= 0` (mod `35`).

So the nilradical of `ZZ_35` is `{ 0 }`.

Footnote

The nilradical of `ZZ_36` is somewhat more interesting.

Since `36 = 2^2 * 3^2` any element that is divisible by `6 = 2 * 3` is nilpotent.

So the nilradical of `ZZ_36` is `{ 0, 6, 12, 18, 24, 30 }`