How to calculate the amount of non-trivial ideals in $\mathbb{Z}/ 60 \mathbb{Z}$?
When we are talking about fields it's always 2, but I have no idea from which side I should start tackle this problem. I think it somehow related to number of divisors of 60.
Ideals are in particular additive subgroups. The additive subgroups of a cyclic subgroup of order $n$ are all cyclic and they are in bijection with the divisors of $n$.
In this case they are the subgroups $\{\overline{kd} | k\in \mathbb Z\}$, where $d$ is a fixed divisor of $60$. It is easy to see they also absorb multiplication, so they are ideals.