Q:

prime ideals of $\mathbb{Z} / 36\mathbb{Z}$

Could I have a good explanation about why the prime ideals of $\mathbb{Z}/36\mathbb{Z}$ are $2\ \mathbb{Z}$ and $3\ \mathbb{Z}$?

A:

The descriptions of ideals in quotients tells you that the ideals of $\mathbb{Z}/36\mathbb{Z}$ are of the form $k\mathbb{Z}/36\mathbb{Z}$, where $k$ divides $36$.
Now an ideal $I$ of $R$ is prime iff the quotient ring $R/I$ is an integral domain, and by the third isomorphism theorem you have :
$$\frac{\mathbb{Z}/36\mathbb{Z}}{k\mathbb{Z}/36\mathbb{Z}}\simeq \mathbb{Z}/k\mathbb{Z}$$
of course $\mathbb{Z}/k\mathbb{Z}$ is an integral domain iff k is prime (check) thus $k$ is a prime dividing $36$, i.e. $2$ or $3$.

