In the ring of integers modulo $n$, $\mathbb{Z}_n$, we have all the ideals of $\mathbb{Z}_n$ are principal and they are $\langle d\rangle$, where $d$ is a divisor of $n$ and prime ideals are $\langle p\rangle$ where $p$ is a prime divisor of $n$.
What about the ring of Gaussian integers modulo $n$, $\mathbb{Z}_n[i]$, In this ring, whether all the ideals are principal and which are the ideals of $\mathbb{Z}_n[i]$ and which of them are principal?
Since these rings are not integral domains in general, I cannot use results about PID or UFD.
Best Answer
Your ring $\mathbb{Z}_n[i]$ can be seen as the quotient $$\Bbb Z_n [i] \cong \Bbb Z[i] / (n).$$ One of the isomorphism theorems, which can be found on every algebra book, says that the ideals of a quotient $R/A$ are in bijection with the ideals of $R$ that contain $A$, and the bijection is given by $$B \mapsto B/A = \{b+A : b \in B\}.$$Because $\Bbb Z[i]$ is a PID, every ideal that contains $(n)$ is of the form $(m)$, where $m \mid n$. If $$n=p_1^{k_1} \ldots p_r^{k_r}$$is the decomposition of $n$ as product of primes (in $\Bbb Z [i]$), then any ideal containing $(n)$ is of the form $(p_1^{s_1} \ldots p_r^{s_r})$, where $s_j \leq k_j$ and, in particular, there are $(k_1+1) \ldots (k_r+1)$ different ideals, which, in turn, correspond to different ideals of $\mathbb{Z}_n[i]$.\ Note that there is nothing particular of $\Bbb Z [i]$ apart from being a PID that we are using here. The same method proves how to find all the ideals in the quotient of a PID by one of its ideals, as you would do with $\Bbb Z _n$, but now taking into account that you need to factorise in $\Bbb Z[i]$ so, for example, $5$ is not prime because $5=(2+i)(2-i)$.
As it has been said in the comments, every ideal of $\mathbb{Z}_n[i]$ is principal. This is a general result. If you look back to the correspondence of ideals in a quotient, if $B$ is generated by $b$, then $B/A$ is generated by $b+A$.