TITLE: How many ideals contain $(x^2+1)$ in $\mathbb{Z}_4[x]$?
QUESTION [5 upvotes]: How many ideals in $\mathbb{Z}_4[x]$ contain $(x^2+1)$?

I noticed that $x^2+1$ is irreducible in $\mathbb{Z}_4[x]$. But $\mathbb{Z}_4[x]$ is not a PID, so $x^2+1$ need not be maximal in $\mathbb{Z}_4[x]$. So I’m a bit stuck here. Can someone help me out? Thanks!

REPLY [5 votes]: I presume $\Bbb Z_4$ means $\Bbb Z/4\Bbb Z$.
The answer to your question is "the number of ideals of the quotient ring
$\Bbb Z_4[X]/(X^2+1)$". But
$$R=\Bbb Z_4[X]/(X^2+1)\cong \Bbb Z[X]/(4,X^2+1)\cong\Bbb Z[i]/(4).$$
The ring $\Bbb Z[i]$ is a PID, and the ideals containing $(4)$
are generated by the factors of $4$. But $(4)=((1+i)^4)$  as ideals
in $\Bbb Z[i]$,
so these ideals are $((1+i)^k)$ for $0\le k\le 4$. There are five of them.