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

Question

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!

Answer 1

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.

Answer 2

I don't know if I understand your question well. Notice $$x^2+1=(x-1)(x+1)+2=(x+3)(x+1)+3\pmod{4}$$ which means $$(x^2+1)\subset (x-1,2)\quad\text{and}\quad (x^2+1)\subset (x+1,2).$$ We may decompose like $$x^2+1=(3x+3)(3x+1)+2\pmod{4}$$ but nothing new comes out.

We may also notice that $$(x+1)^4=x^4+4x^3+6x^2+4x+1=x^4+2x^2+1=(x^2+1)^2\pmod{4}$$ or $((x+1)^4)\subset(x^2+1)$.