TITLE: Maximal ideals of $\mathbb{Z}[x]$ containing $30$ and $x^2 + 1$.
QUESTION [6 upvotes]: I want to find the maximal ideals of the ring $\mathbb{Z}[x]$ containing $30$ and $x^2+1$.
Any such ideal will contain the ideal $(30, x^2+1)$, so we are searching for maximal ideals in the ring
$$\mathbb{Z}[x] / (30,x^2+1) \cong \mathbb{Z}_{30}[x] /(x^2+1) \cong \mathbb{Z}_5[x] /(x^2+1) \oplus\mathbb{Z}_3[x] /(x^2+1) \oplus \mathbb{Z}_2[x] /(x^2+1)$$
Now, we search for the ideals in the summands. Factorizing, $x^2 + 1 = (x+3)(x+2) \bmod 5$, $x^2 + 1$ is irreducible $\bmod 3$, and $x^2 + 1 = (x+1)^2 \bmod 2$.
From this, we see that $\mathbb{Z}_3[x] /(x^2+1)$ is a field - there are no nonzero ideals. Moreover, we get ideals corresponding to the factored components in the other summands.
How do I find the maximal ideals in $\mathbb{Z}_{30}[x] /(x^2+1)$ from here? Furthermore, how do I find the generators of the corresponding ideals in $\mathbb{Z}_{30}$?

REPLY [2 votes]: Using the $\mathbb Z$-algebra isomorphism $\mathbb Z[x]/\langle x^2+1\rangle\cong \mathbb Z[i]$, you can conclude that $\mathbb Z[x]/\langle x^2+1,30\rangle \cong \mathbb Z[i]/\langle 30\rangle $. Now $\mathbb Z[i]$ is a PID. So the maximal ideals of $\mathbb Z[i]/\langle 30\rangle $ correspond to prime factors of $30$ in the UFD $\mathbb Z[i]$. We have the prime factorization $30=(1+i).(1-i).3.(2+i).(2-i)$. So the required maximal ideals in the ring $ \mathbb Z[i]/\langle 30\rangle $ are $\langle 1\pm i\rangle$, $\langle 3\rangle$ and $\langle 2\pm i\rangle$. These correspond to the ideals $\langle1\pm \bar x \rangle$, $\langle 3 \rangle$ and $\langle 2\pm \bar x\rangle$ in the ring $\mathbb Z[x]/\langle x^2+1,30\rangle$. So the required maximal ideals are $\langle2,1\pm x \rangle$, $\langle 3, x^2+1 \rangle$ and $\langle 5,2\pm x\rangle$.