A problem on Correspondence Theorem Question: What does the correspondence  theorem tell us about ideals of $\Bbb{Z}[x]$ that contain $(x^2+1)$?
Attempt: Define a map $\phi:\Bbb{Z}[x]\to\Bbb{Z}[i]$, $x\mapsto i$.
Clearly  kernel of $\phi$ is the ideal generated by the ideal $(x^2+1)$.
By correspondence theorem, ideals of $\Bbb{Z}[x]$ containing $x^2+1$ is the inverse image of $\Bbb{Z}[i]$ under $\phi$.
As $\Bbb{Z}[i]$ is a PID so each ideal $I\subseteq \Bbb{Z}[i]$ can be generated by the ideal $(a+bi)$.
I am afraid what will be the inverse image of $I$ under $\phi$ i.e., $\phi^{-1}(I)=?$
Any help will be appreciated. Thanks!
 A: The correspondence theorem states that for a map $\phi : A \to B$, the ideals of $A$ which contain $\ker(\phi)$ are exactly the inverse images of ideals of $B$.
So an ideal of $\mathbb{Z}[X]$ containing $(X^2 + 1)$ is exactly the inverse image of an ideal of $\mathbb{Z}[i]$ under the projection map $\phi$. As you have noted, $\mathbb{Z}[i]$ is a PID, so its ideals are of the form $(k)$, where $k \in \mathbb{Z}[i]$. Since $\phi$ is a surjection, we can write $k = \phi(j)$ for some $j \in \mathbb{Z}[X]$. Then $\phi^{-1}(\phi(j)) = \{p \in \mathbb{Z}[X] | \phi(p) = c \phi(j)\}$. Now suppose $\phi(p) = c \phi(j)$. Write $c = \phi(d)$. Then $\phi(p) = \phi(dj)$. Then $p = dj + q (X^2 + 1)$. So $\phi^{-1}(\phi(j)) \subseteq (j, X^2 + 1)$. And it's easy to see that the other direction holds as well.
So the ideals of $\mathbb{Z}[X]$ containing $X^2 + 1$ are exactly the ideals of the form $(j, X^2 + 1)$.
Edit: and we may WLOG write $j = aX + b$, since we can always write $j = aX + b + c(X^2 + 1)$ using the division algorithm. So the ideals are $(aX + b, X^2 + 1)$ where $a, b \in \mathbb{Z}$.
