Finding all ideals $T$ in number ring $\mathbb Z[\sqrt{-3}]$ s.t. $\langle 4 \rangle \subset T$. I want to find all ideals $T$ in number ring $\mathbb Z[\sqrt{-3}]$ s.t. $\langle 4 \rangle \subset T$.
My idea：
$  4= 2\times2 = (1-\sqrt{-3}) (1+\sqrt{-3})$ so $\langle4\rangle = \langle 2\rangle \langle2\rangle = \langle(1-\sqrt{-3})\rangle \langle (1+\sqrt{-3})\rangle$
Let $I = \langle2\rangle, J = \langle(1-\sqrt{-3})\rangle, K = (1+\sqrt{-3})\rangle$
Now, note that $\langle4\rangle \subset I, J, K, IJ, JK, IK, IJK, \langle4\rangle, \mathbb Z[\sqrt{-3}]$ and these are the all ideals that contain $\langle4\rangle$.
Now, as easy as this problem may seem, I'm having doubts whether my solution is correct or not, especially I am not able to convince myself that there are no other ideals containing $\langle4\rangle$. Can someone help me figuring this out?
Thanks!
 A: You could use the universial property of the polynomial ring to construct a ring homomorphism $\varphi: \mathbb{Z}[T] \rightarrow \mathbb{Z}[\sqrt-3]$ by mapping $T$ to $\sqrt-3$. As $\varphi(T^2+3)= 0$, this induces a map from the quotient $\bar{\varphi}: \mathbb{Z}[T]/\langle T^2+3\rangle \rightarrow \mathbb{Z}[\sqrt-3]$. Furthermore $\bar{\varphi}$ is clearly surjective and the domain and codomain of $\bar{\varphi}$ are both free $\mathbb{Z}$-modules of rank 2 so this map is even an isomorphism by general theory. As you said, it suffices to compute how many ideals $(\mathbb{Z}[T]/\langle T^2+3 \rangle)/\langle 4 \rangle $ has. Again using universal properties, you can constuct an isomorphism $(\mathbb{Z}[T]/\langle T^2+3 \rangle)/\langle 4 \rangle \cong (\mathbb{Z}[T]/ \langle 4 \rangle) / \langle T^2+3 \rangle \cong \mathbb{(Z/\langle 4\rangle}[T])/\langle T^2 - 1 \rangle$. This ring is now in particular a free $\mathbb{Z}/\langle 4 \rangle$-module of rank $2$ so it only has $16$ elements. Also you can do very explicit calulations in this ring using polynomial division, so it should not be to hard to calculate all ideals of this ring. Finally following the ismorphisms above backwards will give you all ideals above $\langle 4 \rangle$. If you need any more details please let me know!
