Find all ideals of ring $\mathbb{Z}[\sqrt{-3}]$ containing 4.

Question

How to find all ideals of ring $\mathbb{Z}[\sqrt{-3}]$ containing $4$?

Since $\mathbb{Z}[\sqrt{-3}]/(4) \cong (\mathbb{Z}/4\mathbb{Z})[X]/(X^2+3)$, I tried to find all ideals of $(\mathbb{Z}/4\mathbb{Z})[X]/(X^2+3)$. But I was stuck because $\mathbb{Z}/4\mathbb{Z}$ is not a field. Is there any good way?

Answer 1

Since $(\mathbb Z/4\mathbb Z)[X]/(X^2+3) = (\mathbb Z/4\mathbb Z)[X]/(X^2-1)$ is a ring with only 16 elements, one may find all the ideals by hand. It is a local ring with maximal ideal $(X+1,X-1)$. Sitting directly below are $(X+1)$, $(X-1)$, and $(2X)$. $(2X+2)$ is contained in all of them, and $(2)$ is only contained in $(2X)$. Together with the trivial ideals, this gives 8 ideals.

The corresponding ideals in $\mathbb Z[\sqrt{-3}]$ are consequently:

• $(1+\sqrt{-3},1-\sqrt{-3})$,
• $(1+\sqrt{-3})$, $(1-\sqrt{-3})$ and $(2\sqrt{-3},4)$,
• $(2+2\sqrt{-3},4)$ and $(2)$,
• the "trivial" ideals $(4)$ and $\mathbb Z[\sqrt{-3}]$.