$\mathbb Z[\sqrt {-5}]$ is Noetherian

Question

I'm trying to prove that $\mathbb Z[\sqrt {-5}]$ is Noetherian. I already know that $\mathbb Z[X]$ is Noetherian and I'm trying to find a surjective map

$$\varphi: \mathbb Z[X]\to \mathbb Z[\sqrt{-5}]$$

with $\ker\varphi=(X^2+5)$.

If I could find this map I could prove that $\mathbb Z[\sqrt{-5}]\cong \mathbb Z[X]/(X^2+5)$ and then $\mathbb Z[\sqrt{-5}]$ is Noetherian.

Thanks

Answer 1

Two ways:

• Your way with $\varphi(X)$ one of the solutions of the equation $X^2+5=0$.
• Observe that any ideal of $\Bbb{Z}[\sqrt{-5}]$ will also be an additive subgroup of a free abelian group of rank two. Do you recall any results about generators of such a subgroup?