TITLE: $\dfrac{\mathbb{Z}[x]}{(x^2 +5)}$ is isomorphic to $\mathbb{Z}[\sqrt{-5}]$
QUESTION [6 upvotes]: Let $H : \mathbb{Z}[x] \rightarrow \mathbb{Z}[\sqrt{-5}]$ be the evaluation homomorphism given by $H(f) = f(\sqrt{-5})$. We know that $H$ is surjective. I wanna to show that $\ker(H) = (x^2 +5)$. I know that is easy, but I don't find a way to show this. To prove the inclusion $(x^2 +5) \subset \ker(H)$ is easy, so I need help in the other inclusion. I proved this problem using Krull dimension, but I wanna to prove this in a most simple way. Any hint will be helpful.

REPLY [9 votes]: Take $f \in \ker H.$ Then $f(x) = (x^2 + 5)g(x) + r(x),$ for some $g(x), r(x) \in \mathbb Z[x]$ and degree of $r(x)$ is at most one. Now $f(\sqrt{-5}) = 0 \Rightarrow r(\sqrt{-5})=0.$ But $r(x)$ is a polynomial of degree at most one with integer co-efficients. Hence $r(x) =0.$