Finding an isomorphism between $\mathbb{Z}[i]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$ I was having some trouble finding an explicit isomorphism between $\mathbb{Z}[i]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$.
$\textbf{What I have noticed is}:$


*

*7 is a prime element in $\mathbb{Z}[i]$ so $(7)$ is a maximal ideal in $\mathbb{Z}[i]$ and $\mathbb{Z}[i]/(7)$ is  a field. 

*7 is also a prime element in $\mathbb{Z}[\sqrt{-2}]$


$\textbf{What I have been trying to do is this}$


*

*Find a surjective homomorphism between $\mathbb{Z}[i]/(7) \rightarrow \mathbb{Z}[\sqrt{-2}]/(7)$. Since $\mathbb{Z}[i]/(7)$ was a field, the kernel of this homomorphism will be either the whole ring or just $0$. In the latter case it would be a isomorphism. 


I am having trouble finding this surjective homomorphism:
I have noticed that $\bar{{i}}^{2}=-1$ so the image of $\bar{{i}}$ must be sent to something whose square is $-1$. Any help would be appreciated. I may be missing some obvious insight.

Consider the unique homomorphism $f\colon\mathbb{Z}[X]\to\mathbb{Z}[\sqrt{-2}]/(7)$ sending $X\mapsto 2\sqrt{-2}+(7)$. Note that $f(4X)=8\sqrt{-2}+(7)=\sqrt{-2}+(7)$, so $f$ is surjective. Its kernel contains $X^2+1$, as
$$
f(X^2+1)=(2\sqrt{-2})^2+1+(7)=-8+1+(7)=(7)
$$
so $f$ induces a (surjective) homomorphism $g\colon\mathbb{Z}[i]\to\mathbb{Z}[\sqrt{-2}]/(7)$.
What's the kernel of $g$?
