Finding an isomorphism between $\mathbb{Z}[i]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$

Question

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.

Answer 1

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 $g$?

Answer 2

Hint $\ $ We get an image of $\,\Bbb Z[\sqrt{\color{#c00}{-9}}]\,$ in $\,\Bbb Z[\sqrt{-1}]\,$ by mapping $\,\sqrt{-9}\,\mapsto\, 3\sqrt{-1}$

$\!\bmod 7\!:\ \color{#c00}{{-}9\equiv -2}\,$ and the map is onto, since $3$ is a unit so $\,3^{-1}\sqrt{-9}\,\mapsto \sqrt{-1}$