Construct an explicit isomorphism between $\mathbb{Z}[\sqrt{-1}]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$, where $(7)$ denotes that ideal generated by 7.

Question

Construct an explicit isomorphism between $\mathbb{Z}[\sqrt{-1}]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$, where $(7)$ denotes that ideal generated by 7.

My idea is: We know that in the ring homomorphism identity map to identity, i.e $\phi(1)=1$, so fix $\phi(n)=n$. We want to construct isomorphism between these two rings; and let $x+y\sqrt{-1} \in \mathbb{Z}[\sqrt{-1}]/(7)$, and $\phi$ is a morphism then,

$\phi(x+y\sqrt{-1})=\phi(x)+\phi (y\sqrt{-1})=\phi(x)+\phi (y)\phi(\sqrt{-1})$.

If we define $\phi(\sqrt{-1})=\sqrt{-2}$, then it make isomorphism.

Is this idea correct to think this question? Anyone please suggest me some direction to solve this question?

Answer 1

You need $\phi(\sqrt{-1})^2=-1\in\Bbb Z[\sqrt{-2}]/(7)$. So $\phi(\sqrt{-1})=a+b\sqrt{-2}$ where $$a^2-2b^2+2ab\sqrt{-2}=-1\in\Bbb Z[\sqrt{-2}]/(7),$$ that is $a^2-2b^2\equiv-1\pmod7$ and $2ab\equiv0\pmod 7$.

So either $a\equiv0$ or $b\equiv0\pmod7$, but we cannot have $b\equiv0\pmod7$ (why?) so $a\equiv0\pmod7$ and $-2b^2\equiv-1\pmod 7$. What are the solutions to this congruence?

Answer 2

You know that $3^2\equiv 2 \mod 7$. Thus define $$\phi\colon \mathbb Z[\sqrt{-2}]/(7)\to \mathbb Z[\sqrt{-1}]/(7) $$ by $\phi(\sqrt{-2})=3\sqrt{-1}$ and extend as you described.

Then $\phi(\sqrt{-2}\sqrt{-2})=-2$ and $\phi(\sqrt{-2})\phi\sqrt{-2})=-3^2=-2$, so we have a homomorphism.

The homomorphism is surjective as $\phi(5\sqrt{-2})=15\sqrt{-1}=\sqrt{-1}$. Any surjective map between finite sets is a bijection, so we have an isomorphism.