Prove $\ker \phi=\langle 1+\sqrt{-5}, 2\rangle$

Question

Let $\phi:\mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}_2$ $$\phi(p+q\sqrt{-5})=[p+q]_2$$

I want to prove $\ker \phi=\langle 1+\sqrt{-5}, 2\rangle$

So what I did:

$m=p+q\sqrt{5}\in \ker \phi\Longleftrightarrow [p+q]_2=[0]_2\Longleftrightarrow $ $p+q\in 2\mathbb{Z}$

So now how can I continue? Any hint?

Answer 1

You only need to prove that every element $p+q\sqrt{5}$ from the kernel is in $\langle 1+\sqrt{5}, 2\rangle$. Since $p+q$ is even, $p-q$ is also even, $=2n$ for some integer $n$. Then $p+q\sqrt{5}=n\cdot 2+q(1+\sqrt{5})$ as desired.

Answer 2

Clearly $\phi(1+\sqrt{5}) = 0$ and $\phi(2)=0$. Hence $$\langle 1+\sqrt{5}, 2\rangle\subset \ker(\phi).$$ Suppose $a+b\sqrt{5}$ is in kernel. Then $a+b = 2k$ for some $k$. If both $a$ and $b$ are even, then $a+b\sqrt{5}=2\left(a/2+b/2\sqrt{5}\right) \in \langle1+\sqrt{5}, 2\rangle $.

If both are odd, then $a+b\sqrt{5} = 1+\sqrt{5}+(a-1)+(b-1)\sqrt{5} \in \langle1+\sqrt{5}, 2\rangle$ as $(a-1)$ and $(b-1)$ are even. In any case $$\ker(\phi) \subset \langle1+\sqrt{5}, 2\rangle.$$

And the conclusion follows.