# Show that $\ker\phi = (-5+ \sqrt{-7})$

Let $\phi : \mathbb{Z}[\sqrt{-7}] \rightarrow \mathbb{Z}/32$ s.t. $\phi (a+b\sqrt{-7}) = \overline{a+5b}$. Show that $\ker\phi = (-5+ \sqrt{-7})$, where $(-5+ \sqrt{-7})$ is the ideal generated by $-5+ \sqrt{-7}$.

Showing the ideal is in the kernel is easy, but I am struggling to show that the kernel is in the ideal.

It is given that the map is a homomorphism and surjective.

$$\phi (a+b\sqrt{-7}) = 0\iff\overline{a+5b}=0\mod 32$$
We can thus say that $a+5b=32k$ for some integer $k$, i.e.
$$\phi(x)=0\iff x=(32k-5a)+a\sqrt{-7}\,\text{for some}\,a,k$$
$$x=(32k-5a)+a\sqrt{-7}=(32)(k)+a(-5+\sqrt{-7})$$
as well as that $(-5+\sqrt{-7})(-5-\sqrt{-7})=32$, and we are done.