Let $H=(1+i)\mathbb{Z}[i]$. Let $f:\mathbb{Z}\to \mathbb{Z}[i]/H : f(z)=[z]$. Prove $\ker f=2\mathbb{Z}$.
Solution 1:
Using a different approach:
We have $\bar z=\bar0\iff z\in\langle 1+i\rangle$. Therefore, for some $a,b\in\mathbb Z$,
$$z=(a+ib)(1+i)=a-b+i(a+b)$$
Since $z\in\mathbb Z$, we have $a+b=0$. So,
$$z=a-b=a-b+(a+b)=2a.$$
Hence, $z\in 2\mathbb Z$.