Let $H=(1+i)\mathbb{Z}[i]$. Let $f:\mathbb{Z}\to \mathbb{Z}[i]/H : f(z)=[z]$. Prove $\ker f=2\mathbb{Z}$.

Question

Let $H=(1+i)\mathbb{Z}[i]$ and let $f:\mathbb{Z}\to \mathbb{Z}[i]/H$ such that $f(z)=[z]$. I want to prove $\ker f=2\mathbb{Z}$.

(*) I know $[1+i]=[0] $

So $f(z)=0\Longleftrightarrow [z]=[0]\Longleftrightarrow [z]=[1+i]$

Then

$[z][1-i]=[1+i][1-i]$

$ [z][1-i]=[2]$

Now I don't know how to conclude that

$[z]=[2]$

then $\ker f=2\mathbb{Z}$

Can you help me please?

Answer 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, $z\in 2\mathbb Z$.