Quotient of Gaussian Integers via Third Ring Isomorphism It's known that $\mathbb{Z}[i] / (1+i) \cong \mathbb{Z}_2$. 
I'm having difficulties with my attempt. 
So we have $2 \in (1+i)$ since $2 = (1+i)\cdot(1-i)$. Thus $(2) \subset (1+i) \subset \mathbb{Z}[i]$. This looks like a third isomorphism problem so, $\mathbb{Z}[i]/(1+i) \cong (\mathbb{Z}[i]/(2)) / ((1+i)/(2))$. My problem comes from knowing what $(1+i)/(2)$ is. I know I have to get to this being isomorphic to $\mathbb{Z}_2$ and I can see that $\mathbb{Z}[i]/(2) \cong \mathbb{Z}_2[i]$, but I dont know how to get the last reduction. 
 A: $\mathbb{Z}_2[i]\cong\mathbb{Z}[x]/(2,x^2+1)\cong\mathbb{Z}[x]/(2,x^2-1)\cong\mathbb{Z}_2\times\mathbb{Z}_2$ (from the Chinese Remainder Theorem)
$(i+1)/(2)\cong\mathbb{Z}_2$, the last one becomes apparent when you write of the multiplication table. $i$ behaves just like $1$, and $i+1$ behaves just like $0$, so $i=1$. (It's perhaps easier to see this by noticing $i^2\equiv_2-1\equiv_2 1$)
So $\mathbb{Z}_2\times\mathbb{Z}_2/\mathbb{Z}_2\cong\mathbb{Z}_2$
There are however easier ways to prove that $\mathbb{Z}[i]/(i+1)\cong\mathbb{Z}_2$. Consider the map $\phi:\mathbb{Z}[i]\rightarrow \mathbb{Z}_2$ such that $\phi(a+bi)=a+b$. We will show that the map is surjective and that $\ker(\phi)=(i+1)$, the rest will follow from the 1st isomorphism theorem:


*

*Surjective: If $c\in\mathbb{Z}_2$ then $\phi(ci)=\phi(c)=c$.

*$\ker(\phi)=(i+1)$: It is clear that $(i+1)\subseteq\ker(\phi)$, since $\phi(a+ai)=0$. Now suppose that $\phi(a+bi)=0$, then $a\equiv_2 b$ and so $a=b+2k$ for some $k\in\mathbb{Z}$. Therefore $$a+bi=b+bi+2k=b(i+1)+(i+1)(-i+1)k=(i+1)(b-ik+k)$$ so $\ker(\phi)\subseteq (i+1)$

A: There is another isomorphism theorem (maybe the second one in your numeration) which you can use:
If $A\subseteq B$ are rings and $I\subset B$ is an ideal, then we have
$$ (A+I)/I \cong A/(I\cap A) $$
You can apply it for $A=\mathbb{Z}$, $B=\mathbb{Z}[i]$ and $I=(1+i)$. Since $(1+i)\cap \mathbb{Z}=(2)$, $\mathbb{Z}+(1+i)=\mathbb{Z}[i]$ and $\mathbb{Z}/(2)\cong \mathbb{Z}_{2}$, you get what you wanted.
Edit (regarding your attempt and the place where you got stuck):
$(1+i)/(2)$ should mean the ideal in $\mathbb{Z}[i]/(2)$ generated by the element $(1+i)$. I guess from there you can finish the proof your way also.
