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. 

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.
