$\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/\mathbb{2Z}$ When I first looked at this problem, I thought that $\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/5\mathbb{Z}$, but apparently the correct answer is $\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/2\mathbb{Z}$.
Here's where I'm confused: saying that $\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/2\mathbb{Z}$ is saying that there is one element in $\mathbb{Z}[i]$ that is divisible by $1+i$, which is the $0$ element in $\mathbb{Z}[i]/(1+i)$, and there there's only ONE other element in ${\mathbb Z}[i]$ that is not divisible by $1+i$, thus it is isomorphic to $\{0,1\}$.  I'm just not seeing how this isomorphism makes any sense, whatsoever...
A complete proof would be helpful, but I guess I'm more looking for intuition.

Let us construct a bijection
$$
\Bbb Z[i]/(1+i)\ni [x+y i]\longmapsto [x-y] \in \Bbb Z/2\Bbb Z
$$
(note: modulo $(1+i)$ means you let $i=-1$.)
It is direct to check this is well-defined and gives an isomoprhism.
