$\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/\mathbb{2Z}$ 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.

There is a natural ring image of $\,\Bbb Z\,$ in $\,R = \Bbb Z[i]/(1\!+\!i)\,$ by mapping  integer $\,n\,$ to $\ n \pmod{1\!+\!i}$ by composing the natural maps $\,\Bbb Z\to \Bbb Z[i]\to \Bbb Z[i]/(1+i).\,$ This map $\, h\color{#0a0}{ \ {\rm is\ surjective\  (onto)}}$ since $\,{\rm mod}\ 1\!+\!i\!:\ \,1\!+\!i\equiv 0\,\Rightarrow\,i\equiv -1\,\Rightarrow\, a+bi\equiv a-b\in\Bbb Z.\,$ Finally, let's compute the kernel $\,I\,$ of the ring homomorphism $\,h.\,$ By rationalizing a denominator, $\,I = \color{#c00}{\ker h = 2\,\Bbb Z}\,$ as follows
$$ n\in I\iff (1\!+\!i)\mid n\ \ {\rm in}\ \ \Bbb Z[i]\iff \dfrac{n}{1\!+\!i}\in \Bbb Z[i]\iff \dfrac{n(1\!-\!i)}{2}\in\Bbb Z[i]\iff \color{#c00}{2\mid n}$$
Therefore, applying the First Isomorphism Theorem, $\, \color{#0a0}{R = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{2\,\Bbb Z}.$
