$\mathbb{Z}[i]/(1+i) \cong \mathbb{Z}/\mathbb{2Z}$ 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.
 A: 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, $\, \color{#0a0}{R = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{2\,\Bbb Z}.$
A: 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.
A: I find it easiest to do
$${\mathbb Z}[i]/(1+i) \cong {\mathbb Z}[x]/(x^2+1,1+x) \cong ({\mathbb Z}[x]/(x+1))/(x^2+1) \cong {\mathbb Z}/(2).$$
For the final step, note ${\mathbb Z}[x]/(x+1)$ is isomorphic to ${\mathbb Z}$, with $x$ mapping to $-1$. Under that isomorphism, the ideal $(x^2+1)$ of $Z[x]/(x+1)$ corresponds to the ideal $(2)$ of ${\mathbb Z}$.
