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

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 ${\mathbb Z}$.
