Algebraically, why is $\mathbb{Z}[i]/(i + 1)$ isomorphic to $\mathbb{Z}_{2}$?

Question

I understand geometrically why the Gaussian integers modulo $i+1$ is $\mathbb{Z}_{2}$, using lattices. Is there an algebraic isomorphism construction, however?

Answer 1

Consider $a+bi$: we can write $a-b=2q+r$, where $q$ is an integer and $r$ is $0$ or $1$. Then $$a+bi=(b+2q+r)+bi=r+b(1+i)+q(1-i)(1+i)\equiv r\ .$$

Answer 2

You could use $\mathbb{Z}[i] \cong \mathbb{Z}[x]/(x^2+1)$. Then $$ \mathbb{Z}[i]/(i+1) \cong \mathbb{Z}[x]/(x^2+1, x+1) \cong \mathbb{Z}[x]/(-x+1,x+1) \cong \mathbb{Z}[x]/(2,x+1) \cong \mathbb{Z}/2\mathbb{Z}[x]/(x+1)\cong \mathbb{Z}/2\mathbb{Z}. $$

Answer 3

Since $i^2=-1$, if we pass to a quotient where $i+1=0$, we must have $-1=(-1)^2=1$, so such a quotient contains $\mathbb{Z}/2$.

In fact, such a quotient is exactly $\mathbb{Z}/2$, because $1\mapsto 1$ and $i\mapsto 1$, so any $a+bi$ is mapped to $a+b\in\mathbb{Z}/2$.