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

abstract-algebra ring-theory gaussian-integers

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


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


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 $a+b\in\mathbb{Z}/2$.

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