Find $n$ such that $\mathbb{Z}[i]/(-2+2i)\cong \mathbb{Z}/n\mathbb{Z}$ I am trying to find $n$ such that $\mathbb{Z}[i]/(-2+2i)\cong \mathbb{Z}/n\mathbb{Z}$. Here is what I tried:
Consider $\varphi\colon\mathbb{Z}\to\mathbb{Z}[i]/(-2+2i)$ given by $\varphi\colon z\mapsto z+(-2+2i)$. $\varphi$ is a ring homomorphism. Moreover, $\ker\varphi=(-2+2i)\cap\mathbb{Z}$. Any $\alpha\in\ker\varphi$ must have the form $$(-2+2i)(a+bi)=(-2a-2b)+(2a-2b)i$$ for some integer $a,b$. Since $\alpha\in\mathbb{Z}$, we have $2a-2b=0$, i.e. $a=b$. As a result, $$\ker\varphi=\{(-2+2i)(a+ai)\mid a\in\mathbb{Z}\}=\{-4a\mid a\in\mathbb{Z}\}=4\mathbb{Z}.$$ Then by the First Isomorphism Theorem for Rings, $$\mathbb{Z}/4\mathbb{Z}\cong \mathbb{Z}[i]/(-2+2i).$$
However, $-2+2i=(1+i)^3$ and $1+i$ is irreducible over $\mathbb{Z}[i]$, so $$|\mathbb{Z}[i]/(-2+2i)|=|\mathbb{Z}[i]/(1+i)|^3=8.$$ This is clearly a contradiction, what did I do wrong?

$\mathbb Z[i] /(2i-2)$ is not cyclic at all, because $2$ and $1+i$ are two different elements of order $2$.
Since $1$ has order $4$, we obtain that $\mathbb Z[i] /(2i-2)$ is a non-cyclic abelian group with $8$ elements and an element of order $4$. Thus $\mathbb Z[i] /(2i-2) \cong C_2 \times C_4$.
