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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?

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    $\begingroup$ Your homomorphism is not surjective. The residue class $i + (-2+2i)$ is not in its image. $\endgroup$
    – Daniel Fischer
    Commented Feb 18, 2015 at 10:27
  • $\begingroup$ How did you decide that $\;\ker\varphi=(-2+2i)\cap\Bbb Z\;$ ?! We in fact have that, for example, $\;2-2i\in\ker\varphi\;$ , since $\;2-2i=i(-2+2i)\;$ ... $\endgroup$
    – Timbuc
    Commented Feb 18, 2015 at 10:28
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    $\begingroup$ Your mistake is not to verify that $\varphi$ is surjective. In fact it is not. The second argument is correct, so the correct answer is $n= 8$. $\endgroup$
    – Crostul
    Commented Feb 18, 2015 at 10:30

1 Answer 1

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$\mathbb Z[i] /(2i-2)$ 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 $\mathbb Z[i] /(2i-2) \cong C_2 \times C_4$.

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