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Suppose $\alpha = (1+i)\mathbb{Z}[i]$, where $\mathbb{Z}[i]$ denotes the Gaussian integers. I am trying to show that $$\mathbb{Z}[i] \big/ \alpha$$ is a field.

My current argument is this: $\ 2 = (1+i)\cdot(1-i)\ $ so $2 \in (1+i)$. Therefore, in $\mathbb{Z}[i] \big/ \alpha,\ $ $2=0$. With this, $1+i$ is also $0$, implying $1=-i \implies 1^2=(-i)^2=-1$. So we have that the quotient is just the $2$-field, $\mathbb{F}_2$.

Is this sufficient? To me it seems to fall short of a proper proof. Any help would be appreciated!

Joe
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  • Using the theory: $\mathbb{Z}[i]$ is an Euclidean domain and the element $1+i$ is irreducible, then the quotient is a field. – User3773 Mar 28 '17 at 10:58
  • Wouldn't this only work if we were considering the quotient $\mathbb{Z}[i]/(1+i)$? Instead we are looking at $\mathbb{Z}[i]/(1+i)\mathbb{Z}[i]$. Are you arguing that $(1+i)\mathbb{Z}[i]$ is an ideal? Or have I let my notation get the better of me: then $(1+i)=(1+i)\mathbb{Z}[i]$ by the very definition of an ideal? – Joe Mar 28 '17 at 11:02
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    What is $(1+i)\mathbb{Z}[i]$ if not the ideal generated by the element $1+i$? – User3773 Mar 28 '17 at 11:05
  • @Cla Got it, thank you! – Joe Mar 28 '17 at 11:13

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